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BANSAL CLASSES Target IIT JEE 2009 CLASS : XI

Q1. Q2.

Daily Practice Problems

TIME : 40 MIN

DATE : 01-02/05/2008

The no. of gm molecula molecularr of oxygen oxygen in 6.02 x 10 24 CO molecular are (A) 2 (B) 0.2 (C*) 5

DPP. NO.-1

(D) 0.5

How many moles moles of electro electronn weight weight one kilogra kilogram? m? (A) 6.02 x 10 23

Q3.

PHYSICAL CHEMISTRY

(B)

18 gms gms of wate waterr cont contai ains ns (A) 1 gm atom of hydrogen (C) 3 gm atoms of oxygen

(C)

(D*)

(B) 2 gm atoms of hydrogen (D) 2 gm atoms of oxygen

Q4.

Out of 1 gm dioxygen, dioxygen, 1 gm atomic atomic oxygen oxygen and 1 gm of ozone, ozone, the maximum number number of oxygen oxygen atoms atoms are contained in (A) 1 gm of atomic oxygen (B) 1 gm of ozone (C) 1 gm of dioxygen (D*) All contains same number of atoms

Q5.

Which Which has has the highes highestt mass? mass? (A) (A) 50 gm of iron iron (B*) (B*) 5 moles moles of N 2

(C) 0.1 gm atom of Ag (D) 10 23 atoms of carbon

Q6.

4.4 gm gm of an unknown unknown gas occupies occupies 2.24 litre of of volume volume at at STP. STP. The gas may be (A*) (A*) Carb Carbon on diox dioxid idee (B) (B) Carb Carbon on mono monoxi xide de (C) (C) Oxy Oxygen gen (D) Sulp Sulphhure ure dioxi ioxide de

Q7.

Number of of gold atoms in 300 mg of a gold ring of 20 carat gold gold (atomic (atomic mass of gold gold = 197 gm/mol; gm/mol; pure gold is 24 24 carat) carat) are (A) 4.5 x 10 20 (B) 6.8 x 10 15 (C*) 7.6 x 10 20 (D) 9.5 x 10 20

Q8.

One litre of milk weighs weighs 1.032 kg. The butter butter fat fat it contains to the extent of 4% by volume has a density -3 of 865 kg m . The density of fat free “Skimmed milk” is (A) 1000 Kg/m 3 (B*) 1038 Kg/m 3 (C) 1032 Kg/m 3 (D) 167 kg/m 3

Q9.

The mineral mineral quartz quartz and zircon zircon has a density density of 2.65 g/cm 3 and 4.5 gm/cm3 respectively. A rock composed of quartz and zircon has a density of 3 gm/cm 3 % by volume and % by mass of quartz in the sample are (A*) (A*) 81.1 81.1% % by vol, vol,71 71.7 .7% % by mass mass (B) (B) 61.2 61.2% % by vol, vol, 55.3 55.3% % by mass mass (C) (C) 75% by vol, 60% by mass mass (D) Data Insufficient

Q10. Comprehension: Ocean currents are measured in “Sverdrups (sv)” (1 sv = 10 9m3/sec). The gulf stream off the tip of florida, for instance, has a flow of 35 sv. sv. (i) (i)

The flow of the gulf stream stream in milliliters milliliters per minute is 18 (A*) 2.1 x 10 ml/min (B) 3.5 x 10 1020 ml/ ml/min min (C) 3.7 x 1018 ml/min ml/min (D) 5.6 x 1020 ml/min

(ii)

Mass of water water in the the gulf gulf stream stream flows flows past a given given point point in 24 hours hours is is (density (density of water water is 1 gm/ml) 21 21 21 (A) 4 x 10 gm (B*) 3 x 10 gm (C) 8 x 10 gm (D) 7 x 1021 gm

(iii)

Time required required for one petalitre (1 PL = 10 15 ltr) of water to flow past a given point is (approximately) (A) 36 sec (B) 54 sec (C) 15 sec (D*) 28 sec BULLS EYE


TIME : 50 MIN

PHYSICAL CHEMISTRY

Daily Practice Problems DATE : 01-02/05/2008

Q1.

Volume olume at NTP of 0.44 gm of CO 2 is the same as that of (A*) A*) 0.02 .02 gm of hydrog rogen gas (B) 0.085 .085 gm of ammo ammoni niaa gas (C) (C) 320 320 mg of sul sulphur phur di oxid oxidee gas (D) (D) none none of thes thesee

Q2. Q2.

54.4 54.4 gm [Fe( [Fe(H H 2O)5 NO]SO4 contains (A) 3.2 gm oxygen (C) 6.4 gm of oxygen

DPP. NO.-2

(B*) 32 gm of oxygen (D) 64 gm of oxygen

Q3.

Weight eight of oyxge oyxgenn in Fe 2O3 and FeO is in the ratio for the same amount of iron is (A*) 3 : 2 (B) 1 : 2 (C) 2 : 1 (D) 3 : 1

Q4.

The total number number of atoms present present in 25 mg of Camphor Camphor,, C 10H16O are 19 20 (A) 9.89 × 10 (B) 6.02 × 10 (C*) 2.67 × 10 21 (D) 2.57 × 10 20

Q5.

The element element A (at. wt. = 12) and B(at. B(at. wt. = 35.5) combine combine to to form a compound compound X. If 4 mole of B combine with 1 mole of A to give 1 mole of X, the weight of 1 mole m ole of X is: (A) 47.5 gm (B) 74 gm (C*) 154 gm (D) 149 gm

Q6.

Two flask of equal equal volumes volumes are evacuated, evacuated, then one is filled with gas A and other other with gas B at the same temperature and pressure. The weight of gas B at the same temperature and pressure. The weight of gas B was found to be b e 0.8 gm while the weight of gas A is found to be 1.4 gm. The weight of one o ne molecule of B is (A) 1.4 times as heavy as A (B) 0.4 times as heavy as A (C) 0.57 times as heavy as A (D) 0.8 times as heavy as A

Q7.

A sample sample of chalk chalk contain contained ed as impurity a form of clay clay which loses 14.5% 14.5% of its weight weight of water on prolo prolong ng heat heating ing.. 5 gm of chalk chalk on heati heating ng shows shows a loss loss in weig weight ht by 1.507 1.507 gm. gm. Perce Percent ntag agee of chal chalkk in the the sample is (A) 50% (B) 30.14% (C) 70% (D*) 35.55%

Q8.

A pennsyl pennsylvania vania bituminus bituminus coal coal is analys analysed ed as as follows: follows: Exactly Exactly 2.5 gm gm is weighed weighed into a fused fused silica silica crucible weighs 2.415 gm. The crucible next is covered with a vented lid and strongly heated until no volatile matter remains. The residue coke buttons weighs 1.528 gm. The crucible is then heated without the cover until all specks of carbon have disappeared, and the final ash weighs 0.245 0. 245 gm. What are the percen percentag tagee of moistur moisture, e, volati volatile le combus combustib tible le matter matter (VCM), (VCM), fixed fixed carbon carbon (FC) (FC) and and ash? ash? [Ans: 3.4%, 35.5 VCM, 51.3% FC, 9.8%ash]

Q9.

A plant plant virus is found found to consist consist of uniform uniform cyli cylindric ndrical al particles particles of 150Å 150Å in diameter diameter and 5000 Å long. The specific volume of the virus is 0.75 cm 3/gm. If the virus is considered to be a single particle, find its molecular weight. [Ans: 7.1 × 10 7]

Q10. A porous catalyst catalyst for chemical reaction reaction has an internal internal surface surface area of 800 m 2 per cubic centimeter of bulk bulk materi material. al. Fifty Fifty perce percent nt of the bulk bulk volume volume consi consists sts of the pores pores (holes (holes), ), while while the other other 50% of the volume is made up of the soild s oild substance. Assume that the pores are all cylindrical tubules of uniform diameter d and length l , and that the measured internal surface area is the total area of the curved surfaces of the tubules. What is the diameter of the each pore. [25Å] BULLS EYE


TIME : 50 MIN

PHYSICAL CHEMISTRY

Daily Practice Problems DATE : 10-11/05/2008

DPP. NO.-3

Q.1 Q. 1

If water samples are taken taken from from sea, sea, rivers, rivers, clouds, clouds, lake lake or snow, snow, they will be found to contain contain Hydrogen Hydrogen and Oxygen in the approximate ratio of 1 : 8. This indicates the law of (A) Multiple proportion (B*) Definite proportion (C) Reciprocal proportions (D) None of these.

Q.2

The law of multiple multiple proport proportion ion is illustra illustrated ted by (A*) Carbon monoxide monoxide and and carbon carbon dioxide (B) Potassium Potassium bromide bromide and potassiu potassium m chlori chloride de (C) (C) Water ater and and heav heavyy wate waterr (D) Calciu lcium m hyd hydroxi roxidde and barium ium hyd hydroxi roxidde.

Q.3

Hydrog Hydrogen en and oxygen oxygen combine combine to form H 2O2 and H2O containing 5.93% and 11.2% hydrogen respectively. The data illustrates : (A) law law of conse onserv rvaatio tion of mass mass (B) law law of const onstan antt propo roport rtio ionn (C) (C) law law of reci recipr proc ocal al prop propor orti tion onss (D*) (D*) law law of mult multip iple le prop propor orti tion on

Q.4 Q. 4

One of the following combinations combinations illustrate law of reciprocal reciprocal proportions proportions (A) N2O3, N2O4, N2O5 (B) NaCl, NaBr, NaI (C*) CS2, CO2, SO2 (D) PH3, P2O3, P2O5

Q. 5

H2S contains 5.88% H, H 2O contains 11.11% H while SO 2 contains 50% S. This illustrates il lustrates the law of: (A) multiple proportions (B) constant proportions (C) conservation of mass (D*) reciprocal proportions

Q.6

KCl exists exists in two isotopi isotopicc form form KCl 35 and KCl37, these compound follow which law (A) multiple proportions (B) constant proportions (C) conservation of mass (D*) none of these.

Q.7

Weigh eightt of oxyge oxygenn in Fe 2O3 and FeO is in the simple ratio for the same sam e amount of iron is (A*) 3 : 2 (B) 1 : 2 (C) 2 : 1 (D) 3 : 1

Q.8

A sample sample of ammoniu ammonium m phosph phosphate ate,, (NH 4)3 PO4, contains 3.18 mol of hydrogen atoms.The number of moles of oxygen atoms in the sample is : (A) 0.265 (B) 0.795 (C) 1.06 (D) 3.18

Q.9

1 gm-ato gm-atom m of nitrog nitrogen en repres represent entss : 23 (A) 6.02 × 10 N2 molecules (C) 11.2 lit. of N 2at N.T.P.

(B) 22.4 lit. of N 2 at N.T.P. (D) 28 g of nitrogen

Q.10 In an experiment, 2.4 gm of iron iron oxide on reduction with hydrogen hydrogen yield 1.68 gm of iron. In another experiment 2.9 gm of iron oxide give 2.03 gm of iron on reduction with hydrogen. hydrogen. Show that the above data illustrate the law of constant proportions.

BULLS EYE

BANSAL CLASSES Target IIT JEE 2009 CLASS : XI(ALL)

DATE : 16/05/2008

PHYSICAL CHEMISTRY

Daily Practice Problems TIME : 30 MIN

DPP. NO.- 4

Q.1 Q.1

4.6 × 1022 atoms of an element weigh 13.8 gm. The atomic mass of the element is : (A) 120 (B*) 180 (C) 35.5 (D) 108

Q.2

How How many many mole moless of C 2H4O2 contains 6.02 × 10 23 at atoms of hydrogen?

Q.3

Calculate the % Na in a breakfast breakfast cereal which is advertised advertised to contain 110 mg of sodium per 100 g of cereal. [Ans.0.11%]

Q.5

The total number number of moles in a closed closed beaker beaker containing containing : (approx) (approx) 20 58.5 g of NaCl + 602.2 × 10 molecules of O2 + 2 g of hydrogen hydrogen gas + 22.4 litres of SO 2 gas at NTP + 30 ltr of Cl 2 at 0.0821 atm and 27°C. [Ans.3.2] (Assuming no reaction taking place) (R = 0.0821 L atm K –1 mol mol –1)

Q.4 Q.4

1.12L CO2 gas at S.T.P. + 0.44 g CO 2 – 6.02×1021 molecules of CO 2 = ________ moles of CO2.

[Ans. 0.25]

[Ans.0.05]

Q.6

How many neutrons are present in 180 mL water at (i) (i) 4°C 4°C & (ii) 127°C 127°C & 0.821 atm pressure. pressure. [Ans: 80 NA, 0.036 NA]

Q.7

An element element (at. (at. mass mass z) has has a isotopic isotopic masses (z + 2) 2) and (z – 1). 1). Find the the percen percentage tage abundance abundance of of the [Ans.33.3%] heavier isotope.

Q.8

A mixture of gas gas ''X'' ''X'' (mol. wt. 16) and gas gas Y (mol. wt. 28) in in the mole ratio ratio a : b has has a mean mean molecula molecular r weight 20. What would be mean molecular weight if the gases are mixed in the ratio b : a under identical conditions (gases are non reacting). (A*) 24 (B) 20 (C) 26 (D*) 40

Q.9

The abundance abundance of three isotopes isotopes of oxygen oxygen are as follows 16 % of O = 90% % of O17 + % of O18 = 10% Assume at. mass same as mass no. Find out % of o f O 17 and O18, if the isotopic mass is 16.12. [Ans.8, 2]

Q.10 Q. 10 Density Density of dryair containin containingg only N 2 and O2 is 1.146 gm/lit at 740 mm and 300 K. What is % composition of N2 by weight in the t he air. (A) 78% (B) 82% (C) 73.47% (D) 72.42%

BULLS EYE


Q1.

PHYSICAL CHEMISTRY


DATE : 17-18/05/2008

DPP. NO.- 5

Determine Determine the percentag percentagee composition composition of each elements elements of Mg 3(PO4)2. [Mg = 27.5 %; P = 23.7%; O = 48.8]

Q2.

The dot dot at the the end end of this sentenc sentencee has has a mass mass of about about one one microg microgram. ram. Assuming Assuming that that the black black stuff stuff is carbon, calculate the approximate number of atoms of carbon needed to make such a dot. [Ans:5 × 1016]

Q3.

Calculat Calculatee the percen percentag tagee of CaO in (i) CaCO3 (ii) CaC2O4 (iii) Ca3(PO4)2

(iv) Ca(OCl2)·CaCl2·Ca(OH)2·2H2O

Q4.

Caffeine on analys analysis is was found to contain contain 49.48% by wt. C, 5.19% H, 28.85 28.85 % N and and remaining remaining oxygen. What is its simplest formula. If one mole of it weighs 194.1 gm what is its molecular formula.

Q5.

Analysis Analysis of a compoun compoundd yields yields the following following percentag percentagee compositio compositionn by weight: weight: 65.03% 65.03% Ag, 15.68% 15.68% Cr, 19.29% O. The simplest formula of the compound com pound is (A*) Ag2CrO4 (B) Ag2Cr 2O7 (C) AgCr 2O3 (D) Ag2Cr 2O3

Q6.

Cartisone is a molecular substance containing 21 atoms of carbon per molecule. The weight percentage percentage of Carbon in cartisone is 70%. The molecules weight of cartisone is (A) 240 (B*) 360 (C) 450 (D) 840

Q7.

Weight of nitrogen gas can be obtained from 720 kg of urea (A) 226 kg (B) 350 kg (C*) 336 kg

(D) 420 kg

Q8.

A polys polysty tyren rene, e, having having formula formula Br 3C6H3 (C (C3H8)n, was prepared by heating styrene with tribromobenzoyl peroxid peroxidee in in the the absenc absencee of air. air. If If it was found found to contain contain 10.46% 10.46% bromine bromine by weight weight,, the the value value of n is (A) 19 (B*) 45 (C) 38 (D) 56

Q9.

A certain oxide of iron contains contains 2.5 gm of oxygen oxygen for every 7 gm of iron. If it is regarded regarded as as a mixture of FeO and Fe 2O3 in the weight ratio x : y. What is x : y

Q10. Q10 . Amphitamine Amphitamine (also (also called dexedrine) dexedrine) is a solid with molecula molecularr formula formula C 9H13 N. Its Its dens density ity is 0.949 0.949 g ml (i) Total Total number of atoms in 18.04 gm of amphitanine are (A) 7 × 10 22 (B) 8 × 1022 (C) 7.5 × 10 22

(D) 6 × 10 22

(ii) Volume of 4.5 mol of dexedrine is (A) 300 ml (B) 350 ml

(D) 640 ml

(C) 450 ml

(iii) The weight of amphitemine that contains exactly the same number of H atoms as there are in 171.1 gm of water is (A) 197.4 gm (B) 167.7 gm (C) 217.4 gm (D) 205.7 gm

BULLS EYE


DATE : 19/05/2008

PHYSICAL CHEMISTRY


DPP. NO.- 6

Q.1

The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of the compound will be (A) C2H2 (B) C3H3 (C) C4H4 (D*) C6H6

Q.2

On analysis, a certain compound was found to contain iodine and oxygen in the ratio of 254 gm of iodine (at.mass 127) and 80 gm oxygen (at. mass 16). What is the formula of the compound. (A) IO (B) I2O (C) I5O3 (D*) I2O5

Q.3

A given sample of pure compound contains 9.81 gm of Zn, 1.8 × 10 23 atoms of chromium and 0.60 mol of oxygen atoms. What is the simplest formula. (A) ZnCr 2O7 (B*) ZnCr 2O4 (C) ZnCrO4 (D) ZnCrO6

Q.4

‘X’ is molecular substance containing 21 atoms of carbon per molecule. The weight % of C in ‘X’ is 50.4%. What is molecular weight of X. (A) 400 amu (B) 176.5 amu (C) 287.6 amu (D*) 500 amu

Q.5

What is the empirical formula of a compound which contains 60.0% oxygen and 40.0% sulfur by mass? [Ans. SO3]

Q.6

A compound gave on analysis the following percent composition: K = 26.57%, Cr = 35.36%, O = 38.07%. Derive the empirical formula of the compound. [Ans. K 2Cr2O7]

Q.7

A chloride of sulphur was found to have a molecular mass of 135. A 5.4 gm sample was also found to contains 2.84 gm of chlorine. Deduce the molecular formula of the chloride. [Ans.S2Cl2]

Q.8 (a) (b)

One of the earliest methods for determining the molecular weight of proteins was based on chemical analysis. A hemoglobin preparation was found to contain 0.335% iron. If the hemoglobin molecule contains one atom of iron, what is its molecular weight? If it contains 4 atoms of iron, what is its molecular weight?

BULLS EYE


DATE : 19/05/2008

PHYSICAL CHEMISTRY


DPP. NO.- 7

Q1.

Calculate the number of moles of ammonia required to produce 2.5 mole of [Cu(NH 3)4]SO4. [10]

Q2.

Calculate the mass of BaCO 3 produced when excess of CO 2 gas passes through the solution containing 0.205 mole of Ba(OH) 2. [40.4]

Q3.

0.6 moles of Cu2S is roasted with excess of Oxygen to produce copper metal & SO 2. Calculate the [76.2] weight of Cu metal produced?

Q4.

Purest form of Carbon is formed by heating sugar (C 12H22O11) in an inert atmosphere. What is the maximum amount of Carbon which can be obtained by heating 68.4 gm of sugar. [28.8 gm]

Q5.

A sample of K 2CO3 weighing 27.6 gm was treated by a series of reagents to convert all of its Carbon to [11.6 gm] K 2Zn3[Fe(CN)6]2. How many grams of product were obtained.

Q6.

What molar concentration of K 4[Fe(CN)6] should be used so that 40 ml of the solution liberates 130.8 [1/30] mg Zn (dissolved) by forming K 2Zn3[Fe(CN)6]2. (Zn = 65.4)

Q7.

A piece of plumber’s solder weighing 3.00 gm was dissolved in dilute nitric acid, then treated with dilute 2SO4. This precipitated the lead as PbSO 4, which after washing and drying weighed 2.93 gm. The solution was then neutralized to precipitate stannic acid, H 4SnO4, which was decomposed by heating yielding 1.27 gm SnO 2. What is the analysis of the solder as % Pb & % Sn. [Pb = 66.7%; Sn = 33.3%]

Q8.

The chemical formula of a compound Verseneis C 2H4 N2 (C2H2O2 Na)4. If each mole of this could bind 1 mole of Ca 2+, what would be the rating of pure Versene, expressed as mg CaCO 3 bound per gm of the compound?

Q9.

A mixture containing onlyFeCl 3 and AlCl3 weights 5.95 gm. The chlorides are converted into the anhydrous oxides and ignited to Fe 2O3 and Al2O3. The oxides mixture weighs 2.62 gm. Calculate the percent Fe & [Fe = 18.22%; Al = 9.53%] Al in the original mixture.

Q10. A Metalurgical Engineer, during his project in HINDALCO industry he was working in the unit of recovery of alulminium. When the Baeyer Process is used for recoveringAluminium from Siliceous ores, some Silica is persent in the recovered metal. He discovered that when all silicon in the ore is converted to the form of mud having the average formula; 3Na 2O·3Al2O3·5SiO2·5H2O, the precipitation of Silicon in the mud is complete. A certain ore contained 13% Kaolin (Al 2O3·2SiO2·2H2O) and 87% Gibbsite (Al 2O3·3H2O). What % of total aluminium in this ore is recoverable in the Baeyer’s process. [90 %]

BULLS EYE


PHYSICAL CHEMISTRY


DATE : 22/05/2008

DPP. NO.- 8

Q.1

The weight of 350 ml of diatomic gas at 0°C and 2 atm pressure is 1 gm. What is the atomic weight of the gas ? (A) 17.7 (B*) 15.99 (C) 27.7 (D) 38

Q.2

The vapour density of gas A is four times that of B. If molecular mass of B is M, then molecular mass of A is (A) M (B*) 4M (C) M/4 (D) 2M

Q.3

When 10 gm of Na2SO4. xH2O is heated to dryness 5.035 gm water vapour is produced, therefore the value of x is (A) 10 (B) 7 (C) 6 (D*) 8

Q.4

What is fertilizer rating of NH 4 NO3 (Fertilizer rating = N content by mass).

Q.5

An element X forms oxides having percentages of X equal to 75%, 50% & 80%. Show that if proves law of Multiple Proportions. [Ans.Ratio]

Q.6

A compound contains 8% sulphur by mass. What will be its least molecular weight.

Q.7

If pure Mohr’s salt crystals contain 15% iron and 30% water. How much Fe and how much water [ Ans.Fe = 3gm ; H2O = 6 gm ] should be taken to get 20 gm of its crystals.

Q.8

How many kg of sodium and liquid chlorine can be obtained from 585 metric ton of NaCl ?

[Ans.35%]

[Ans.400]

[Ans.230, 355 MT]

Q.9

A sample of impure cuprite Cu 2O contains 66.6% copper. What is the % of pure Cu 2O by mass in the sample. [Ans.75%]

Q.10 A clay was partially dried and then contained 50% silica and 7% water. The original clay contained 12% [Ans.47.31%] water. What is the % of silica in the original sample?

BULLS EYE


PHYSICAL CHEMISTRY


DATE : 23/05/2008

DPP. NO.- 9

Q1.

What mass of Oxygen gas, O 2 from the air is consumed in the combustion of 702 gm of octane C 8H18, one of the principal components of gasoline? 2C 8H18 + 25 O2  16 CO2 + 18 H2O [2.46 kg]

Q2.

What volume of a 0.75 M solution of hydrochloric acid can be prepared from the HCl produced by the reaction of 43.875 gm og NaCl with an excess of sulphuric acid? NaCl(s) + H2SO4(l)  HCl (g) + NaHSO4(s) [1 litre]

Q3.

A mordant is a substance that combines with a dye to produce a stable fixed colour in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide. 2 CH 3CO2H + Ca(OH)2  Ca(CH3CO2)2 + 2H2O What mass of Ca(OH) 2 is required to react with the acetic acid in 25.0 ml of a solution that has a density of 1.065 gm/ml and contains 60% acetic acid by mass? [9.85 gm]

Q4.

Automotive air bags inflat when a sample of sodium azide, NaN 3 is very rapidly deocmposed 2NaN3 (s)  Na(s) + 3N2(g) What mass of sodium azide is required to produce 358.4 lt of nitrogen gas with a density of 1.25 gm/lt. [693.33 gm]

Q5.

Citric acid C 6H8O7, a component of jams, jellies and fruity soft drinks, is prepared industrially via fermentation of sucrose by the mold Aspergillus niger. The overall reaction is C12H22O11 + H2O + 3O2  2C6H8O7 + 4H2O What mass of citric acid is produced from exactly 1 metric ton of sucrose if the percentage yield is 92.3 [1036 kg] %?

Q6.

Toluene, C6H5CH3 is oxidised by air under carefully controlled conditions to benzoic acid, C 6H5CO2H, which is used to prepare the food preservative sodium benzoate, C 6H5CO2 Na. What is percent yield of a reaction that converts 1 kg of toluene to 1.21 kg of benzoic acid? 2C6H5CH3 + 3O2  2C6H5CO2H + 2H2O [91.24%]

Q7.

Silicon nitride is a very hard, high temeprature resistance ceramic used as a component of tubine blades in jet engines. It is prepared according to the equation 3Si + 2 N 2  Si3 N4 Which is the limiting reactant when 1.96 gm of Si and 1.4 gm of N 2 combine? How much silicon nitride will the reaction produce. [3.27 gm]

Q8.

The phosphorus pentoxide used to produce phosphoric acid for cola soft drinks, under the trade name of “coke”, is prepared by burning phosphorus in oxygen. Determine the limiting reagent when 0.2 moles of P 4 and 0.2 mol of O 2 react according to the chemical equation. P 4 + 5O2  P4O10 Calculate the percent yield if 10.0 gm of P 4O10 is isolated from the reaction.

(a) (b) Q9.

Uranium can be isolated from its ores by dissolving it as UO 2(NO3)2 and then separating it as solid UO2(C2O4)· 3H2O. Addition of 0.4031 gm of sodium oxalate, Na 2C2O4, to a solution contianing 1.481 gm of uranyl nitrate, UO 2(C2O4)·3H2O. Determine the limiting reagent and the percent yield of this reaction. Na 2C2O4 + UO2(NO3)2 + 3H2O UO2(C2O4).3H2O + 2NaNO3

Q10. How many moles of the anesthetic gas halothane

can be prepared from 30 mol of C atoms,

20 mol of H 2 molecules, 12 mole of F2 molecules, 12 moles of Cl 2 molecule & 10 mole of Br 2 molecules? BULLS EYE


PHYSICAL CHEMISTRY


DATE : 24/05/2008

DPP. NO.- 10

Q.1

12 mol of H2 and 11.2 mol of Cl 2 are mixed and exploded. The composition by mol of mixture is (A) 24 mol of HCl (B) 0.8 mol Cl 2 and 20.8 mol HCl (C*) 0.8 mol H 2 & 22.4 mol HCl (D) 22.4 mol HCl

Q.2

In an experiment 2.847 gm of pure MOCl 3 was allowed to undergo a set of reactions as a result of which all the Cl was converted to AgCl. The weight of AgCl was 7.2 gm. Find at. wt. of M. [Ans.47.72]

Q.3

An unidentified bivalent metal M reacts with an unidentified halogen X to form an unknown compound of halogen gas.When 1.12 g of it is heated, 0.72 g of MX is obtained along with 56 ml of halogen gas at STP. Identify the metal. [Ans.Cu]

Q.4

Window glass is made by mixing soda ash , limestone & silica & then heating to 1500°C to drive off CO2 from the mixture. The resultant glass contains 12% Na 2O, 13% CaO & 75%. Silica by mass. How much of each reactant would you start to prepare 0.35 kg of glass? [Ans.0.072 g, 0.081 g, 0.26 g]

Q.5

How many ml of bromine will have the same mass as 12.5 ml of Hg if the densities of bromine and Hg are 3.12 g/ml and 13.6 g/ml respectively : (A*) 54.5 ml (B) 50.5 ml (C) 25.4 ml (D) 45.5 ml

Q.6

One mole of a mixture of CO and CO 2 requires exactly 20 gram of NaOH in solution for complete conversion of all the CO 2 into Na2CO3. How many moles more of NaOH would it require for conversion into Na2CO3 if the mixture (one mole) is completely oxidised to CO 2: (A) 0.2 (B) 0.5 (C) 0.4 (D*) 1.5

Q.7

A 0.596 gm sample of a gaseous compound containing only B and H occupies 484 cm 3 at STP. When the compound is ignited in excess oxygen, all its hydrogen is recovered as 1.17 g H 2O, and all the B is left over as B 2O3. What are the empirical formula and molecular formula, and the molecular weight of B and H compound. What weight of B 2O3 is produced by the combustion.

[Ans.(B2H6), 27.6 gm, 1.5 gm]

Q.8

In an electric arc tube A, 140 g N 2 is taken and in tube B, 20 g H 2 is taken . When half the mass of substance in tube A is transferred to B, reaction took place in B. After the reaction is complete, half of the mass of substance in B is now transferred to A. Find out the final composition of gases in moles in A and B.[no reversibility] [Ans.A (2.084 mol N2 ; 3.33 mol NH3) ; B ( 1.25 mol H2 ; 2.5 mol NH3)]

Q.9

According to following reaction : A + B O3  A3O4 + B2O3 Find the number of moles of A 3O4 produced if 1 mole of A is mixed with 1 mole of B O 3. [Ans.1/3]

Q.10 Methyl benzoateis prepared bythe reaction between benzoic acidand methanol, according tothe equation C6H5COOH + CH3OH  C6H5COOCH3 + H2O Benzoic acid Methanol Methyl benzoate In an experiment 24.4 gm of benzoic acid were reacted with 70.0 mL of CH 3OH. The density of CH3OH is 0.79 g mL –1. The methyl benzoate produced had a mass of 21.6 g. What was the percentage yield of product ? (A) 91.7% (B*) 79.4% (C) 71.5% (D) 21.7% BULLS EYE


DATE : 25/05/2008

PHYSICAL CHEMISTRY

Daily Practice Problems DPP. NO.- 11

Q1.

1 gm of EuCl2 is treated with excess of aqueous AgNO 3 and all the chlorine is recovered as 1.29 g gm of AgCl. Calculate the atomic weight of Eu. (Cl = 35.5, Ag = 108) [152.5]

Q2.

1 gm of a chloride of an element contains 0.835 gm of chlorine. If the vapour density of the chloride is [Ans: 28, 4] 85, find the atomic weight of the element and its valency.

Q3.

The percentage of chlorine in a chloride of an element is 44.71%, 158.5 gm of this chloride on vapourisation occupies a volume of 22.4 litre at NTP. Calculate atomic weight and valency of the element.

Q4.

Cu2S and Ag2S are isomorphous in which percentage of sulphur are 20.14 % and 12.94% respectively. Calculate the atomic weight of silver and sulphur if atomic weight of Cu is 63.5. [Ans: 107.7]

Q5.

The natural titanium oxide known as rutile containing 39.95% of oxygen is isomorphous with SnO [Ans: 48.1] known as cassiterite. Calculate the atomic weight of titanium.

Q6.

116 gm of Fe3O4 has 1.5 mol of Fe. Calculate the molecular weight of Fe 3O4 without using atomic weights of Fe & O. [Ans: 232]

Q7.

100 ml of a 2M solution contains 11.7 gm of a substance. Calculate the molecular weight of the compound.

2

[Ans: 58.5]

Q8.

0.701 gm of silver salt of a dibasic acid on ignition yielded 0.497 gm of metallic silver. Calculate the [Ans: 90.7] molecular weight of the acid. (Ag = 108)

Q9.

0.532 gm of chloroplatinate of a diacid base on ignition left 0.195 gm of residue of Pt. Calculate the molecular weight of the base. (Pt = 195) [Ans: 122]

Q10. A solution containing 3 gm of a monobasic organic acid was just neutralised by 40 ml of 0.5 M NaOH solution. Calculate the molecular weight of the acid.

BULLS EYE


Q1.

DATE : 26/05/2008

PHYSICAL CHEMISTRY


0.16 gm of an organic compound, on complete combustion produced 0.44 gm of CO 2 and 0.18 gm of H2O. Calculate percentage of Carbon and hydrogen in the organic compound. [Ans: 75%C, 25%H]

Q2.

0.66 gm of an organic compound containing C, H & O gave on combustion 0.968 gm of CO 2 and [Ans: 46.67%] 0.792 gm of H 2O. Calculate percentage of O in the compound.

Q3.

0.2033 gm of an organic compound in Duma method gave 31.7 ml of nitrogen at 14 0C & 744 mm of Hg pressure. Calculate percentage of nitrogen in the compound. [Ans: 18.16 %]

Q4.

In the estimation of chlorine in a given compound, it was found that 0.0811 gm of a compound gave 0.2368 gm of silver chloride. Calculate percentage of chlorine in the given compound. [Ans: 72.27%]

Q5.

0.36 gm of an organic compound containing sulphur produced H 2SO4 by Carius method. Which on treatment with BaCl 2 produced quantitatively 0.233 gm of BaSO 4. Calculate percentage of S in the compound? [Ans; 8.89%]

Q6.

In a gravimetric determination of phosphorus, 0.248 gm of an organic compound was stronglyheated in a Carius tube with concentrated nitric acid. Phosphoric acid so produced was precipitated as MgNH 4PO4 which on ignition yielded 0.444 gm of Mg 2P2O7. Find the percentage of phosphorus in the compound. [Ans: 50%] [Mg = 24, P = 31, O = 16]

Q7.

1.525 gm of an organic compound was heated with NaOH solution and ammonia so produced was passed into 30 ml of M HCl solution. The remaining HCl was further neutralised by 120 ml of M/10 NaOH solution. Calculate percentage of Nitrogen in the compound. [Ans: 16.52%]

Q8.

Find the percentage of nitrogen in an organic compound analysed by Kjeldahl method. 1.61 gm of the compound produced NH 3 which was absorbed in 250 ml of M/4 H 2SO4 solution. The remaining acid was diluted to one litre, 25 ml of which required 25.5 ml of M/10 NaOH for exact neutralisation. [Ans: 20%]

Q9.

0.123 gm of an organic compound produced 0.099 gm of CO 2 and 0.0507 gm of H 2O. 0.185 gm of the same compound produced 0.319 gm of AgBr. Find the percentages of C, H and Br in the compound. Also calculate empirical formula of the compound. (21.96%C, 4.48% H, 73.37% Br, C 2H5Br)

Q10. 0.42 gm of an organic compound containing C, H, O & N gave on combustion 0.924 gm of CO 2 and 0.243 gm of water. 0.208 gm of the substance when distilled with NaOH gave NH 3, which required 30 ml of M/40 H2SO4 solution for neutralisation. Calculate the amount of each element in 0.42 gm of the compound. (0.252 gm C, 0.027 g H, 0.042 g N & 0.099 gm O)

BULLS EYE


PHYSICAL CHEMISTRY


DATE : 28/05/2008

DPP. NO.- 13

Q.1

A certain compound hasthemolecular formula X 4O6. If 10 gm of X 4O6 has 5.72 g X, atomic mass of X is (A*) 32 amu (B) 37 amu (C) 42 amu (D) 98 amu

Q.2

Xenon is 0.0000087% by volume of the dry atmosphere. Express it in ppb (vol.)

Q.3

600 ml of a mixture of ozone and oxygen at STP weighs one gm. Calculate the volume of ozone at STP in the mixture. [Ans.200 ml]

Q.4

An evacuated glass bulb was weighed, then filled with oxygen and reweighed, the difference in weight was 0.25 gm. The operation was repeated under identical conditions of (P, V, T) with unknown gas “X”. The difference was 0.375 gm, find molecular weight. [Ans. Mwt.= 48]

Q.5

If ratio of mole fraction of solute to solvent is unity, what would be % by wt. (concentration of solute) (M solute = molecular mass of solute, M solvent = M molecular mass of solvent) (A*)

× 100

(C) 66.67 % Q.6

[Ans.87 ppb]

(B) 50% (D)

× 100

Molarity of pure ethanol (C 2H5OH) with density d g/ml. (A)

(B) 25 d

(C*) 21 d

(D)

Q.7

To a sample of an element X (at. wt. 70) element Y (at. wt. 120) is added as impurity. The ratio of atoms of X to Y in the mixture is 1 : 10 –7. How many gms of Y will be required for 100 gm of X for this ratio. [Ans.1.714×10 –5 gm]

Q.8

A ‘w’ gm mixture of AgCl &AgBr undergoes a loss in wt. by 5% if it is exposed to excess chlorination. Calculate the percentage composition of the mixture. [Ans. % AgBr = 21.12 % AgCl = 78.88% ]

Q.9

On analysis of blood sample of a driver suspected of being drunk over than the permissible value, it was obtained that 60 gm sample reacted with 30 ml of 8 M K 2Cr 2O7 (acidic soln.). If the permissible value for the alcohol content in the blood is 1 % by mass, will the driver be prosecuted for drunken drining. [Assume K 2Cr 2O7 reacts only with the alcohol present in blood.] Reaction : 2K 2Cr 2O7 + 8H2SO4 + C 2H5OH  2Cr 2(SO4)3 + 11H2O + 2K 2SO4 + 2CO2

Q.10 It is known that when 1 mole of H + reacts with 1 mole of OH – x kJ of energy is released. Suppose 200 ml of 1 M XOH is mixed with 100 ml 2M H 2Y [dibasic acid]. Calculate the total amount of energy released. Assume that dissociation of the acid & base is complete & requires no energy. What should be the volume of each solution that should be taken to obtain 100 ml of solution & maximum release of energy. [Ans. 0.2x kJ; 80 ml, 20 ml]

BULLS EYE


DATE :08/06/2008

PHYSICAL CHEMISTRY


Q1.

15 ml of a gaseous hydrocarbon was required for complete combustion in 357 ml of air (21% of oxygen of volume) and the gaseous products occupied 327 ml (all volume being measured at NTP). What is the formula of the hydrocarbon? [Ans: C 3H8]

Q2.

A sample of gaseous hydrocarbon occupying 1.12 litre at NTP when completely burnt in air produced 2.2 gm of CO 2 and 18 gm of H 2O. Calculate the weight of compound taken and the volume of O 2 at NTP required for its burning. Find the molecular formula of the hydrocarbon? [Ans: CH4, 0.8 gm, 2.24 litre]

Q3.

9 volumes of a gaseous mixture consisting of a gaseous organic compound A and just sufficient amount of oxygen required for complete combustion yielded on burning 4 volumes of CO 2, 6 volumes of water vapour & 2 volumes of N 2, all volumes measured at the same temperature and pressure. If the compound A contained only C, H and N (i) how many volumes of oxygen are required for complete combustion and (ii) what is the molecular formula of the compound A? [Ans: (i) 7 volumes; (ii) C 2H6 N2]

Q4.

20 ml of a mixture of C 2H2 and CO was exploded with 30 ml of oxygen. The gases after the reaction had a volume of 34 ml. On treatment with KOH, 8 ml of oxygen remained. Calculate the composition of the mixture. [Ans: C 2H2 = 6 ml, CO = 14 ml]

Q5.

40 ml of a mixture of hydrogen, CH 4 and N2 was exploded with 10ml of oxygen. On cooling, the gases occupied 36.5 ml. After treatment with KOH, the volume reduced by 3 ml and again on treatment with alkaline pyrogallol, the volume further decreased by 1.5 ml. Determine the compoistion of the mixture. [Ans:H2 = 12.5%, CH4 =- 7.5%, N2 = 80%]

Q6.

1 litre of a mixture of CO and CO 2 is taken. This mixture is passed through a tube containing red hot charcoal. The volume now becomes 1.6 litres. The volumes are measured under the same condition. Find the composition of the mixture by volume. [Ans. CO 2 = 0.6 lt, CO = 0.4 lt]

Q7.

40 ml of ammonia gas taken in an endiometer tube was subjected to sparks till the volume did not further change. The volume was found to increase by 40 ml. 40 ml of oxygen gas then mixed and the mixture was further exploded. The gases remained were 30 ml. Deduce the formula of ammonia. [Ans: NH 3]

Q8.

When 100 ml of a O 2 –O3 mixture was passed through turpentine, there was reduction of volume of 20ml. If 100 ml of such a mixture is heated, what will be increase in volume? [Ans: 10 ml]

Q9.

One lt. of sample of ozonised oxygen weighs 1.5g at 0 oC & one atm pressure. 100 ml of this sample reduced to 90 ml, when treated with turpentine under the same conditions. Find the molecular wt. of ozone. [Ans:48]

Q10. Crude calcium carbide is made in an electric furnace by the following equation:CaO + 3C CaC2 + CO The product contains 85% of CaC 2 & 15% of unreated CaO. (a) How much CaO is to be added to the furnace charge for each 1000 kg of CaC 2 (Crude) product? (b) How much CaO is to be added to the furnace charge for each 1000 kg of CaC 2 (pure) product? [Ans: (a) 893.75kg, (b) 1051.47 kg] BULLS EYE


PHYSICAL CHEMISTRY


DATE : 08/06/2008

DPP. NO.- 15

Q1.

On passing 25 ml of a gaseous mixture of N 2 & NO over heated copper, 20 ml of the gas remained calculate the % of each in the mixture. [Ans: N 2 = 60%, NO = 40%]

Q2.

A mixture of oxygen & hydorgen is analysed by passing it over hot copper oxide & through a during tube. Hydrogen reduces the CuO according to the equation, CuO + H 2 Cu + H2O, oxygen then oxidises the copper formed: Cu +

O2

CuO. 100 cm3 of the mixture measured at 25 oC & 750 mm

yields 84.5 cm 3 of dry oxygen measured at 25 oC & 750 mm after passing over CuO & drying agent. What is the mole % of H 2 in the mixture? [Ans: 10.3%] Q3.

At high temperature the compound S 4 N4 decomposes completely into N 2 & sulphur vapour. If all measurement are made under the same conditions of temperature and pressure, it is found that for each volume S4 N4 decomposed 2.5 volumes of gaseous products are formed. What is the molecular formula of sulphur? [Ans: S 8]

Q4.

Determine the formula of ammonia from the following data : volume of ammonia = 25 ml volume on addition of O 2 after explosion = 71.2 ml volume after explosion with O 2 on (cooling) = 14.95 ml volume after being absorbed by alkaline pyrogallol = 12.5 ml

[Ans: NH 3]

Q5.

20 ml of a mixture of methane and a gaseous compound of Acetylene series were mixed with 100 ml of oxygen and exploded. The volume of products after cooling to original room temperature and pressure, was 80 ml and on treatment with potash solution a further contraction of 40 ml was observed. Calculate (a) the molecular formula of hydrocarbon (b) the percentage composition of the mixture. [Ans. (a) C3H4, (b) 50]

Q6.

At 300 K and 1 atm pressure a gaseous mixture having mass ratio 1 : 2 : 2 of C 2H4, CO & N2 gas respectively is taken in an eudiometer tube. This mixture is exploded, resulting gas mixture cooled to initial conditions was passed through KOH solution, volume decreased by 9.852 lts, find the molar mass of initial and final gaseous mixture. [Ans: 28, 28]

Q7.

A 50 ml sample of hydrogen and oxygen mixture was placed in a tube at 18 oC and confined at barometric pressure. A spark was passed through the sample so that the formation of water would go to completion. The resulting pure gas had a volume of 10 ml at the barometric pressure. What was the initial mole percent of hydrogen in the mixture if the residual gas after sparking was hydrogen. [Ans: 73.4%]

Q8.

A mixture of H 2 and acetylene (C 2H2) was collected in a eudiometer tube. Then 60 ml of oxygen were also introduced. The resulting mixture of all the gases was exploded. On cooling it was found to have undergone a contraction of 45 ml. On introducing caustic potash solution a further contraction of 32 ml occurred and 13 ml of oxygen alone were left behind. Calculare the percentage composition of the mixture of hydrogen and acetylene.

Q9.

[Ans:

,

]

A mother cell desintegrate into sixtyidentical cells and each daughter cell further disintegrate into twenty four smaller cells. The smallest cells are uniform cylindrical in shape with diameter of 120Å & each cell is 6000Å long. Determine molar mass of the mother cell if density of the smallest cell is 1.12 gm/cm 3. [Ans: 6.6 x 1010 gm/mol]

BULLS EYE

Q10. Question No. (a) to (b) (2 questions) A mixture of H 2 and Acetylene (C2H2) was collected in a Eudiometer tube. Then, 60 ml of oxygen were also introduced. The resulting mixture of all the gases was exploded. After cooling a resulting gaseous mixture passes through Caustic potash solution a contraction of 32 ml occurred and 13 ml of oxygen alone were left behind. (a).

After explosion, on cooling of resulting mixture, contraction in volume will be (A) 21 ml (B) 30 ml (C*) 45 ml (D) none

(b).

Percentage composition of the gaseous mixture of H 2 & acetylene are (A) 53.3, 46.7 (B*) 46.7, 53.3 (C) 15.7, 84.3 (D) 84.3, 15.7


PHYSICAL CHEMISTRY


DATE : 08/06/2008

DPP. NO.- 16

Q1.

A definite amount of gaseous hydrocarbon having (carbon atoms less than 5) was burnt with sufficient amount of O 2. The volume of all reactants was 600 ml, after the explosion the volume of the products [CO2(g) and H2O(g)] was found to be 700 ml under the similar conditions. The molecular formula of the compound is (A*) C3H8 (B) C3H6 (C) C3H4 (D) C4H10

Q2.

A mixture (15 mL) of CO and CO 2 is mixed with V mL (excess) of oxygen and electrically sparked. The volume after explosion was (V + 12) mL. What would be the residual volume if 25 mL of the original mixture is exposed to KOH. All volume measurements were made at the same temperature and pressure (A) 7 mL (B) 12 mL (C*) 10 mL (D) 9 mL

Q3.

One gram of the silver salt of an organic dibasic acid yields, on strong heating, 0.5934 g of silver. If the weight percentage of carbon in it 8 times the weight percentage of hydrogen and one-half the weight percentage of oxygen, determine the molecular formula of the acid. [Atomic weight of Ag = 108] (A) C4H6O4 (B*) C4H6O6 (C) C2H6O2 (D) C5H10O5

Q4.

A mixture of C 3H8 (g) & O2 having total volume 100 ml in an Eudiometry tube is sparked & it is observed that a contraction of 45 ml is observed what can be the composition of reacting mixture. (A*) 15 ml C 3H8 & 85 ml O2 (B*) 25 ml C3H8 & 75 ml O2 (C) 45 ml C 3H8 & 55 ml O2 (D) 55 ml C3H8 & 45 ml O2

Q5.

Carbon can react with O 2 to form CO & CO2 depending upon amount of substances taken. If each option is written in an order like (x, y, z, p) where x represents moles of C taken, y represents moles of O2 taken z represents moles of CO formed & p represents moles of CO 2 formed, then which options are correct. (A*) (1, 0.75, 0.5, 0.5) (B) (1, 0.5, 0, 0.5) (C) (1, 0.5, 0.5, 0) (D) (1, 2, 1, 1)

Q6.

One mole mixture of CH 4 & air (containing 80% N 2 20% O2 by volume) of a composition such that when underwent combustion gave maximum heat (assume combustion of only CH 4). Then which of the statements are correct, regarding composition of initial mixture.(X presents mole fraction)

Q7.

(A*)

(B)

(C)

(D) Data insufficient

A excess of potassium permanganate solution was added to 20 ml of an aqueous solution of H 2O2. Oxygen liberated was added to a mixture of H 2 & CH4 in an eudiometer tube & exploded. The contration in volume was found to be 45 ml was then treated with KOH and further contration of 5 ml was observed. The residual gas was found to be 14.26 ml of oxygen. (Eudiometer tube is mantained to 1 atm & 300 K). Balanced chemical equation:. 2KMnO4 + 6H+ + 5H2O2 2K + + 2Mn2+ + 8H2O + 5O2 (a) Calculate molar ratio of H 2 : CH4 in the mixture. (b) Calculate total volume of O 2 released by H 2O2 on reaction with KMnO 4 (at 1 atm & 300 K). (c) Calculate volume strength of H 2O2. [Ans: (a) 1 : 15, (b) 49.26 ml, (c) 1.12V]

BULLS EYE

Q8.

Question No. (a) to (b) are based on the following Passage. Read it carefully & answer the questions that follow

A monobasic acid of weight 15.5 gms is heated with excess of oxygen & evolved gases when passed through KOH solution increased its weight by 22 gms and when passed through anhydrous CaCl 2, increased its weight by 13.5 gms. When the same mass of this organic acid is reacted with excess of silver nitrate solution form 41.75 gm silver salt of the acid which on ignition gave the residue of weight 27 gm. (a). (b).

Q9.

The molecular formula of the organic acid is (A) C2H6 (B) C2H5O2

(C*) C2H6O2

The molar masses of the acid & its silver salt respectively are (A) 60, 168 (B) 167, 60 (C*) 60, 167

(D) C2H4O (D) 168, 60

Question No. (a) to (c) (3 questions)

N2O5 and H2O can react to form HNO 3, according to given reaction N2O5 + H2O  2HNO3 the concentration of a mixture of HNO 3 and N2O5 (g) can be expressed similar to oleum. Then answer the following question. (a).

Find the percentage labelling of a mixture containing 23 gm HNO 3 and 27 gm N 2O5. (A) 104.5% (B*) 109% (C) 113.5% (D) 118%

(b).

Find the maxmum and minimum value of percentage labelling (A) 133.3 % (B) 116.66%, 0% (C*) 116.66%, 100% (D) None

(c).

Find the new labelling if 100 gm of this mixture (original) is mixed with 4.5 gm water (A) 100 +

(B*) 100 +

(C) 100 +

(D) 100+


PHYSICAL CHEMISTRY


DATE : 17/06/2008

DPP. NO.- 17

Q1.

If a one litre sample of a gas at 760 torr is compressed to 0.8 litre at constant temperature. Calculate the final pressure of the Gas. [950 torr]

Q2.

At what temperature does a sample of gas occupy 4.0 litre at 1.11 atm, if it initially occupied 2.22 litre at 1.0 atm and 60 0C. [393oC]

Q3.

To what temperature must a neon gas sample be heated to double its pressure if the initial volume of gas at 750C is decreased by 15%. [318.6 o]

Q4.

A meteorological balloon has a radius of 1m. when released from sea level at normal pressure (1 atm) and 250C and expanded to a radius of 3m. When it has risen to its maximum height altitude where the temperature was –20 0C. What is the pressure inside the balloon at that height. [0.03144 atm]

Q5.

A refrigeration tank holding 5.0 litre Freon (C 2Cl2F4) at 250C and 3 atm. Pressure developed a leak and repaired, the tank has lost 76 gm of gas. What was the pressure of the gas remaining in the tank. [0.828 atm]

Q6.

If the maximum volume of a cylinder in an automobile engine is 0.63 lt. the volume after the air fuel mixture is compressed in 0.068 lt, and the fuel mixture is drawn at 1 atm pressure, what is the pressure in the cylinder at maximum compression? [9.26 atm]

Q7.

A large irregularly shaped closed tank is first evacuated and then connected to a 50 litre cylinder containing compressed nitrogen gas. The gas pressure in the cylinder originally at 21.5 atm, falls to 1.55 atm after it is connected to the evaculated tank. Calculate volume of the tank. [643.55 lt]

Q8.

A gaseous system has a volume of 580 cm 3 at a certain pressure. If its pressure is increased by 0.96 atm, its volume becomes 100 cm 3. Determine the pressure of the system. [0.2 atm]

Q9.

Liquefied natural gas (LNG) in mainly methane (MW = 16). A 10 m 3 tank is constructed to store LNG at –164 0C and 1 atm pressure under which conditions its density is 415 kg/m 3. calculate the volume of a storage tank capable of holding the same mass of LNG as a gas at 20 0C and 1 atm pressure. [6231.74 m3]

Q10. In the estimation of molar mass of an organic base by chloroplatinate salt method, different masses of residue obtained (w 2 gm) are plotted against known amount of its salt (w 1 gm), and following curve is plotted. With the help of given graph, calculate molar mass of organic base. Given that molar mass of salt of the base is 1408. (Pt = 195) [89]

BULLS EYE


Q1.

PHYSICAL CHEMISTRY


DATE : 17/06/2008

DPP. NO.- 18

According to the ideal gas law, the molar volume of a gas equals: (A*)

RT p

(B)

RT pV

(C) 22.4L

(D)

RT g pV

Q2.

A fire extinguisher contains 4.4 kg of CO 2. The volume of CO 2 delivered by this fire extinguisher at room temperature is: (A) 24.5 litres (B) 10 x 24.5 litres (C*) 100 x 24.45 litres (D) 1000 x 24.5 litres

Q3.

What is the conclusion you would draw from the following graphs?

(A) As the temperature is reduced, the volume as well as the pressure increase. (B) As the temperature is reduced, the volume becomes zero and the pressure reaches infinity (C*) As the temperature is reduced, both volume and the pressure decrease. (D*) A point is reached, where theorectically, the volume as well as the pressure becomes zero. Q4.

At the top of the mountain the thermometer reads 0 0C and the barometer reads 710 mm Hg. At the bottom of the mountain the temperature is 30 0C and pressure is 760 mm Hg. Density of air at the top with that at the bottom is (A) 1 : 1 (B*) 1.04 : 1 (C) 1 : 1.04 (D) 1 : 1.5

Q5.

A quantity of gas is collected in a graduated tube over the mercury. The volume of the gas at 20 0C is 50.0 mL and the level of the mercury in the tube is 100 mm above the outside mercury level. The barometer reads 750 mm. Volume at STP is (A*) 39.8 mL (B) 40 mL (C) 42 mL (D) 60 mL

Q6.

I, II, III are three isotherm respectively at T 1, T2 and T3. Temperature will be in order (A) T1 = T2 = T 3 (B) T1 < T2 < T3 (C*) T 1 > T2 > T3 (D) T1 > T2 = T 3

Q7.

A certain mountain is 14,100 feet above sea-level. The pressure at the top is 17.7 inches of Hg. If you blow up a balloon at sea level, where the pressure measured to be 29.7 inches and carried it to the top to the top of the mountain, by what factor would its volume change (A) 29.7 –17.7 (B*) 29.7/17.7 (C) 17.7/29.7 (D) no change

BULLS EYE

Q8.

1.509 g sample of an osmium oxide (OsO x) which melts at 40 0C and boils at 130 0C is placed in a cylinder with a movable piston that enables the cylinder to expand against the atmospheric pressure of 745 torr. When the sample is heated to 200 0C, it is completely vaporized and the volume of the cylinder expands to 235 mL. Choose the correct alternatives (A*) the molar mass of the oxide is 254 g/mol (B) the molar mass of the oxide is 238 g/mol (C*) x = 4 (D) x = 3

Q9.

Choose the correct statements (A*) The density of a gas varies with temperature (B*) The density of a gas varies with pressure (C) The volume of a gas always increases with the temperature increases (D*) The volume of a gas may not increase with the temperature increases

Q10. At a temperature T K , the pressure of 4.0 g argon in a bulb is p. The bulb is put in a bath having temperature higher by 50K than the first one. 0.8 g of argon gas is to be removed to maintain original pressure. The temperature T is equal to (A) 510 K (B*) 200 K (C) 100 K (D) 73 K


PHYSICAL CHEMISTRY


DATE : 23/06/2008

DPP. NO.- 19

Q1.

Equal weights of ethane and hydrogen are mixed in an empty container at 25 oC. Total pressure exerted by hydrogen is: (A) 162 (B) 1 : 1 (C) 1 : 16 (D*) 15 : 16

Q2.

A 0.5 litre flask contains gas ‘A’ and a one litre flask contains gas ‘B’ at the same temperature. The density of gas ‘A’ is 3.0 gm/l and that of ‘B’ 1.5 gm/l. The molecular mass of gas ‘A’ is one half that of gas ‘B’. The ratio of pressures P A/P B exerted by the gases is: (A*) 4 (B) 3 (C) 2 (D) 1

Q3.

A 1 : 1 mixture (by weight) of hydrogen and helium is enclosed in a one litre flask at temperature 0oC. Assuming ideal behaviour, the partial pressure of helium is found to be 0.42 atm then concentration of hydrogen would be (A*) 0.0375 (B) 0.028 (C) 0.0562 (D) 0.0187

Q4.

Daltions’s law of partial pressure is not applicable to: (A) O 2 + O3 (B) CO + CO2 (C*) NH3 + HCl

Q5.

(D) I2 + O2

To which of the following gaseous mixture’s Dalton’s law not applicable (A) Ne + He + SO 2 (B*) NH3 + HCl + HBr (C*) NO + O2 + CO2

(D) N2 + H2 + O2

Q6.

A 1.0 g sample of air consists of approximately 0.76 g of nitrogen and 0.24 g of oxygen. This sample occupies a 1.0L vessel at 20 0C. Then (A) the partial pressure of N 2 is 1.45 atm (B) the partial pressure of O 2 is 0.36 atm (C*) the total pressure is 0.83 atm (D) the total pressure is 1.05 atm.

Q7.

Weight of 112 mL of oxygen at NTP on liquefaction would be: (A) 0.32 g (B) 0.64 g (C*) 0.16 g

(D) 0.96 g

Q8.

A flask contain 36% nitrogen, 39% oxygen and 25% hydrogen. If the total pressure of the mixutre of gases is 500 mm, the partial pressure of oxygen is (A) 120 mm (B) 145 mm (C) 195 mm (D) 210 mm

Q9.

A gas in an open container is heated from 27 oC to 127 oC. The fraction of the original amount of gas escaped from the container will be (A) 3/4 (B) 1/2 (C) 1/4 (D) 1/8

Q10. A gas in an open container is heated from 27 oC to 127o. The fraction of the original amount of gas remaining in the container will be (A) 3/4 (B) 1/2 (C) 1/4 (D) 1/8 Q11. Qualitative analysis of an unknown acid found only carbon, hydrogen and oxygen. Quantitative analysis gave the following data. A 10.46 mg sample, when burned in oxygen, gave 22.17 mg CO 2 and 3.4 mg H 2O. Its molecular weight was determined to be 166. When 0.168 gm sample of the acid was titrated with 0.125 M NaOH, the end point reached after 16.18 ml of the base had been added (i) What is the empirical formula? (ii) What is molecular formula? (iii) Is the acid mono–, di– or triprotic? [Ans: (i) C 4H3O2 (ii) C8H6O4 (iii) Diprotic] BULLS EYE


PHYSICAL CHEMISTRY


DATE : 23/06/2008

DPP. NO.- 20

Q1.

A mixture of three gases A (density = 0.82), B (density = 0.26) and C (density = 0.51) is enclosed in a vessel at constant temperature when equlibrium is established: (A) Gas A will be at the top in the vessel (B) Gas B will be at the top in the vessel (C) Gas C will be between gases A and B in the vessel (D*) The gases will mix homogeneously throughout the vessel.

Q2.

Two gases A and B having same volume diffused through a pin hole in 10 and 5 seconds respectively. The molecular weight of A is 48. What would be the molecular weight of B? (A*) 12.0 (B) 15.5 (C) 18.0 (D) 14.5

Q3.

Under the similar conditions of P and T the rate of diffusion of hydrogen is about: (A) Twice that of He (B) Four times that of He (C) One half that of He (D*) 1.4 times that of He

Q4.

A commercial gas cylinder contained 50 litres of helium at 27 oC and 10 atmospheric pressure. Assuming ideal gas behaviour of helium find the number of 5 litre balloon that can be filled up at N.T.P. (A) 91 (B) 81 (C) 10 (D) 273

Q5.

Vaporisation of 0.24 gm of a volatile substance gave 45 ml of vapour at NTP. What will be the vapour density of the substance (density of H 2 at NTP = 0.089 gram per litre) (A) 95.39 (B) 5.993 (C) 95.93 (D) 59.93

Q6.

A vessel contains 1.7 g NH 3 gas and 5.1 g H 2S gas. The value of effective molar mass of this gaseous mixture at 27 oC will be (A) 34 (B) 27.2 (C) 25.5 (D) 68

Q7.

One mole of N 2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N 2O4(g) decomposes to NO 2(g). The resultant pressure is (A) 1.2 atm (B) 2.4 atm (C) 2.0 atm (D) 1.0 atm

Q8.

The rate of diffusion of a gas having molecular weight just double of molecular weight of nitrogen gas, is 56 ml per second. The rate of diffusion of nitrogen will be (A) 56 ml/sec (B) 28 ml/sec (C) 112 ml/sec (D) 79.19 ml/sec

Q9.

50 ml of hydrogen diffuses out through a small hole from a vessel in 20 minutes. Time needed for 40 ml of oxygen to diffuse out is (a) 12 minutes (b) 64minutes (c) 8 minutes (d) 32 minutes

Q10. 16 ml of hydrogen was found to diffuse in 60 seconds. What volume of suphur dioxide would diffuse in the same time under the same conditions (a) 2 ml (b) 4 32ml (c) 2 2ml (d) 60 ml Q11. Two grams of H 2 diffuses from a container in 10 minutes. How many gram of oxygen would diffuse through the same container in the same time under similar conditions (a) 0.5 gms (b) 4gms (c) 6gms (d) 8gms Q12. (a) How much 85% pure salt cake (Na 2SO 4) could be produced from 250 kg of 95% pure salt NaCl in the reaction, 2NaCl + H 2SO4  Na2SO 4 + 2HCl (339 kg) (b) A mixture of FeO & Fe2O3 is reacted with acidified KMnO 4 solution having a concentration of 0.25 M, 100 ml of which was used. This solution was then titrated with Zn dust which converted Fe3+ of the solution to Fe 2+, the Fe 2+ required 1000 ml of 0.1 M K 2Cr 2O7 solution. Find % of FeO & Fe2O3 in the mixture. [Ans: 19.15% FeO & 80.85% Fe 2O3] BULLS EYE


PHYSICAL CHEMISTRY


DATE : 27/06/2008

Q.1

Which of the following statements is (are) true. Justify your answers.

(a)

Graph between

(b) (c) (d)

DPP. NO.- 21

1 vs P for 1 mole of gas at 300 K has a slope between 0 & 1. V Isobar between log V vs log T has varying slope depending upon the values of n & P. Isobars between log V vs log T has varying intercepts on y & x axis depending upon the values of n & P. Isochore between PT vs P for constant n is a straight line.

Q.2

P vs V curves were plotted for three different samples containing same masses of H 2, O2 & N2 at same temperature. Markout which graph is applicable for which sample.

Q.3

A gas at a pressure of 5.0 atm is heated from 0° to 546°C and simultaneously compressed to one-third of its original volume. Hence final pressure is (A) 10.0 atm (B) 30.0 atm (C) 45.0 atm (D) 5.0 atm

Q.4

A quantity of hydrogen gas occupies a volume of 30.0 mL at a certain temperature and pressure. What volume would half this mass of hydrogen occupy at triple the absolute temperature if the pressure were one-ninth that of the original gas? (A) 270 mL (B) 90 mL (C) 405 mL (D) 135 mL

Q.5

A sample of air contains only N 2, O2 and H2O. It is saturated with water vapours and total pressure is 640 torr. The vapour pressure of water is 40 torr and the molar ratio of N 2 : O2 is 3 : 1. The partial pressure of N2 in the sample is (A) 540 torr (B) 900 torr (C) 1080 torr (D) 450 torr

Q.6

Calculate the total pressure in a mixture of 4g O 2 & 2gH2 confined in a bulb of 1 litre at 0°C. What will be the pressure if the mixture is sparked to cause complete reaction & final temperature is kept at 27°C. What will happen if temperature is kept at 127°C.[Aq. tension of water at 27°C = 12 mm of Hg] [25.2 atm, 12.33 atm, 32.84 atm]

Q.7

A mixture of H 2 and CH4 has a density 0.25 times as the density of O 2 at the same T and P. Find the % of CH4 by mole in the mix. [Ans.85 %]

Q.8

NH3 gas when sparked continuously for a long time in a closed rigid container to cause its complete dissociation it was found that pressure inside the container was 4 atm. In another similar container (same volume and same temperature) same amount of NH 3 gas was sparked for small time and the pressure was 3 atm. Calculate Initial pressure of NH 3 taken. Pressure of each gas in the second container at the given condition.

(i) (ii)

[Ans. (i) 2 atm, (ii) NH3 – 1 atm, N2 – Q.9

1 3 atm, H2 – atm] 2 2

A mixture of NH3 and N2H4 in a closed vessel at 300 K. The total pressure is 0.5 atmp. The vessel is heated to 1200 K at which both decompose completely 2NH3  N2 + 3H2; N2H4  N2 + 2H2; and final pressure of mixture becomes 4.5 atmp. Find mol % of hydrazine in the original mixture. [Ans.25%]

Q10. In a 100 lt flask 1 mole of A 2B2 are taken initially at a temperature of 300K. At 300K, A 2B2 does not dissociate. However when the temp is raised to 600 K. A 2B2 starts dissociating at the rate of 10 -2 mole /minute. Calculate the partial pressure at A 2B2, A2 & B2 after (a) 10 minutes (b) 100 minutes respectively.

BULLS EYE


DATE : 30/06/2008

PHYSICAL CHEMISTRY


Q1.

Temperature of a fixed amount of a gas is decreased from 20 0C to -20 0C. The pressure of the as is doubled during this process. Find out the new volume of the gas in terms of precentage of the original volume. [43.17%]

Q2.

An open vessel at 270 is heated unitil 3/5 parts of the air in it as expelled. Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated. [477 oC]

Q3.

The volume of a hot-air balloon is 100 m 3 , the temperature of the air outside the balloon is 25 0C, the pressure is 1.00 bar and the average molecular weight of air is29g/mol.If the balloon is to be used to lift a mass of 1.00 kg. what should be the minmum temperature of air inside the balloon? [300.53K]

Q4.

Consider a hot-air balloon that appoximates a sphere 5.0 m in diameter and contains air heated to 65 0C. The surrounding air temperature is 21 0C.The pressure in the balloon is equal to the atomspheric pressure, which is 745 torr. (i) What total mass can the balloon lift ? Assume the average molecular weight of air as 29.0. (ii) If the balloon is filled with helium at 21 0Cand all other conditions are as stated above, what total mass can the ballon lift ? (iii) What mass could the hot-air balloon in part (i) lift if it were on the ground where the atmospheric pressure is 630 torr [(i) 10.041 Kg, (ii) 66.5 kg, (iii) 8.491 Kg]

Q5.

What is the approximate mass of the atmosphere of the earth. Assume the radius of the earth to be 6370 km. [5.264 x 10 18 Kg]

Q6.

250 ml nitrogen at 720 mm pressure and at a particular temperature and 380 ml of oxygen at 650 mm pressure and at the same temperature are taken in a 1litre flask at the same temperature .find out the pressure of the gaseous mixture in the flask . [427 ml]

Q7.

A one litre flask contains oxygen at 950 mm pressure . A two litre flask contains oxygen at 100 mm pressure What would be the pressure of the gas if the two flasks are joined by a very small and thin tube? Consider the volume of the tube to be negligible. Assume temp. to be same. [383.33]

Q8.

Calculate the total pressure in a 10 L cylinder which contains 0.4 g of helium 1.6 g of oxygen and 1.4 g of nitrogen at 27 0C. Also calculate the partial pressure of heliyum gas in the cylinder. Assume the ideal behaviour for gases. [0.492 atm, 0.246 atm]

Q9.

Vaporisation of 0.168 g of a substance displaced 49.4 cm 3 of air at 200C and 740 mm of Hg. The air was collected by downward displacment of water. If the vapour pressure of water at 20 0C is equal to the pressure of 1.8 cm of Hg,find out the moleular weight of the substance. [86]

Q10. A piece of soild carbon di-oxide, with a mass of 5.6g, is placed in a 4.0 litre otherwise empty container at 270C.What is the pressure in the container after all of the carbon di-oxide vaporises If 5.6 g of solid carbon di-oxide were placed in the same container but it already contained air at 740 torr. what would be the partial pressure of carbon di-oxide and the total pressure in the container after the corbon dioxide vaporises? [0.783 atm, 0.446 atm, 1.156 atm]

BULLS EYE


PHYSICAL CHEMISTRY


DATE : 30/06/2008

DPP. NO.- 23

Q1.

Uranium isotopes have been separated by taking advantage of the different rates of effusion of the two isotopic forms of UF6. One form contains Uranium of atomic weight 238, and the other of atomic weight 235. What are the relative rates of effusion of these two molecules.

Q2.

The pressure in a vessel that contained pure oxygen dropped from 2000 torr to 1500 torr in 55 min as the oxygen leaked through a small hole into a vacuum. When the same vessel was filled with another gas the pressure dropped from 2000 torr to 1500 torr in 85 min. What is the molecular weight of the second gas? [76]

Q3.

A large cylinder of helium has a small thin Orifice through which helium escaped into an evacuated space at the rate of 6.4 mmol/hr. How long would it take for 10mmol of CO to leak through a similar Orifice if the CO were confined at the same pressure? [4.134 hr]

Q4.

A porous cup filled with hydrogen Gas at atmospheric pressure is connected to a glass tube which has one end immersed in water as shown in Figure. Explain why the water rises in the Glass tube.

Q5.

20 dm3 of SO2 diffuse through a porous partition in 60 s.What volume of O 2 will diffuse under similar conditions in 30 s? [14.14 ml]

Q6.

150 ml of a gas diffuse through a porous pot in 25 seconds. 250 ml of another gas diffuse through the same vessel in 50 seconds. If the vapour density of the first gas is 25, Find out the vapour density of the second gas. [36]

Q7.

At room temperature, ammonia gas at 1 atm pressure and hydrogen chloride gas at P atm pressure are allowed to effuse through identical pin holes from opposite ends of a glass tube of one metre length and of uniform cross section. Ammonium chloride is first formed at a distance of 60 cm from the end through which HCl gas is sent in. What is the value of P? [P = 2.2 atm]

Q8.

At 270C, hydrogen is leaked through a tiny hole into a vessel for 20 minutes as that of hydrogen leaked through the same hole for 20 minute. After the effusion of the gases the mixture exerts a pressure 6 atm. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litre. What is the molecular mass of the unknown gas? [1088]

Q9.

A teacher enters a classroom from front door while a student from back-door. There are 13 equidistant rows of benches in the classroom. The teacher releases N2O, the laughing gas, from the first bench while the student releases the weeping gas (C 6H11OBr) from the last bench. At which row will the students start laughing & weeping simultaneously? [9th from front]

Q10.

Pressure in a bulb dropped from 2000 to 1500 mm in 47 minute when the contained Oxygen leaked through a small hole. The bulb was then completely evacuated. A mixture of oxygen and another Gas of molecular weight 79 in molar ratio 1 : 1 at a total pressure of 4000 mm was introduced. Find the molar ratio of two gases remaining in the bulb after a period of 74 minute. [1/1.24]

BULLS EYE


PHYSICAL CHEMISTRY


DATE : 11/07/2008

DPP. NO.- 24

Q1.

Kinetic energy of gases presumes collision between molecule to be perfectly elastic beacuse (A) collisions will not split the molecule (B) the molecules are tiny (C*) the molecules are rigid (D) the temperature remains constant irrespective of collisions

Q2.

According to thekinetic theory of gases, in an ideal gas between two successive collisions a gas molecules travels (A*) in a straight path (B) with an accelerated velocity (C) in a circular path (D) in a wavy path

Q3.

If the ratio of molecular weight of two gases is 1:4, what would be the ratio of their root mean square velocities at the same temperature (A) 1:2 (B*) 2:1 (C) 1:4 (D) none of these

Q4.

The temperature of an ideal gas is increased from 140 K to 560 K. If at 140 K the root-mean square velocity of the gas molecles is V, at 560 K it becomes (A) 5V (B*) 2V (C) V/2 (D) V/4

Q5.

Translational kinetic energy for 2 moles of a gas at 27 0C is (A*) 7.48 x10 3 J (B) 6.48 x103 J (C) 5.48 x103 J

Q6.

(D) 4.48 x103 J

The average kinetic of a molecule of SO 2 at 270C is E. What would be the average kinetic energy of a molecule of CO2 at the same temperature (A)

E  64

(B)

44

E  44

(C) E

64

(D*) 22  E

Q7.

Two flasks X and Y have capacity 1 litre and 2 litre respectively. Each of them contains 1 mole of a gas. The temperature of the flask are so adjusted that the average speed of molecules in X is twice as that in Y. The pressure in flask X would be (A) 8 times of that in Y (B) twice of that in Y (C) same as that in Y (D*) half of that in Y

Q8.

For a monoatomic gas kinetic energy = E. The relation with rms velocity is 1/ 2

 2 E   (A*) u    m 

Q9.

1/ 2

 3 E   (B) u    2m 

1/ 2

 E   (C) u    2m 

1/ 2

 E  (D) u     3m 

If a gas is expanded at constant temperature (A*) the pressure decreases (B*) the kinetic energy of the molecules remains the same (C) the kinetic energy of the molecules decreases (D) the number of molecules of the gas increases

Q10. The volume of two gases A and B are the same under identical conditions of temperature and pressure. They would differ in (A) kinetic energy (B*) rate of effusion (C) number of molecules (D*) boiling point BULLS EYE


Q1.

PHYSICAL CHEMISTRY


DATE : 15/07/2008

DPP. NO.- 25

At what temperature would the most probable speed of CO 2 molecules be twice that at 50 0C.

[10190C]

Q2.

At what temperature would N 2 molecules have the same average speed as He atoms at 330? [2310 K]

Q3.

Calculate the avg. & total K.E. of 0.5 mole of an ideal gas at 0 0C.

Q4.

Oxygen at 1 atmosphere and 0 0C has a density 1.429 g/litre. Find the rms speed of oxygen molecule.

[K avg = 3404.6 J/mol; K total = 1702.3]

[461.2 m/s]

Q5.

A gas bulb of 1 litre capacity contains 2 x 10 21 molecules of Nitrogen exerting a pressure of 7.57 x 10 3 N/m2. Calculate the rms velocity and temperature of the gas molecules. If the ratio of most probable speed to the rms speed is 0.82, calculate the most probable speed of these molecules at this temperature. [C = 494.22 m/s,

= 405.3 m/sec]

Q6.

A flask contains 1 mol of N 2 molecules at 100K. How many molecules have a velocity in the range of 300 to 300.1 m/sec.? Calculate the ratio of the number of molecules within the 0.1 m/sec. range of velocity at 300m/sec. to that it 500m/sec.

Q7.

According to the figure, calculate the value of speed at which the two curves cross each other.

Q8.

Calculate the no. of molecules in one mole of an ideal gas that have energies greater than four times the average thermal energy at 25 0C and 500C.

From 9 to 11

For a gaseous molecular system the probability of finding a molecule with velocity between v and v + dv is given by

Q9.

where m = mass of gas molecule ; k = Boltzmann constant T = temperature of gas ; dN = no. of molecules with velocity between v to v+dv N = total no. of molecules At some temperature the fraction of molecules with kinetic energies between E and E + dE is given by (A)

(B)

(C)

(D)

Q10. Fraction of molecules with K.E. greater than E is given by (A)

(B)

(C)

(D) BULLS EYE

Q11. Ratio of fraction of molecules with K.E. greater than and less than average K.E. is (A)

1 e1/ 2  1

(B)

1 e3/ 2  1

(C)

e1/ 2

1  e1/ 2

(D)

e3/ 2 e3 / 2  1

Q12. Assuming ideal gas behaviour, how many atoms of Ar are contained in a typical human breath of 0.5 lt at 1 bar and 37 0C. Air consists of 1% Ar atoms. Assuming that the Ar atoms from the last breath of Plato have been distributed randomly throughout the atmosphere (5 × 10 18m3), how long would it take to breath one of these atoms? A typical adult breath rate is 10min –1. Q13. Consider a 20 litre sample of moist air at 60 0C under a total pressure of 1 atm in which the partial pressure of water vapour is 0.12 atm. Assume the composition of dry air given is 78 mol% N 2, 21 mol% O2 and 1 mol% Ar. (a) What are the mole percentages of the gases in the sample ? (b) At 600C, equilibrium vapour pressure is 0.2 atm. What volume must the mixture occupy at 6 0C if the relative humidity is to be 100%? (c) What fraction of the water will be condensed if the total pressure of the mixture is increases isothermally to 200 atm? Q14. A water gas mixture has the composition by volume of 50% H 2, 45% CO and 5% CO 2. (i) Calculate the volume in litres at STP of the mixture which on treatment with excess steam will contain 5 litres of H 2. The stoichiometry for the water gas shift reaction is CO + H2O  CO2 + H2 (ii) Find the density of the water gas mixture in kg/m 3. (ii) Calculate the moles of the absorbants KOH, Ca(OH) 2 and ethanolamine (HO–CH2 –CH2 –NH2) required respectively to collect the CO 2 gas obtained. Q15. Suppose you are a manager of a fish drying unit. You are to dry 200 kg of fish containing 40% water on wet basis. The drier specification is mentioned in the figure:

SVP at 2000C = 40 mm of Hg SVP at 1000C = 25 mm of Hg What is the process time of the drier?

[12.57 hr]


Q.1

PHYSICAL CHEMISTRY


DATE : 20/07/2008

DPP. NO.- 26

For a nitrogen gas at STP, which of the following graphs would be obtained on plotting the proportion of gas molecules with a particular speed (Y-axis) against the speed of the molecules (X-axis) (A)

(B)

(C*)

(D)

Q.2

Most probable velocity, average velocity and root mean square velocity are related as: (A*) 1 : 1.128 : 1.224 (B) 1 : 1.234 : 1.128 (C) 1.128 : 1 : 1.234 (D) 1.128 : 1.234 : 1

Q.3

Let the most probable velocityof hydrogen molecules at a temperature t 0C is V0. Suppose all the molecules dissociate into atoms when temperature is raised to (2t + 273) 0C then the new r.m.s velocity is (A) 2 / 3 V0 (B) 3(2  273 / t ) V0 (C) 2 3 V0 (D*) 6 V0

Q.4

The root mean square velocity of one mole of a monoatomic gas having molar mass M is U relation between the average kinetic energy (E) of the gas and U rms is

rms. The

3E E 2E 2E (B) Urms= (C*) Urms= (D) Urms= 3M 2M 3M M Two bulbs A and B of equal capacity are filled with He and SO 2 respectively at same temperature. How will the U RMS be affected if volume of B becomes 4 times that of A at constant temperature. How will the U RMS be affected if half of the molecules of SO 2 are removed at constant temperature. [Ans.No change in both] (A) Urms =

Q.5 (i) (ii) Q.6

Q.7

Find the ratio of molecules possesing velocities in the given range From Ump to Ump + 0.005 Ump and From Ump – 0.04 Ump to Ump + 0.04 Ump.[Ans. 1/16]   Mu 2  What is the fraction of molecules possessing velocities [Given exp  2RT  = 0.2797]   (i) From U avg to Uavg + 0.005 Uavg of a gas having molecular mass 100 g at 400 K. (ii) From U avg – 0.04 Uavg to Uavg + 0.04 Uavg.

Q.8

What are the number of collisions that occur in a sample of He per second at 1 bar & 27 0C? The collision diameter of He is 0.22 nm.

Q.9

Repeat the calculations of above problem for He at 10 bar & 27 0C and also at 1 bar & 2980 K. Which has the greater effect on the collision frequency, temperature or pressure?

Q.10 A simple sketch of a molecular velocity selector as shown in figure. The relation between the rotation frequency v and the velocity of the molecules that sucessfully pass through the slits is given by V = (3600/)l  Where l is the distance between the slits and  is the slit displacement angle. What is the rotation frequency needed to pass N 2 molecules at 298 K with V mp = 421 ms –1 through a selector with l = 10.0 cm and  = 2.000?

BULLS EYE


PHYSICAL CHEMISTRY


TIME : 100 MIN

DATE : 28/07/2008

DPP. NO.- 27

THIS IS A TEST PAPER WHICH IS HELD ON 23.07.06 (ACME)

Q.1

Calculate number of neutrons in 140 g of

.

Q.2

What volume of H 2S(g) measured at 1.0 atm and 0°C is produced when 16.60 g of KI reacts with excess of H 2SO4 according to equation 8KI + 5H 2SO4  4K 2SO4 + 4I2 + H2S + 4H2O.

[4]

[4]

Q.3

Copper is obtained from ores containing the following minerals: Azurite : Cu3(CO3)2(OH)2 Malachite : Cu 2 CO3 (OH)2 Chalcocite : CuFeS 2 Which mineral has (a) the lowest, (b) the highest copper content on a percentage by mass basis? [4]

Q.4

What mass of MgCl2 would contain the same total number of ions as 292.5 g of NaCl.

Q.5

An aqueous solution of ethanol has density 1.025 g/ml and it is 2 M. Determine Molalityof this solution.

[4]

[4]

Q.6

1.8 gm of copper reacted with excess of O 2 and produced x gram of non-stoichiometric compound Cu1.8O. Find the value of x. [4]

Q.7

A certain public water supply contains 0.10 ppb (part per billion) of chloroform (CHCl 3). How many molecules of CHCl 3 would be obtained in 0.478 ml drop of this water? (assume d = 1 g/ml) [PPb =

×109] HCl(aq) solution into % by mass, if density of solution is x g/ml.

[4]

Q.8

Convert 20%

Q.9

A compound of formula XCl3 reacts with aqueous AgNO3 to yield a precipitate of solid AgCl according to the following equation XCl3(aq) + 3AgNO3(aq)  X(NO3)3 (aq) + 3AgCl(s) When a solution containing 1 g of XCl 3 was allowed to react with an excess of aqueous AgNO 3, 2.65 g of solid AgCl was formed. What is the identity of the atom X? [5]

[4]

Q.10 A compound X contains 63.3 percent Manganese (Mn) and 36.7 percent oxygen (O) by mass. When X is heated O 2 is evolved and a new compound Y containing 72.0 % Mn & 28% oxygen is formed. (a) Determine the empirical formula of X and Y. [4] (b) Write a balanced equation for the conversion of X to Y. Q.11 26.2 g of MgCl 2.2H2O is mixed with 92.8 g of H 2O. Calculate (i) Molarity of Cl – ion (ii) Molality of Cl – ion density of resultant solution is 2.38 g/ml.

[5]

BULLS EYE

Q.12 Consider the following unbalanced reaction: MnI2 + F2  MnF3 + IF5 (i) Balance the reaction (ii) What is the minimum mass of F 2 that must be used to react with 30.9 g of MnI 2 is overall [6] yield of MnF 3 is no more than 80%. Q.13 On combustion analysis, a 0.450 g sample of Caproic acid (contained only C,H & O) gives 0.418 g of H 2O and 1.023 g of CO 2. What is the empirical formula of Caproic acid? If the molecular mass of Caproic acid is 116 amu, what is the molecular formula? [5] Q.14(a) The vapour density of a mixture of gas A (Mol. mass = 40) and gas B (Mol. mass = 80) is 25. Calculate mole % of gas B in the mixture. Molar mass = 2 × vapour density. (b) How many ml of 0.1 M HCl are required to react completely with 1 g mixture of Na 2CO3 & NaHCO3 containing equimolar amounts of two? [6] Q.15(a) An enzyme isolated from human red blood cells was found to contain 0.316% Selenium. What is the minimum molecular weight of enzyme? (b) 54 g of aluminium is to be treated with a 10% excess of required H 2SO4. The chemical equation (unbalanced) for the reaction is Al + H 2SO4  Al2(SO4)3 + H2. What volume of concentrated sulphuric acid of density 1.80 g/cm 3 and containing 90% H 2SO4 by [6] weight must be taken? Q.16 A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of 13.2% by weight. 2.58 g of the mineral on heating lost 1.232 g of CO 2. Calculate the % by weight of the other metal. [6]


Q.1

PHYSICAL CHEMISTRY


TIME : 100 MIN

DATE : 01/08/2008

DPP. NO.- 28

State whether the following statement are true or false. Justify your answer briefly: (i) A mixture of ideal gas is cooled upto liquid helium temperature (4.22K) to from an ideal solution. (ii) The rate of diffusion of a gas is inversely proportional to its density.  n 2 a  (iii)  p  V 2 v  nb   nRT   In the van der Waals’equation given above, the constant ‘a’ reflects the actual voluem of the gas molecules.

Q.2

van der Waals’ equation explains the behaviour of (A) all gaseous substances (B) ideal gases (C) real gases (D) none of the above

Q.3

van der Waals’contains a and b correspond to the following respectively: (A) intermolecular attraction and bond energy. (B) actual volume of the molecules and intermolecular repulsion. (C) shapes of molecules and intermolecular repulsion. (D) intermolecular attraction and actual volume of the molecules.

Q.4

In van der Waals’ equation of state of the gas law, the constant b is a measure of (A) volume occupied by the molecules. (B) intermolecular attraction. (C) intermolecular repulsion. (D) intermolecular collisions per unit volume.

Q.5

Which of the following statements is incorrect of ideal gaseous behaviour. (A) All the molecules are identical and have intermolecular forces of attraction. (B) The gaseous molecules exchange energies during molecular collision. (C) The collisions are inelastic. (D) All of the above.

Q.6

At constant volume, for a fixed number of moles of a gas the gas increases with rise of temperature due to (A) increases with rise of temperature due to. (B) increased rate of collision amongst molecules. (C) increases in molecular attraction. (D) decrease in mean free path.

Q.7

Which of the following equation may represent the variation of d/p vs pressure for a real gas O 2 at 273 K (density in gm/litre & pressure in atm) is (A) (C)

Q.8

d p d p

= 0.02 p + 1 = 0.4 p2 + 1.43

(B) (D)

d p d p

= 0.4 p2 + 22.4 = 0.02 p + 32

At 1000C and 1 atm, if the density of liquid water is 1.0 g cm 3 and that of water vapour is 0.0006 g cm –3 , then volume occupied by water molecules in 1 litre of steam at that temperature is (A) 6 cm 3 (B) 60 cm 3 (C) 0.6 cm3 (D) 0.06 cm 3 BULLS EYE

Q.9

A real gas tends to behave more ideally when (A) it is compressed to a smaller volume at constant temperature. (B) temperature is raised keeping the volume constant (C) more gas is introduced into same volume and at the same temperature. (D) temperature is raised and volume allowed to increase.

Q10. In van der Waals’ equation of state for a non-ideal gas, the term that accounts for intermolecular forces is: (A) V – b

(B) RT

(C) P 

a

V2

(D) (RT) –1


TIME : 100 MIN

PHYSICAL CHEMISTRY

Daily Practice Problems DATE : 01/08/2008

DPP. NO.- 29

Questions given below consist of two statements each printed as Assertion (A) and Reason (R); while answering these questions you are required to choose any one of the following four responses: (A) if both (A) and (R) are true and (R) is the correct explanation of (A) (B) if both (A) and (R) are true but (R) is not correct explanation of (A) (C) if (A) is true but (R) is false (D) if (A) is false and (R) is true (E) Both (A) and (R) are false. 1.

Assertion: Reason:

CH4, CO2, has volume Z (compressibility factor) less than one. Z < 1 is due to the repulsive forces among the mole-culsive.

2.

Assertion: Reason:

van der Waals’equation explains the behaviour of ideal gases. Ideal gases can only be compressed.

3.

Assertion: Reason:

Greater the value of van der Waals’constant ‘a’ greater is the liquefaction of gas. Excluded volume or co-volume equals to (v–nb) for n moles.

4.

Assertion: Reason:

Excluded volume of co-volume equals to (v-nb) for n moles. Co-volume depends on the effective size of gas molecules.

5.

Assertion: Reason:

Gases like N 2, O2 behave as ideal gas at low temperature and high pressure. Molecular interaction is not the responsible factor.

6.

Assertion: Reason:

Compressibility factor (Z) for non-ideal gases is always greater than 1. Non ideal gases always exert higher pressure than expected.

7.

Assertion: Reason:

CH4, CO2, has volume Z (compressibility factor) less than one. Z < 1 is due to the repulsive forces among the mole-culsive.

8.

Assertion: Reason:

Gases like N 2, O2 behave as ideal gas at low temperature and high temperature Molecular interaction is not the responsible factor.

9.

Assertion: Reason:

vander Waal’s equation is applicable only to non–ideal gases. Ideal gases obey the equation PV = nRT.

10.

Assertion:

Slope of Z Vs. P curve for a real gas is equal to

Reason:

1 a  Z    T = const.  (b ) which represents the slope of Z Vs.P curve. RT RT  P 

11.

Assertion: Reason:

Z is always greater than 1 for H 2. As Ini slope of Z Vs. P curve is +ve, ‘b’ factor dominates the behavior of the gas.

12.

Assertion: Reason:

CO2 gas has relatively larger dip is Z values than CH 4 gas at 0oC in low pressure region. Size of CO2 molecules is greater than that of CH 4.

13.

Assertion:

CO2 gas always has dip in Z values whereas H 2 gas always has coatinues increase in Z values. BULLS EYE ‘b’ factor dominates for H 2 and ‘a’ factor dominates for CO 2 generally.

Reason:

1 a (b ). RT RT

14.

Assertion:

As temp. decreases ini slope of Z Vs. P curve becomes -ve.

Reason:

As temp. decreases the term

15.

Assertion: Reason:

a 1 a > b and the ini slope is (b). RT RT RT At T >>>>> TC, PV-isotherm almost approaches rectangular hyperbola. At large temp. K.E. of gas molecules largely increases.

16.

Assertion:

TC for an ideal gas cannot be determined.

Reason:

TC =

17.

Assertion: Reason:

An ideal gas cannot be liquified and compressed. a = 0 for an ideal gas

18.

Assertion:

For (Z > 1), repulsive forces may dominate the behavior of the one particular gas.

Reason:

Z=

19.

20.

Assertion:

8a , both a & b are zero for an ideal gas so T C mathematically becomes 27 Rb indeterminate form.

 0     0 

PVm , as P increases, Z increases & finally becomes greater than one. RT For different gases under the same conditions as Z increases, gas become more compressible. PVm and as ‘T’ decreases, ‘Z’ increases. RT

Reason:

Z=

Assertion:

For two different gases under the same conditions, as Z decreases, gas become more compressible

Reason:

Z=

PVm , as Z decreases, when ‘a’ increases. RT


TIME : 100 MIN

PHYSICAL CHEMISTRY


DPP. NO.- 30

Q.1

Choose the correct alternative (more than one may be correct) (B.M.C.= Bimolecular collision) (at constant P) (n- is constant throughout) (A)  is constant (B) BMC for 1 collision is directly proportional to T. (C) BMC for all the molecules per unit volume is directly proportional to T 2. (D*) None of these

Q.2

Suppose that we change U RMS of gas in a closed container from 5 × 10 –2 cm/sec to 10 × 10 –2 cm/sec, which one of the following might correctly explain how this change was accomplished (A) By heating the gas we double the temperature (B) By removing 75% of the gas at constant volume we decrease the pressure to one quarter of its original value (C) By heating the gas we quadruple the pressure (D) By pumping in more gas at constant temperature we quadruple the pressure.

Q.3

The Vander Waal's constant for some gases are given Gas a (L2 atmp mol –2) b (L mol –1) NH3 4.17 0.037 CO 2 3.59 0.043 CH4 2.25 0.043 O2 1.36 0.032 The gas with highest critical temperature is (A) NH3 (B) CO2 (C) CH4

Q.4 Q.5

(D) O2

For a real gas, behaving ideally the pressure is: a V (A) abVmolar (B) molar (C*) V b (D) None ab molar Calculate the final pressure in a container when 12 gm C (s) is burnt with 24 gm O 2 in a container of capacity 24.65 litre & temperature 300K, no. reactants left behind (A*) 1 atmp (B) 1.75 atmp (C) 1.375 atmp (D) none

Q.6

With the increase in temperature of a gas, the fraction of molecules having velocities within a given range around the most probable velocity, would (A) increase (B*) decrease (C) remain unchanged (D) initially increase and then decrease

Q.7

Which of the above statements is/are correct? If the Van der Waal's parameters of two gases are given as: a/dm6 bar mol –2 b/dm3 mol –1 Gas A 6.5 0.056 Gas B 18.0 0.011 then (A) Critical volume of A < Critical volume of B. (B*) Critical pressure of A < Critical pressure of B. (C) Critical temperature of A > Critical temperature of B. (D) Critical temperature of A = Critical temperature of B.

BULLS EYE

Q.8

For oxygen at 25°C, the collision diameter is 0.361 nm. What is the mean free path for oxygen molecules at (a) 1 atm pressure and (b) 0.1 Pa pressure?

Q.9

Choose the correct alternative (more than one may be correct) (B.M.C.= Bimolecular collision) (at constant V) (n- is constant throughout) (a)  is constant (b) BMC for 1 collision is directly proportional to . (c) BMC for all the molecules per unit volume is directly proportional to T 3/2.

Q.10 Choose the correct alternative (more than one may be correct) (B.M.C.= Bimolecular collision) (at constant T) (n- is constant throughout) (a)  is directly proportional to V (b) BMC for 1 collision is directly proportional to V 2. (c) BMC for all the molecules per unit volume is directly proportional to P 2. Q.11 At 27°C the density of a gas in g l –1 is given by the equation d = 4.00 P + 0.020 P 2. Calculate the molecular weight of the gas. [Ans. 100] Q.12 A certain gas has the following values for its critical constants. P c = 45.6 atm., Vc = 0.0987 l mol –1 and Tc = 190.6 K. Calculate the van der Waals’ constants of the gas. Estimate the radius of the gas molecules assuming that they are spherical. [Ans. b = 0.0329 l mol –1, a = 1.333 l 2 atm mol –2, 2.35 Å] Q.13 Calculate the Boyle’s temperature T B for CO2 (g) assuming it to be a van der Waals’ gas. (a = 3.59 l 2 atm mol –2; b = 0.0427 l mol –1) [Ans. 1026 K] Q.14 Suppose a gas contains 5 molecules with a velocity of 2 m/s, 10 molecules with a velocity of 3 m/s and 4 molecules with a velocity of 6 m/s. Calculate the values of the average, most probable and r.m.s. velocities. Q.15 The critical temperature (T c) of ethane is 32.3 °C : the critical pressure is 48.2 atm. Calculate the critical volume using (i) ideal gas law (ii) the Van der Waal’s equation.

BANSAL CLASSES Target IIT JEE 2009 CLASS : XI 1.

PHYSICAL CHEMISTRY


TIME : 100 MIN

DATE : 07/08/2008

DPP. NO.- 31

Read the following assumption and answer the question no. 1 to 4 at the end of it. Analysis of distribution of molecular speeds of an ideal gas can be mathematically represented by the equation

 N

 M   4   N  2 RT 

3/ 2

e  Mu

2

/ 2 RT 2

u du

where N is the total number of particles, M molecular mass,  N number of particles having velocity between speed u & u + du, T is absolute temperature. The graph of ‘fraction o f molecules’ vs ‘speed’ when plotted was found to be ‘unsymmetrical’ with very few molecules with very high or low speed & mostly molecules possessing average speeds of the range of 300-400 m/s. The plotting of the graph & the comparison of fraction of molecules with mass of the gas & temperature of the gas can be compared by identifying the ‘downward’ terms. At higher speeds the nature of

  Mu  2 RT  & at lower speed it is influenced by the graph is influenced by the exponential term e   2

 M   parabolic term   2 RT 

Q.1

3/ 2

u 2 . Using the above expression values of U , U and U can be mps avg rms

established. Mark the correct statement

(I) Number of molecules having speed around 300 m/s will be substantial greater than number of molecules possession either speed close to zero or very high speed. (II) The speed possessed by most of the molecules & average speed has the same value. (III) At lower speed the fraction of the molecules increase with the speed. (IV) At higher speed the fraction of molecules increases with speed. (A) Only 2 & 3 are correct statements (B*) Only 1 & 3 are correct statements (C) Only 4 is incorrect statement (D) Only 1 is correct Q.2

47 gm of initial mixture of O 3 and N2O4 is heated at 400 K in a closed rigid container and all ozone is dissociated in O 2 at this temperature but N 2O4, 20% dissociated in NO 2. Initially moles of ozone are doubled of N 2O4, then find initial and final molar mass of the mix. (A*)

188 , 44.76 3

(B)

47 , 44.76 1.05

(C) 62.67, 4.476

(D) None of these

Q.3

Find initial and final translational kinetic energy of the mixture at 400 K (A) 300 R, 400 R (B) 450 R, 630 R (C*) 300 R, 420 R (D) 420 R, 630 R

Q.4

If at the time of heating no dissociation of anycompound, then find the total heat energy required to raise the temperature of gas mixture by 10 0 C (vibrational degree of freedom not excited) (A) 20 R (B)22.5 R (C)15 R (D*) 45 R

2.

Read the following assumption and answer the question no. 1 and 2 at the end of it. Assume that you have a sample of gas in cylinder with a movable piston, as shown in the following drawing: BULLS EYE

Q.1

Cylinder will retain its original form if: (A) the temperature is increased from 300 K to 450 K at constant pressure. (B) the pressure is increased from 1 atm to 2 atm at constant temperature. (C) the temperature is decreased from 300 K to 200 K and the pressure is decreased from 3 atm to 2 atm (D) the temperature is increased from 200 K to 300 K and the pressure is decreased from 3 atm to 2 atm

Q.2

Temperature of one mole of an ideal gas contained in it is increased from 300 K to 301 K. Pressurevolume work obtained is (A) 8.314 J (B) 2.494 J (C) 0.8314 J (D) 83.14 J

3.

Read the following assumption and answer the question no. 1 to 5 at the end of it. The apparatus shown consists of three temperature jacketed 1.000 L bulbs connected by stopcocks. Bulb A contains a mixture of H 2O(g), CO2(g) and N 2(g) at 250C and a total pressure of 564 mm Hg. Bulb B is empty and is held at a temperature of –70 0C. Bulb C is also empty and is held at a temperature of –190 0C. The stopcocks are closed, and the volume of the lines connecting the bulb is zero. CO 2 sublimes at –780C, and N 2 boils at –196 0C.

Q.1

The stopcock between A and B is opened and the system is allowed to come to equilibrium. The pressure in A and B is now 219 mm Hg. Select the correct alternate: (A) A contains CO 2(g) and N 2(g) and B contains CO 2(g), N2(g) and H 2O(s) (B) A contains CO 2(g) and B contains N 2(g) and H2O(l ) (C) A contains CO 2(g), N2(g) and H 2O(s) and B contains CO 2(g) and N 2(g) (D) A contains H 2O(g) and B contains H 2O(g), N2(g) and CO2(g)

Q.2

How many moles of H 2O are in the system? (A) 0.026 mol (B) 0.0013 mol

(C) 0.013 mol

(D) 0.13 mol

Q.3

Both stopcocks are opened and the system is again allowed to come to equilibrium. The pressure throughout the system is 33.5 mm Hg. Select the correct alternate. (A) Each bulb contains N 2(g), H2O(s) and CO2(g) (B) Each bulb contains N 2(g), H2O(g) and CO2(g) (C) Each bulb contains N 2(g), H2O(s) and CO 2(s) (D) A contains N 2(g), B contains N 2(g), H2O(s); C contains N 2(g) and CO 2(s)

Q.4

How many moles of N 2 are in the system? (A) 0.022 mol (B) 0.011 mol

(C) 0.018 mol

(D) 0.036 mol

How many moles of CO2 are in the system? (A) 0.022 mol (B) 0.011 mol

(C) 0.018 mol

(D) 0.036 mol

Q.5

4.

Read the following assumption and answer the question no. 1 to 5 at the end of it.

The constant motion and high velocities of gas particles lead to some important practical consequences. One such consequence is that gases mix rapidly when they come in contact. Take the stopper off a bottle of perfume, for instance, and the odour will spread rapidly through the room as perfume molecules mix with the molecules in the air. This mixing of different gases by random molecular motion and with frequent collisions is called diffusion. A similar process in which gas molecules escape without collision through a tiny hole into a vacuum is called effusion.

Q.1

Select the correct alternate(s) (A) All gases spontaneously diffuse into one another when they are brought into contact. (B) Diffusion into a vacuum will take place much more rapidly than diffusion into another gas. (C) Both the rate of diffusion of a gas and its rate of effusion depend on its molar mass. (D) All of the above statements are correct.

Q.2

X mL H2 effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is: (A) 10 s : He (B) 20 s : O 2 (C) 25 s : CO (D) 55 s : CO 2

Q.3

When CO2 under high pressure is released from a fire extinguisher, particles of solid CO 2 are formed despite the low sublimation temperature (-77 0C) at 1 atm. It is due to: (A) the gas does work pushing back the atmosphere using kinetic energy of molecules and thus lowering the temperature. (B) volume of the gas is decreased rapidly hence temperature is lowered. (C) both of the above (D) none of the above

Q.4

The stopcocks of the bulbs X (containing NH 3) and Y (containing HCl) are opened simultaneously; white fumes of NH 4Cl and formed at point B. If AB = 36.5 cm, then BC is approximately:

(A) 18.0 cm Q.5

(B) 25.0 cm

(C) 20.0 cm

(D) 36.5 cm

Select the incorrect statement(s): (A) Larger the size of the molecules, smaller the mean free path (B) Greater the number of molecules per unit volume, smaller the mean free path (C) Larger the temperature, larger the mean free path (D) Larger the temperature, smaller the mean free path


PHYSICAL CHEMISTRY


TIME : 100 MIN

DATE : 07/08/2008

DPP. NO.- 32


Homogeneous mixing and compressibility both result from the fact that the molecules are far apart in gases. Mixing occurs because individual gaseous molecules have little interaction with their neighbours and, assuming that no reaction takes place, the chemical identities of those neighbours are irrelevant. Compressibility is possible in gases because only about 0.1% of the volume of a typical gas is taken up by the molecules themselves under normal circumstances; the remaining 99.9% is empty space.

Q.1

Compressibility factor of a gas is given by: (A)

Q.2

PV R

(B)

PV R

2

(C)

R PV

For the gas under the critical state, compressibility factor is: (A) 1 (B) 3/8 (C) 8/3

(D)

PV nRT

(D) always greater than 1

Q.3

Assume molecules are spherical of radius 1 Å, volume occupied by molecules in one mole of a gas at N.T.P. is: (A) 2.52 m 3 (B) 2.52 × 10 –4m3 (C) 2.52 ×10 –6m3 (D) 2.52 × 10 –2m3

Q.4

For H2 and He, force of attraction is negligible, hence compressibility factor is: (A)

Q.5

a    (C) 1   RTV 

 RT   (D) 1   Pb 

PV RT

 Pb   (B) 1   RT 

a   1    (C)  RTV 

 RT   (D) 1   Pb 

Compressibility factor for a real gas under high pressure region is: (A)

2.

RT

 Pb   (B) 1   RT 

Compressibility factor for a real gas under low pressure region is: (A)

Q.6

PV

PV RT

(B)

RT PV

(C) (1 + RT)

 Pb   (D) 1   RT 


Variation of volume with temperature was first studied by French chemist, Jacques Charles, in 1787 and then extended by another French Chemist Joseph Gay Lussac in 1802. For a fixed mass of a gas under isobaric condition, variation of volume V with temperature t 0C is given by

V = V0[1+t] Where V 0 is the volume at 0 0C and at constant pressure.

Q.1

For every 1 0 change in temperature, the volume of the gas changes by ............. of the volume at 0 0C: (A) 1/237 unit (B) 1 unit (C) 273 unit (D) 22.4 unit BULLS EYE

Q.2

Select the incorrect statement (A)  (given above) is called volume coefficient. (B) Value of  is 3.66 ×10 –3 0C –1 for all gases. (C) 273 K is the lowest possible temperature attained (D) Absolute zero is the temperature reached when all possible thermal energy has been removed from substance.

Q.3

Graph between volume V and temp t ( 0C) is of the type with given specifications (A) indicating zero volume at point O

(B) indicating OA = 22.4 L mol –1 and temperature –273 0C at point B

(C) indicating OA = 22.4 L mol –1 and temperature 273 K at point B

(D) indicating OB = 22.4 L mol –1 and temperature –273 0C at point A

Q.4

Under isochoric condition, graphs between P and T are shown below. Volume order is : (A) V1 = V2 = V3 (B) V1 > V2 > V3 (C) V1 < V2 < V3 (D) V2 < V1 < V3

3.


The distribution of speeds for helium atoms at different temperature is given below:

Q.1

This distribution is referred to as: (A) Maxwell–Boltzmann distribution (C) Gay–Lussac distribution

(B) Graham–distribution (D) Maxwell–Planck distribution

Q.2

Select the incorrect statement (A) There is only a small number of molecules with very low velocities or very high velocities (B) Largest number of molecules have a velocity corresponding to the maximum of the curve, which is called most probable velocity. (C) Most probable velocity decreases on increasing temperature. (D) Distribution of the molecules in the given velocity range decreases on increasing temperature.

Q.3

Most probable velocity is: (A)

Q.4

3 RT M

(B)

8 RT M 

8 RT

Relation for a given velocity is

 M

(C)

2 RT M

(D)

PV M

. At 300K, value of this velocity for O 2 gas is:

(A)

8  8.314  300 –1 ms   32

(B)

8  8.314  300 –1 ms  16

(C)

8  8.314  27 –1 ms   16

(D)

8  8.314  300 –1 ms   32

Q.5

The root mean square velocity of a gas at 300K is 30 R . The molar mass of the gas, in kg mol –1, is: (A) 1.0 (B) 1.0 × 10 –1 (C) 1.0 × 10 –2 (D) 1.0 × 10 –3

4.


Gases tend to behave non-ideally at low temperatures and high pressure, and to explain deviation from ideal behaviour two types of corrections volume correction and pressure correction are introduced.

Q.1

Select the incorrect statement (s) (A) Volume correction is due to finite size of molecules and pressure correction is due to force of attraction between molecules. (B) At high temperatures, molecules have greater kinetic energy, and attractive forces are smaller and the behaviour of gases is close to the ideal gas behaviour. (C) Volume correction is also called co-volume or excluded volume and is four times the volume of spherical molecules present in one mole of the gas. (D) At very low pressure, force of attraction is effective and pressure correction needs further resolution.

Q.2

Application of pressure and volume correction gives vander Waals equation, and other related equations. Which of the following is not the correct equation for the given name? (A) vander Waals

:

 2 n2a  T  p  V 2  [V – nb] = nRT  

(B) Virial

:

 B C  PV = RT 1   2  ........  V V 

(C) Clausius

:

(D) Dieterici

:

a   P  [V – b] = RT T  TV 2  P(V – b) = RTe –a/RTV

Q.3

Following represents equation of state for n moles of a real gas  n2a   P  2  [V – nb] = nRT T V  

Select the correct statements: (A) Constant a is a measure of force of attraction between gas molecules and

n 2a V 2

is also called internal

pressure. (B) a is expressed in atm L 2mol –2, b is expressed in L mol –1.  Pb   (C) At high pressure, compression factor is 1   RT  (D) All of the above are correct statements.

Q.4

Which assumption of kinetic theory is not followed when a real gas shows a non-ideal behaviour? (A) Gas molecules move at random with no attractive forces between them. (B) The velocity of gas molecules is dependent on temperature. (C) The amount of space occupied by a gas is much greater than the space occupied by the actual gas molecules. (D) In collisions with the walls of the container or with another molecules, energy is conserved.

Q.5

One way of writing the equation of state for real gas is:  B   .......  V  where B is constant. B in terms of the van der Waals constants a and b is written as: P V  RT 1 

a    (A)  b   RT 

Q.6

a    (B)  b   RT 

b    (C)  a   RT 

b    (D)  a   RT 

Table given below indicates the value of van der Waals constant a (in L 2 atm mol –2)

The gas which can most easily be liquefied is: (A) O2 (B) N2

(C) NH3

(D) CH4

Q.7

The internal pressure of a van der Waals gas is: (A) independent of the molar volume (B) inversely proportional to the molar volume (C) inversely proportional to the square of the molar volume (D) directly proportional to the molar volume.

5.


Any gas can be liquefied by decreasing its temperature and increasing its pressure. Decreasing of the temperature decreases the average kinetic energy of the molecules, and increasing of the pressure decreases the average distance between molecules. When the molecules are close together, their kinetic energy is lowered, they do not possess enough energy to overcome the force of attraction between molecules and liquid if formed. For each substance, however there exists a temperature, above which the substance cannot be liquefied, no matter how great the applied pressure.

Q.1

Temperature above which gas cannot be liquefied is called: (A) critical temperature T c =

8a 27 Rb

(C) inversion temperature T i =

2a Rb

(B) Boyle’s temperature T b =

a Rb

(D) triple point T p = 273 K.

Q.2

An ideal gas obeying kinetic gas equation can be liquefied if: (A) its temperature is more than critical temperature. (B) its pressure is more than critical pressure. (C) its pressure is more than critical pressure but its temperature is less than critical temperature. (D) it cannot be liquefied at any value of P and T.

Q.3

Critical temperature of CO 2 is 310C. CO2 is (A) a gas at 35 0C and vapour at 25 0C (C) vapour at 35 0C as well as at 25 0C

(B) a gas at 35 0 C as well as at 25 0C (D) a gas at 35 0C and liquid at 25 0C

Q.4

To liquefy a gaseous substance whose critical temperature is below room temperature, requires: (A) high pressure and lowering of temperature (below T c) (B) low pressure and raising of temperature (above T c) (C) high pressure and raising of temperature (above T c) (D) low pressure and lowering of temperature (below T c)

Q.5

Which of the following statements on critical constants of gases are correct: A: B: C:

 T c  Larger the value of  P  of a gas, larger would be excluded volume  c  Critical temperature (T c) of a gas is greater than its Boyle temperature (T b) At the critical point in the van der Waals gas isotherm   P    = 0 V   m T c

Select the correct statements: (A) A and B (B) B and C

(C) B and C

(D) A, B and C


TIME : 100 MIN

PHYSICAL CHEMISTRY


DPP. NO.- 33

Read the following assumption and answer the question no. 1 to 3 at the end of it. How can the ideal gas law be modified to yield an equation that will represent the experimental data more accurately ? We begin by correcting an obvious defect in the ideal has law, namely the prediction thatunder any finite pressure the volume of the gas is zero at the absolute zero of

temperature : V = RT /p. On cooling real gases liquefy and arrange the new equation so that it predicts a finite, positive for the gas at 0 K by adding a positive constant b to the ideal volume : V  b 

RT p

According to the above equation the molar volume at 0 K b, and we expect that b will be roughly comparable with the molar volume of the liquid or solid . The above equation also predicts that as the pressure becomes infinite the molar volume approaches the limiting value b. This prediction in accord with experience than the prediction of the ideal gas law that the molar volume approaches zero at very high pressure.

Q1.

From the above information, if we calculate compressibility factor then it comes out to be equal to that of a real gas as per vanderwaal’s equation at  (A*) high pressure (B) low pressure (C) high temp (D) low temp.

Q2.

This calculated compressibility factor cannot represent the behavior of which of the following gases at 0oC & const. volume  (A*) N2 (B) H2 (C) He (D) CH4

Q3.

A : This calculated ‘Z’ cannot explain the behavior of some gases in low pressure range. R : In low pressure range, attraction forces become more influential.

2.

Read the following assumption and answer the question no. 1 to 3 at the end of it. We have already noted that the worst offenders in the matter of having values of Z less than unity are methane and carbon dioxide which are easily liquefied. Thus we begin to suspect a connection between ease of liquefaction and the compressibility factor, and to ask why a gas liquefies. First of all energy, the heat of vaporization, must be supplied to take a molecule out of the liquid and put it into the vapour. This energy is required because of the forces of attraction acting between the molecule and its neighbors in the liquid. The force of attraction is molecules are together, as they are in a liquid, and very weak if the molecules are far apart, as they are in a gas. The problem is to find an appropriate way of modifying the gas to take account of the effect of these weak attractive forces.

The pressure exerted by a gas on the walls of a container acts in a outward direction. Attractive forces between the molecules tend to pull them together, thus diminishing the outward thrust against the wall and reducing the pressure below that exerted by the ideal gas. This reduction in pressure should be proportional to the force of attraction between the molecules of the gas.

Q1.

Vanderwaal’s const. ‘a’ may be proportional to  (A*) heat of vaporisation (B*) intermolecular forces (C*) heat of sublimation (D*) mol. wt. of a gas molecule BULLS EYE

Q2.

Reduction in pressure of a real gas relative to that of an ideal gas is directly proportional to  (A*) vanderwaal’s const. ‘a’ (B) N* 2 (C*) (N*) (D) None N*  no. of molecules per unit vol.

Q3.

Vanderwaal’s const. ‘a’ is related to intermolecular forces of attraction which are  (A) strong forces (B*) weak forces (C) avg. magnitude forces (D) none

3.

Read the following assumption and answer the question no. 1 to 6 at the end of it. Consider the two small volume elements v1 and v2 in the following figure 

suppose if each volume element contains one molecule and that the attractive force between the two volume elements is some small value f. then 

Q1.

If another molecule of the same gas is added to v 2, then the force acting between two elements should be  (A) f (B*) 2 f (C) 3 f (D) 4 f

Q2.

If again a third molecule of the same gas is added to v 2, then the force acting between two elements should be  (A) f (B) 2 f (C*) 3 f (D) 4 f

Q3.

The force of attraction between these two volume elements keeping no. of molecules fixed in v 1 is  (A*) directly proportional to no. of molecules per unit vol. in v 2. (B) _______, _________________________ v 1. (C) always same (D) none

Q4.

If none of these v 1 or v2 is kept fixed with respect to the no. of molecules per unit volume respectively given by C 1 and C2 then  (A*) Force of attraction is directly proportional to C 1C2 (B) ________, __________________ is always fixed and retains a const. value. (C) can’t be predicted. (D) none.

Q5.

Net force on vol. element v 2 is  (A) greater than that of on v 1 (C*) equal to _____, __________

(B) less ________, ________________ (D*) equal to zero

Q6.

Vanderwaal’s forces are very short range forces, therefore these should not be effective for large distances i.e. large containers. But in practice it is not so because  (A*) ‘a’ is assumed to be const. through out and thus a drawback of vanderwaal’s real gas behavior. (B) At some places ‘a’ starts dominating the behavior of the gas. (C) On increasing pressure molecules come closer so actually ‘a’ increases. (D) None.

4.


To calculate Z for the vander Waals gas we multiply by V and divide by RT this yields Z 

pV RT



V



a

V  b RT V

The numerator and denominitor of the first term on the right - hand side are divided by V : Z 

1



a

1  b / V RT V

At low pressure b / V is small compared with unity , so the first term on the right may be developed into a power series in 1 / V by division ; thus 1(1  b / V )  (b / V ) 2  .... Using this result in the 2

3

a  1  b   b  precending equation for Z and collecting terms, we have Z  1   b          ..., RT   V  V   V 

Which expresses Z as a function of temperature and molar volume . It would be preferable to have Z as a function of temperature and pressure ; however, this would entail solving the vander Waals equation for V as a function of T and p, then multiplying the result by p/RT to obtain Z as a function of T and Since the vander Walls equation is a cubic equation in V , the solution are too complicated to be particular informative. We content ourselves with an approximate expression for z(T,p) which we obtain from Eq. the term in p 2 , is

Z  1 

1  a  a  a  2 2 b  b   p    p  ... 3 RT  RT  RT  RT    

Q1.

Considering only volume correction term in vanderwaal’s gas equation. By this size effect P (A*) increases than that of ideal value. (B) decreases ________, _____________ (C) remains same as that of ideal value (D) none

Q2.

At Boyle’s temp. the real gas behaves over a wide range of pressure because  (A*) In Z Vs. P curve slope of the curve becomes zero. (B*) Real gas becomes tangent to ideal gas line in Z Vs. P curve (C*) Effect of size and intermolecular forces roughly compensate each other (D) none

Q3.

A : In low pressure region there is a large dip in ‘Z’ value for CO 2 at 0oC. Whereas in high pressure Z again increases. R : ‘a’ can be influential only in low pressure region not in high pressure region.

5.

Read the following assumption and answer the question no. 1 to 4 at the end of it. If the pressure - volume relations for a real gas are measured at various temperatures, a set of isotherms such as are shown in fig. 3.5 are obtained. At high temperatures the isotherms look much like those of an ideal gas, while at low temperatures the curves have quite a different appearance. The horizon portion of the low -temperatures curves is particular striking. Consider a container of gas in a state described by point A in fig. Imagine one wall of the container to be

real

movable (piston); keeping the temperature at T 1 , we slowly push in thus decreasing the volume . As the volume becomes smaller, the pressure rises slowly along the curve until the volume v2 is reached . Reduction of the volume beyond v2 produces no change in pressure until v3 is reached. The small reduction in volume from v3 to v4 produces a large increase in pressure from pe to p’. This is a rather remarkable sequence of remains a large increase in pressure in volume over a wide range in which the pressure remains at the constant value pe .

Q1. Q2.

First drop of the liquid appears at  (A*) T1 (B*) V2

(C) V1

(D) T2

Last trace of gas disappears at  (A*) V3 (B) V4

(C*) T1

(D) T2

Q3.

A : range change is pressure is observed for a very small change in volume from V 3 to V4. R : Pressure through out remains const. on the horizontal line from V 2 to V3. and therefore at end of the line abrupt change in pressure in observed for the earlier large volume change from V 2 to V3.

Q4.

Specify T/F  (A) Range of volume over which condensation occurs is smaller at high pressure (T/F) (B) Exactly at critical point the isotherm is very close to that of an ideal gas. (T/F) (C) There is lesser possibility that gas will have nearly ideal behavior at T < T c. (T/F) (D) Most nearly ideal behavior may be observed at T = T 4 K. (T/F) (E) The line from V 2 to V3 represent two phase region. (T/F)

( T) ( F) (T) (T) ( T)


Q1.

Q4.

TIME : 100 MIN

DATE : 08/08/2008

DPP. NO.- 34

(B)

(C*)

(D)

When an electron ray deflected by magnetic field (strength B) and brought back to its original position by an electric field (strength E) then its momentum is (A)

Q3.


The final speed of the electron accelerated from rest of mass m and charge e through a potential difference V in vacuum is (A)

Q2.

PHYSICAL CHEMISTRY

(B*)

es – is an atom are held by (A) coloumbic forces (B) nuclear forces

(C)

(D)

(C) gravitation forces (D) Vander-waal’s forces

As per Rutherford the space between proton and e - is hydrogen atom is (A) filled with air (B) empty (C) filled with magnetic radiation

(D) none

Q5.

Rutherford created a theoritical picture of the atom based on (A) stars in galaxy (B) model of planets revolving round the sun (C) behaviour of waves in the oceans (D) clouds in the sky that move & mix with changing shapes

Q6.

In Rutherford’s experiment, a few in millions of -particle suffer 180 o concluded that (A) e-s revolve removed the nucleus (B) space between es - & nucleus is empty (C) nucleus occupies much smaller volume than that of atom (D) all the above

Q7.

When -particles are sent thru a thin metal foil most of them go straight thru the foil because: (A) -particles are much heavier than es (B) -particles are the charged (C) most of the part of atom is empty space (D) -particles more with high speed

Q8.

To what maximum distance will an -particle with kinetic energy 0.4 Mev approach in case of a head on collision with stationary lead nucleus Z = 82.

Q9.

With what vel. should an -particle travel towards the nucleus for a copper atom so as to arrive at a distance of 10 -13m. from the nucleus of copper atom.

Q10. According to Rutherford all the positive charge is present in the nucleus because only one out of (20,000) -particles could retrace its path. (T/F)

BULLS EYE


Q1.

PHYSICAL CHEMISTRY


TIME : 100 MIN

DATE : 08/08/2008

The velocity of photon is: (A) Independent of its wavelength (C) Depends on its source

DPP. NO.- 35

(B) Depends on its wavelength (D) Equal to the square of its amplitude

Q2.

Which is not characterishics of planek’s quantum theory of radistion: (A) Radiation is associated with energy (B) Energy is not absorbed or emitted in whole no. or multiples of quantum (C) The magnitude of energy associated with a quantum is proportional to the frequency. (D) Radiation energy is neither emitted nor absorbed continuously but in small packets called quantam.

Q3.

Which is not the property of the photons: (A) Momentum (B) Energy

Q4.

The rest mass of a photon of wavelength (A) Zero

(B)

(C) Velocity

(D) Rest mass

(C)

(D)

is:

Q5.

Calculate the mass of a photon having wavelength 1 nm. is

Q6.

Which of the following statements is true: (A) Energy associated with one photon is directly proportional to the frequancy of radiation (B) Total energy of radiation is inversely related to the of radication (C) Graph between wavelength & energy of one photon is non linear (D) Curve between energy of one photon & wave no. of radiation is linear

Q7.

Planck’s constant has the same dimensions as: (A) Work (B) Wavelength (C) Power

(D) Angular momentum

Q8.

If wavelength of photon is 2.2 x 10 -11 m, than momantum of photon is: (A) 3 x 10 -23 kg. m/sec. (B) 3.33 x1022 kg. m/sec. (C) 1.452 x 10 -44 kg. m/sec (D) 6.89 x 10 43 kg. m/sec.

Q9.

Calculate the frequancy of a yellow light having wavelength 600 nm. :

Q10. The relation E = Q11.

relates particle & wave nature of radiation. (T/F)

A per electromagnetic theory the energy of a wave depends only on its amlitude and not on its frequency or wavelength. (T/F)

Q12. The velocity of photon is: (A) Independent of its wavelength (C) Depends on its source

(B) Depends on its wavelength (D) Equal to the square of its amplitude

BULLS EYE

Q13. Which is not the characteristic of planck’s Quantum Theory of Radiation: (A) Radiation is associated with energy (B) Energy is not absorbed or emitted in whole no. multiples of Quantum (C) The magnitude of energy associated with a Quantum is proportional to the frequancy. (D) Radiation energy is nether emitted nor obsorbed continuously but in small packets called Quanta. Q14. The dynamic mass/equivalent mass of a photon of wavelength (A) Zero (B) he/ (C) h/c

is: (D) h/

Q15. Calculate the no. of photons emitted in 10hrs. by a 60W sodium lamp  = 5893Å.

qb-dpp..

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