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CH-207 PHYSICAL CHEMISTRY PRACTICALS – I I. THERMOCHEMISTRY
Water Equivalent of Calorimeter
Determination of Heat of Ionisation
Integral and Differential Heat of Solution
II. ADSORPTION
Verification of Freundlich and Langmuir adsorption isotherms
III. PHASE DIAGRAMS
Simple Eutectic - “Napthalene - “Napthalene – Biphenyl” System
Partially miscible liquids – “Phenol “Phenol - Water” Water ” system system
Effect of electrolytes and non electrolytes on CST (+2 expt)
Three component systems (+4 expt )
IV. PARTITION COEFFICIENTS
Iodine between carbon tetra chloride and water
Benzoic acid between benzene and water.(Graphically water.(Graphically also)
Equilibrium constant, KI + I 2KI2
V. VISCOMETRY
Verification of Kendall‟s equation
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THERMOCHEMISTRY
___________________________________________________________________ __________________________________________ _________________________
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Experiment 1: Water Equivalent of Calorimeter
___________________________________________________________________ __________________________________________ _________________________ Aim: To determine the water equivalent of the gi ven calorimeter Principle:
When a known weight of hot water is added to known volume of th water at room temperature, the calorimeter and the cold water will take u heat from the hot water. The heat gained by the cold water and calorimeter equal to heat lost by hot water added. Using this principle the water equiva the calorimeter can be determined. Requirements: Calorimeter, Sensitive thermometer. Procedure:
A known volume (100ml) of distilled water is taken in a beaker a temperature at an interval of 1 minute is recorded. Some water (more than 1 is heated to about 90 C and then 100ml of this hot water is poured calorimeter. As soon as hot water is added to calorimeter, the calorimeter is and its content is stirred with the stirrer attached. Simultaneously the tempe of calorimeter is recorded at an interval of 1min till it attains a constant After getting 3 consecutive constant temperature values for hot water, 10 cold water is added to calorimeter. Soon after the addition the calorimeter closed and it is stirred continuously. Simultaneously the temperature of calor is recorded at an interval of every 30 seconds till a constant value is achieve constant temperature is noted and calculations are made. °
Observation: Time (min)
Temperature of Cold water Hot water
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Time (sec)
30 60 90 120
Temperature after mixing C °
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THERMOCHEMISTRY
___________________________________________________________________ __________________________________________ _________________________
Sheet Music
Experiment 1: Water Equivalent of Calorimeter
___________________________________________________________________ __________________________________________ _________________________ Aim: To determine the water equivalent of the gi ven calorimeter Principle:
When a known weight of hot water is added to known volume of th water at room temperature, the calorimeter and the cold water will take u heat from the hot water. The heat gained by the cold water and calorimeter equal to heat lost by hot water added. Using this principle the water equiva the calorimeter can be determined. Requirements: Calorimeter, Sensitive thermometer. Procedure:
A known volume (100ml) of distilled water is taken in a beaker a temperature at an interval of 1 minute is recorded. Some water (more than 1 is heated to about 90 C and then 100ml of this hot water is poured calorimeter. As soon as hot water is added to calorimeter, the calorimeter is and its content is stirred with the stirrer attached. Simultaneously the tempe of calorimeter is recorded at an interval of 1min till it attains a constant After getting 3 consecutive constant temperature values for hot water, 10 cold water is added to calorimeter. Soon after the addition the calorimeter closed and it is stirred continuously. Simultaneously the temperature of calor is recorded at an interval of every 30 seconds till a constant value is achieve constant temperature is noted and calculations are made. °
Observation: Time (min)
Temperature of Cold water Hot water
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30 60 90 120
Temperature after mixing C °
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Calculations: Weight of Cold water
= W 1
Contant temperature for cold water
= T 1 C
Weight of Hot water
= W 2
Contant temperature for Hot water
= T 2 C
Resultant temperature
= T 3 C
°
°
°
If e is the water equivalent of the calorimeter, Heat gained by water and calorimeter
= (e + W 1 )(t 3 – t 1 )
Heat lost by hot water
= W 2(t 2 – t 3 )
Heat gained = Heat lost (e + W 1 )(t 3 – t 1 ) = W 2(t 2 – t 3 ) e = [W 2(t 2 – t 3 )/ (t 3 – t 1 ) ] - W 2 Result:
___________________________________________________________________ __________________________________________ _________________________
Experiment 2: Determination of Heat of Ionisation __________________________________________________________________ ______________________ _ ____________________________________________ Aim: To determine the heat of ionisation ioni sation of a weak acid with sodium hydroxide. Principle:
Weak acids do not ionise completely in dilute soluti ons much, when th neutralised by strong alkalies. This heat of neutralisation will be accompan the heat of ionisation also.(ie) If either the acid or base is weak or if bo Read Free Foron 30this Days Sign up to vote title weak, the heat of neutralisation is not constant quantity equal to 13.7Kcal b than this value. In such a case neutralisation process consist not only in the Not useful Useful Cancel anytime. + of $4.99/month. H or OH but the ionisation of weak acid or weak base. The measured h Special offer for students: Only neutralisation will be equal to the sum of heat evolved in the union of H +
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Procedure:
The calorimeter is taken and its water equivalent is determined. Now the same calorimeter the temperature for 100ml 1N NaOH solution is deter at an interval of 1min till a constant temperature is noted. The alkali is removed from the calorimeter to beaker and washed. Next transfer 100ml o acid into the calorimeter and its temperature at an interval of 1min is deter till a constant reading is noted. After getting 3 consecutive constant reading 100ml 1N NaOH kept in beaker is poured into the calorimeter having 10 weak acid in it. The mixture is stirred well continuously and temperature is at an interval of 30 seconds till 3 consecutive constant values are obtained constant value is noted. Observation: (i)Standardisation of NaOH solution Volume of NaOH (ml)
Burette Reading Initial Final
Volume of Acid (ml)
V1N1 = V2N2
You're Reading Normality of Stock Solution = a Preview Time (sec)
Temperature of a free trial. Unlock full access with Acid
Alkali
Mixture
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Mass of weak acid, M 2 Special offer for students: Only $4.99/month. Temperature of weak acid, T 2
= =
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___________________________________________________________________
Experiment 3: Integral and Differential Heat of solution Sheet Music
___________________________________________________________________ Aim: To determine the integral and differential heats of the solution of KCl. Principle:
Integral heat of solution at a particular concentration is the heat c when 1 mole of solute is dissolved in pure solvent to produce a solution given concentration. From the heat change for the dissolution of m moles of the solute, th change per mole can be determined. When heats of solution at different so are plotted against the number of moles of the solute added, the slope of the gives the differential heat of solution. The differential heat of solution m defined as the heat of solution of a mole of solute in such a large quan solution of known concentration that addition of one mole of the solute do change the concentration appreciably. Requirements: Calorimeter, Sensitive thermometer, KCl, Procedure:
100ml of water is taken in the Dewar flask. The temperature of water About 2gm of KCl (weighted accurately) is added into the flask and dissolve stirring. The resultant temperature is noted. The experiment is repeated with You're Reading a Preview gms of KCl and dissolving in 100ml of water each time. Unlock full accessiswith a free trial. using the relati on. The integral heat of solution calculated Integral heat of solution = (C + W + m) ΔT x M/m.
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Where C is the water equivalent, W is the wt of water, m the wt added and M the molecular weight of KCl. Observation:
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Weight of KCl
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Temperature
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ADSORPTION
___________________________________________________________________
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Experiment 1: Adsorption of Oxalic acid on Charcoal
___________________________________________________________________ Aim:
To determine the adsorption isotherm of oxalic acid from aqueous so by activated charcoal and to calculate the unknown concentration of oxalic ac Principle:
The extend of adsorption depends on temperature i.e. decrease increase of temperature, for the process absorption is an exothermic pr However at a fixed temperature for a given amount of adsorbent, the amo adsorption increases with concentration of adsorbent (in solution) or pressu expression representing the variation of the amount adsorbed with the equil pressure or concentration of the adsorbate or concentration at a fixed tempe is known as adsorbtion isotherm. An empirical equation relating the amount of adsorption t concentration of the adsorbate or to the pressure of the gas at a fixed tempe is known as Freundlich Adsorption Isotherm. When Charcoal powder is adde solution of oxalic acid, a fraction of the acid will be adsorbed by charcoal .Acc to Freundlich, α x/m = kP 1/ x/m = kc 1 / n
You're Reading a Preview
Where, „x‟ is the number of moles of oxalic acid absorbed by „m‟ gra Unlock full access with a free trial. charcoal and „c‟ is the equilibrium concentration of the acid, „k‟ and „ constants, former depends upon the nature of both adsorbent and adsorbate greater than unity. The plot of With x/m Free values Download Trialagainst corresponding press concentrations is practically a straight line at low concentration, indicating t the direct proportionality of the amount of adsorption with pressure for low v Taking logarithms of above equations we get:log x/m = log k + 1/ α log P log x/m = log k + 1/n log c
Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Therefore the plot of log(x/m) against log c should be straight lin & The New York Times Not useful Useful
anytime. slope equal to 1/n and intercept equal to log k.Cancel Using the above expressi Special offer for students: Only $4.99/month. strength of the given oxalic acid solution can be found out .It should be note such a graph is linear for a narrow range of concentration or pressure.
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conical flask is titrated against standard N/10 KMnO 4.The experiment is re with the unknown acid and its concentration is calculated. In order to test the validity of Freundlich adsorption isotherm, plot lo values against log c e and by putting ce /(x/m) values versus c e .A line obtained shows the applicability of isotherm. Calculate the constants „a‟ a from the slope and the intercept on the ordinate axis. Precautions:
Fresh activated charcoal should be used and the filter paper should be The temperature of each flask should be constant since the r adsorption increases with temperature.
Observations: Weight of Oxalic Acid,W = Normality of Oxalic acid ,N 1 = Standardisation of KMnO 4 KMnO4 x Oxalic acid Trial Volume of oxalic no. acid (V1)
Burette readings Initial Final
Volume of KMnO4 (V2)
You're Reading a Preview a free trial. Normality of KMnO4, N2Unlock = (V1full X access N1)/Vwith 2 = = Download With Free Trial
Volume of oxalic acid
Volume of water
Initial concentration
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Volume of filtrate(V2)
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Eqilibrium conc. of acid; Ce
Oxalic acid adsorbent(x)
x/m
Log x/m
Log Ce
Ce /(x/m
I. From log Ce Vs log x/m graph log x/m = therefore,x/m = But m = 2
therefore, x =
Co = x x 1000 VxE =
+ Ce
From langmuir adsorption isotherm Ce /(x/m) = Ce /b
+
Y = mx + c
1/ab
You're Reading a Preview
Intercept, 1/ab
= Unlock full access with a free trial.
Slope m, = dy/dx
=1/b =
a
=
b
=
1/a
=
Download With Free Trial
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Special offer for students: Only $4.99/month. From log Ce vs log x/m graph, concentration of given Oxalic acid =
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PHASE DIAGRAMS
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Two component system of solids
Introduction The most common method employed to obtain the freezing point diagram of mixture is thermal analysis method. The method consists in the cooling of a m mixture of the components and a cooling curve is drawn between temperature and The cooling curve ia a continuous one as long as there is no phase change during co When a solid starts seperating the rate of cooling is decreased due to the libera heat. Consequently the cooling curve exhibits a break at the freezing point.
Suppose a liquid mixture of two components X and Y is cooled slowly. At first its temperature would fall regularly along the curve AB, until the freezing point of the solution is reached at which the liquid become saturated with X. with crystallisation of X there would be a reduced rate of fall of temperature owing to You're Reading a Preview evolution of latent heat, but since the solution would be Unlock full access with a free trial. constantly flowing more concentrated in Y, the freezing point would fall along C practice the transition from AB to CD is rarely ideal, more often a solution super below its true freezing point Download B to some temperature With Free Trial B and when crystallisation commences there is a sudden evolution of heat which temporarily sends the tempe up to C‟ .Eventually the solution reaches saturation w.r.t Y as well as X.Then solid solid Y crystallise outside by side in the same proportions as they are present solution the temperature remains constant –the eutectic point.
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(eutectic temperature) are determined. The freezing points are plotted again composition. The freezing points are plotted against the temperature. Fro liquids curve ACB obtained the composition of the given mixture of A and B c determined. Requirements: Flat bottomed test tubes, sensitive thermometer, air jacket, water Napthalene, Biphenyl. Procedure:
Prepare the following mixtures of the components by taking w amounts in properly labeled flat bottomed test t ubes. Napthalene % Biphenyl %
100 00
90 10
70 30
50 50
30 70
10 90
00 100
Using a cork fit up a sensitive thermometer and a ring stirrer in th tube. Melt the contents in a water bath, wipe it clean and place it in an air j Allow the temperature to fall , stir thoroughly and record the temperature half minute continue to record temperature until it remain constant and the mass solidifies. Similarly repeat the determination with other mixtures. Draw the c curve for each of the mixture and from each of the mixture determin maximum temperature of first arrest. Plot the value of freezing point again percentage composition. From the phase diagram eutectic temperatur eutectic compositions are found out. a Preview You're Reading
Unlock full access with a free trial. Composition of unknown mixture: The cooling curve of the unknown mixture is constructed. A line is corresponding to its freezing pointWith in the phase Download Free Trial diagram. The line cuts the di at two points. Now one of the components is added to mixture and aga freezing point is determined. The shift in the freezing point is found out a correct point, out of the two i s selected and the unknown composition is foun
Observation:
Master your semester with Scribd Composition I II III & The New York Times Biphenyl % 00 10 30 Special offer for students: Only $4.99/month. Napthalene %
100
90
70
IV
V
50
70
50
30
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10
00
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___________________________________________________________________
Experiment 2: Critical Solution Temperature: Phenol – Water system _____________________________________________________________ Aim: To determine the critical solution temperature of phenol water system. Theory:
In general liquids become more soluble with increasing temperatur eventually a temperature will be reached when the liquids become com miscible. This temperature is known as critical solution temperature or con point. The solubility of phenol in water increases on increasing temperature particular temperature (upper consolute temperature) the two liquids are m in each other and the solution becomes homogeneous.
Requirements: Flat bottomed test tubes, sensitive thermometer, air jacket, water bath, P Water. Procedure:
5ml of phenol is measured out into a flat bottomed test tube in w thermometer is also inserted. 3ml of water is added and placed in a water ba stirred continuously. At a certain temperature the turbidity disappears an slightly lower temperature turbidity reappears. The average of the You're a Preview temperatures is found out Reading and taken as miscibility temperature a composition. Now 1ml water is added and the corresponding temperatu fullout. accessThis with aisfree trial. complete miscibility areUnlock found repeated by adding 1ml each. A g plotted with temperature on Y-axis against composition (% by weight) of phe X-axis. The maxima of parabolic give Trial CST of Phenol water system. Downloadcurve With Free Observation: Volume of water (ml)
Temperature of Turbidity Disappearance Appearance
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__________________________________________________________________
Experiment 3: Effect of Electrolytes on CST “KCl – Phenol” system
___________________________________________________________________ Aim:
To determine the effect of impurity on CST and to find the concentrat unknown KCl solution Theory:
Impurities have a regular effect on the CST of Phenol-water system. KCl solutions of different concentrations are used. The CST is found to in with concentration of KCl solution (electrolyte). A straight line graph can be p and from the graph concentration of the given electrolyte can be found.
Requirements: Flat bottomed test tubes, sensitive thermometer, air jacket, water Phenol, KCl. Procedure:
The CST of Phenol water system is found out using 0.002M KCl (elect instead of pure distilled water. The same is repeated with 0.04M, 0.06M, and 0.1M solution of electrolyte. A graph is plotted with CST against concen of electrolyte solution. The CST corresponding is found out and from the gra concentration of KCl is determined. Observation:
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Volume of phenol = Volume of KCl (ml)
Download With Trial Temperature of Free Turbidity Disappearance Appearance
Miscibility Temperature
% Comp of phen
0.02M KCl
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___________________________________________________________________
Experiment 4: Effect of Electrolytes on CST “Succinic acid – Phenol” system
___________________________________________________________________ Aim:
To determine the effect of immpurity on CST to find the concentra unknown succinic acid solution. Theory:
Impurities have a regular effect on the CST of Phenol water syst succinic acid solutions of different concentration are used, the CST is fou decrease with concentrations of acid solution. A straight line graph can be p and from the graph concentration of given acid solution is found out. Procedure:
The CST of Phenol water system is found out using 0.5 of succin solution instead of pure distilled water. The same is repeated using 1%, 1.5% 4% solution of succinic acid. A graph is plotted with CST against the concen of succinic acid. The CST corresponding to succinic acid solution of un concentration is found out and from the graph concentration of acid is determ Observation: Volume of phenol = Volume of Acid
0.5%
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__________________________________________________________________
Experiment 5: Three Component systems - Determination of Unknow Concentration Water –Toluene -Acetic acid system
___________________________________________________________________ Aim:
To construct a phase diagram of CH 3COOH C6H5-CH3 and H2O. A determine concentration of unknown mixture. Theory:
The mutual solubility of a pair of partially miscible liquids may be ma altered by the addition of a third component. In general when the third comp is soluble is in only one of the other two components; the mutual solubility two liquids is decreased. Then the third component is soluble in bo components their mutual solubility is increased. On mixing water and toluene we get two conjugate solutions of dif composition. When CH3COOH acid is added to these, the mutual solubility of and toluene increases and a point is reached at which the mixture b homogenous which depend on the relative composition of water and t mixture. Procedure:
Toluene and water is taken in different composition such that the volume is kept constant (say10ml) in different small conical flasks. Acetic You're Reading a Preview added from the burette to each bottle with constant shaking till the tu disappears. The percentage composition by weight of each compon Unlock full access with a free trial. represented on the three corners of triangular graph. From the amount of H 2O added and the initial amount of toluene and acid, the percentage composition of the when the turbidity appe Download With Freemixture Trial plotted in the triangular diagram. The points obtained are joined by a smooth curve. A definite volume mixture of unknown composition consisting of water and acetic acid i s taken density is found out. Now toluene is added form the burette till turbidity ap The percentage by weight of toluene is calculated and a line corresponding point is drawn parallel to AC (acetic acid-water axis). This cuts the grap Read Free Foron 30this Days SignBup vote title point. A line is drawn from the opposite apex totothis point. And extrapola useful Useful touch the line AC. This experiment is repeated toCancel getanytime. theNot composition of the mixture. Special offer for students: Only $4.99/month.
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Weight Weight Weight Weight Weight
of of of of of
empty bottle bottle + water water toluene acetic acid
= = = = =
Relative density of water Relative density of CH 3COOH Relative density of toluene
= = =
Weight of bottle + unknown Weight of unknown
= =
Relative density of unknown Weight of toluene Relative density of toluene Weight of toluene
= = = = Relative density of toluene × volume = = =
Weight of mixture % Composition of toluene Result:
___________________________________________________________________ You're Reading a Preview
Experiment 5.1: Tie-line for H 2O-C6H5CH3-CH3COOH system
Unlock full access with a free trial. ____________________________________________________________
Aim:
Download With Free Trial
To construct the tie line for Water- Toluene- Acetic acid system. Theory:
In this a mixture of three components are prepared. The three laye then analyzed separately. The composition of each of the layer by a point i separated and a curve known as bimodal curve is drawn. Any mixture with Read Free For 30 Days Sign up to vote on this title area enclosed by the curve and the base of the triangle will itself resolve in useful liquid layers and ay mixture outside this area will form one liquid area. An Useful Not Cancel anytime. connecting only one conjugate solution, which is i n equilibrium with one anot Special offer for students: Only $4.99/month. called a tie line. When the composition of two layers becomes identical the t
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Observation : Sl.No
Liquid Layer
Volume
Weight of Water Weight of Organic layer Weight of Aqueous layer
Burette Reading Initial Final
Volume of NaOH
= = =
Relative density of Aqueous layer Relative density of Organic layer
= Weight of Aqueous layer/ Weight of Wa = Weight of Organic layer/ Weight of Wat
Normality of Organic layer Weight/liter of CH3COOH in Organic layer Weight of acetic acid in 1ml of Organic layer % By weight of acetic acid in 2ml organic layer
= = = =
Normality of aqueous layer = Weight/liter of CH3COOH In aqueous layer = You're Reading a Preview Weight of acetic acid in 1ml of aqueous layer = % By weight of CH3COOH in 2ml of aqueous layer = Unlock full access with a free trial.
Result:
Download With Free Trial
___________________________________________________________________
Experiment 7: Three Component systems - Determination of Unkn Concentration Water – Chloroform – Acetic acid sys ____________________________________________________________ Read Free Foron 30this Days Sign up to vote title
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Procedure:
Chloroform and water is taken in different composition such that th volume is kept constant (say10ml) in different small conical flasks. Acetic added from the burette to each bottle with constant shaking till the tu disappears. The percentage composition by weight of each compon represented on the three corners of triangular graph. From the amount of H 2O added and the initial amount of chlorofor acetic acid, the percentage composition of the mixture when the turbidity ap is plotted in the triangular diagram. The points obtained are joined by a smooth curve. A definite volume mixture of unknown composition consisting of water and acetic acid i s taken density is found out. Now chloroform is added form the burette till tu appears. The percentage by weight of chloroform is calculated and corresponding to this point is drawn parallel to AC (acetic acid-water axis cuts the graph at a point. A line is drawn from the opposite apex B to this and extrapolated to touch the line AC. This experiment is repeated to g composition of the given mixture .
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Observations: Sl no:
Volume H2O CHCl3
CH3COOH
Weight H2O CHCl3
CH3COOH
% Composition H2O CHCl3
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Weight Weight Weight Weight Weight
of of of of of
empty bottle bottle + Water Water Chloroform Acetic acid
= = = = =
Special offer for students: Only $4.99/month. Relative density of Water
=
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___________________________________________________________________
Experiment 8: Three Component systems – Tie line of Water – Chloroform – Acetic acid system _____________________________________________________________ Aim: To construct the tie line for water-chloroform- acetic acid system. Theory:
In this a mixture of three components are prepared. The three laye then analyzed separately. The composition of each of the layer by a point i separated and a curve known as bimodal curve is drawn. Any mixture with area enclosed by the curve and the base of the triangle will itself resolve in liquid layers and ay mixture outside this area will form one liquid area. An connecting only one conjugate solution, which is i n equilibrium with one anot called a tie line. When the composition of two layers becomes identical the t shrinks to a point. Procedure:
A homogenous mixture of three components having a composition bimodal curve is prepared. The mixture was shaken thoroughly and allow settle to two layers. Separate the layers using a separating funnel. A small a You're a Preview of each layer is taken out intoReading a previously weighed dried conical flask and ti using NaOH using phenolphthalein as indicator. A line is drawn connecting Unlock full access with a free trial. points on the curve and the point corresponds to the system.
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Observations: SL no:
Liquid layer
Volume
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Burette reading Initial Final
Volume of NaOH
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% By weight of CH3COOH in 2ml of aqueous layer = Result:
PARTITION COEFFICIENTS
___________________________________________________________________
Experiment 1: Partition Coefficient of Iodine between CCl 4 and Water
___________________________________________________________________ Aim: To determine the partition coefficient of Iodine between CCl4 and Water. Theory:
If a solute is shaken up with two immiscible solvents, it will distribute between the two solvents in such a way that the ratio of concentration of the in two solvents remains constant, provided the solute remain in the molecular state in both the solvents. If C1 is the concentration of the solute in solvent I and C 2 in in sol then C1 /C2 = k Where, K is called the partition coefficient. Requirements: I2 in CCl 4, K2Cr2O7, Na2S2O3.5H2O,10% KI, starch, Distilled H2O, CCl Procedure:
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In three 250ml stoppered conical flask the solutions are mixed Unlock full access with a free trial. following order. Reagents Bottle 1 Bottle 2 Bottle 3 I2 in CCl4 10 20 Download With Free Trial15 CCl4 10 5 0 Water 100 100 100 The contents of the bottles are shaken well for about ½ hr to equilibrium. The bottles are kept in constant temperature bath or left alone hr. When the equilibrium is reached 2ml of CCl 4 layer of each bottle is t Free Foron 30this Days Sign up to vote title against N/10 sodium thiosulphate solution Read after adding 10ml of 10% KI and water using starch solution as indicator. The of water helps in the ti Useful Not useful addition Cancel anytime. by gradually extracting I2 into water layer, where the reaction with s Special offer for students: Only $4.99/month. thiosulphate occurs. 20ml of aqueous layer from each bottle is titrated with sodium thiosulphate 10 times in a standard flask.
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Normality of K2Cr2O7
= mass/lit Eq. wt =
Normality of Na2S2O3 Estimation
N/10 Na2S2O3 Vs I2 in Organic layer (starch) Bottle No Volume of organic Burette reading layer pipetted Initial Final
Vol of N/10 Na2S2O3
Estimation N/100 Na2S2O3 Vs I2 in aqueous layer (starch) Bottle No Volume of aq Burette reading Vol of N/10 Na2S2O3 layer pipetted Initial Final
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Volume of N/10 Na 2S2O3 solution ≡ 2ml of CCl 4 layer = V1 ≡ Volume of N/100 Na 2S2O solution 20ml of H O Solution = V2 Download With Free Trial 3 2 Volume of N/100 Na 2S2O3 solution ≡ 20ml of CCl4 layer = V1 X 10 X 10 Partition Coefficient,K = (V 1 X 100)/V2 Bottles
V1
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V2
K
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If Iodine is added to an aqueous KI solution the following equlibrium e KI + I2 → KI3 or I- + I2 → I3-
The equilibrium constant of the reaction is given by K= [ KI3 ] [ KI ] [ I 2 ] Since molecular iodine is soluble in both aqueous and organic phase, added to a system of CCl 4 and aqueous KI solution of known concentration, distributed between the two phase. In the system consisting of iodine, K CCl4, the total concentration of I 2 in aqueous layer is given by the sum concentration of the free iodine in aqueous layer and that of KI 3 of I3-.Th concentration of Iodine in CCl 4 layer can be found by titration with sta Na2S2O3 Knowing the partition coefficient of iodine between CCl 4 and wat concentration of free iodine (uncombined) in aqueous layer can be calculate concentration of KI 3 is obtained by subtracting the concentration of free iod aqueous layer. Then calculate the difference of initial concentration of K concentration of KI3 formed from the concentration of KI that has reacted gives the concentration of KI at equilibrium. This reciprocal of equilibrium co „k‟ gives the instability constant of the complex KI3-Instability Constant = [ KI ] [ I2 ] [ KI3 ] Requirements: I2 in CCl 4, K2Cr2O7, Na2S2O3.5H2O,10% KI, starch, Distilled H2O, CCl Procedure:
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Take 3 stoppered bottles and label them. Prepare approximately 0 Unlock full access with a free trial. solution and in the 3 bottles in the solution are taken as f ollows.
Download With Free Trial Bottle No KI solution I2 /CCl 4 I 50ml 20ml II 50ml 15ml III 50ml 10ml
CCl4 0ml 5ml 10ml
All the above bottles are stoppered and shaken well for half an hour a Master your semester with Scribd aside in a thermostate. 5ml of organic layer isFree titrated with N/10Na S Read Foron 30 Days Sign up to vote this title adding 10ml 10%KI solution and water using starch as indicator. Similarly 1 & The New York Times Useful Not useful the aqueous layer is titrated against the same thiosulphate solution di 2
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Special offer for students: Only $4.99/month. Repeat the titration for the organic and the aqueous layer from bottle II a also.
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Normality of K2Cr2O7 Normality of Na2S2O3 Normality of KI
= mass/lit Eq. wt = =
Estimation N/10 Na2S2O3 Vs I2 in Organic layer (starch) Bottle No Volume of organic Burette reading layer pipetted Initial Final
Vol of N/10 Na2S2O3
Estimation N/10 Na2S2O3 Vs I2 in aqueous layer (starch) Bottle No Volume of aq Burette reading Vol of N/10 Na2S2O3 layer pipetted Initial Final
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Download With Free Trial Calculation: Partition Coefficient = Bottle 1: Concentration of I 2 in CCl4, C2 N/2 Concentration of I 2 in aq layer,C 1 N/2 Special offer for students: Only $4.99/month. Concentration of I2 ,C4 Combined Iodine = C - C /D
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= Free Foron 30this Days =Read Sign up to vote title = Useful Not useful Cancel anytime. = = C2 /D =
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Bottle 3: Concentration of I 2 in CCl4, C2 = N/2 = Concentration of I 2 in aq layer,C 1 = N/2 = Concentration of I2 ,C4 = C2 /D = Combined Iodine = C 1 - C2 /D = Concentration of KI in equation = Normality of KI – Combined Iodine K3 = = [ KI3 ] = C1 [ KI ] [ I 2 ] [ KI ] C4 Result:
___________________________________________________________________
Experiment 3: Determination of concentration of KI solution
___________________________________________________________________ Aim:
To determine the concentration of given KI solution from the equil constant of the reaction KI + I 2 → KI3 Principle:
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In a separate experiment using standard solution of KI, the equil Unlock.Then full access with athe free trial. constant „k‟ is determined using given KI solution. The experim repeated using the value of „k‟ obtained before and the strength of the gi solution is calculated. Download With Free Trial Instability Constant, k, = [ KI ] [ I 2 ] [ KI3 ] [ KI ] = k [ KI 3 ] [ I2 ] From the experiment the equilibrium constant of KI is obtained. The concentration of KI solution is obtained by adding the concentration of KI Read Free Foron 30this Days Sign up to vote title equilibrium constant.
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From the title value, instability constant and hence the concentrat given KI is obtained. Observation: Weight of K2Cr2O7 = Standardisation of Na 2S2O3.5H2O N/10 Na2S2O3 x K2Cr2O7 (starch) Volume of K2Cr2O7 pipetted Burette reading Initial Final
Normality of K2Cr2O7 Normality of Na2S2O3
Vol of N/10 Na2S2O3
= mass/lit = Eq. wt =
Estimation N/10 Na2S2O3 Vs I2 in Organic layer (starch) Bottle No Volume of organic Burette reading layer pipetted Initial Final
Vol of N/10 Na2S2O3
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Unlock full access with a free trial.
Download With Free Trial Estimation N/10 Na2S2O3 Vs I2 in aqueous layer (starch) Bottle No Volume of aq Burette reading Vol of N/10 Na2S2O3 layer pipetted Initial Final
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[ KI ] = K1 C1 C4 Bottle 2: Concentration of I 2 in CCl4, C2 = N/2 = Concentration of I 2 in aq layer,C 1 = N/2 = Concentration of I 2 ,C4 = C2 /D = Combined Iodine = C 1 - C2 /D = Concentration of KI in equation = Normality of KI – Combined Iodine K2 =
[ KI3 ]
=
[ KI ] [ I 2 ]
C1 [ KI ]
[ KI ] = K2 C1 C4 Bottle 1: Concentration of I 2 in CCl4, C2 = N/2 = Concentration of I2 in aq layer,C 1 = N/2 = Concentration of I 2 ,C4 = C2 /D = Combined Iodine = C 1 - C2 /D = Concentration ofYou're KI in equation Normality of KI – Combined Iodine Reading a=Preview K3 =
[ KI3 ]
=
C1
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[ KI ] [ I 2 ]
[ KI ]
[ KI ] = KDownload With Free Trial 3 C1 C4 Result:
Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title ___________________________________________________________________ & The NewExperiment York Times Useful Not useful 4: Partition-Coefficient of Benzoic acid between C H and Cancel anytime.
6
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using phenolphthalein indicator. The partition Coefficient of benzoic acid be water and benzene are determined.
Contents Benzoic acid Water Benzene
Bottle I 1g 50ml 50ml
Bottle II 2g 50ml 50ml
Bottle III 3g 50ml 50ml
Observation: Normality of NaOH used for titration of benzene layer,N1 = N/10 Normality of NaOH used for titration of aqueous l ayer,N1‟ = N/100 Estimation Benzoic acid in Benzene layer Vs N/10 NaOH (phenolphthalein) Bottle No Volume of benzene Burette reading Vol of N/10 NaOH layer pipetted, V2 N1 Initial Final
Concentration of Benzoic acid C org = (V1 x N1)/V2 = You're Reading a Preview Estimation Benzoic acid in AqueousUnlock layerfullVs N/100 (phenolphthalein) access with NaOH a free trial. Bottle No Volume of Aqueous Burette reading Vol of N/100 NaOH layer pipetted, V2 N1 Initial Final
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Concentration of Benzoic acid C = (V Master your semester with Scribd= Partition Coefficient, K = √(C )/C & The New York Times K = √(C )/C aq
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1
1
org
aq
2
org
aq
K3 = √(Corg)/Caq
x N1‟ )/V2 = = =
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Physical pharmacy I-II
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VISCOMETRY
___________________________________________________________________
Experiment 1: Verification of J-Kendall’s equations
___________________________________________________________________ Aim: To verify Kendall‟s equation from viscosity measurements Principle: Kendall‟s equation is Log фAB = xA log фA + xB log фB Where, фAB XA XB фA фB
= fluidity of mixture = mole fraction of compound „A‟ = mole fraction of compound „B‟ = fluidity of compound „A‟ = fluidity of compound „B‟
but xA + xB =1 ; log фAB
You're Reading a Preview
therefore xB =1-xB
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= xA(log фA-log фB )+ log фB
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Now фAB is plotted against mole fraction x .The slope of the graph giv value of (log фA –log фB). The value of the slope is compared with t calculated value. Procedure:
The Viscometer is cleaned and a constant volume of the Master your semester with Scribd mixt Read Free Foron 30this Days Sign up to vote titleThen a num introduced into it and its time of flow is determined separately. mixtures are prepared and their times of flow are determined. & The New York Times Useful Not useful Mole fractio particular component is plotted against the fluidity of the mixture. Then fro Cancel anytime.
Special offer for students: Only $4.99/month. graph, slope is found and compared with the theoretical value.
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Weight of
Number of moles of „A‟
Calculations:
Number of moles of „B‟
Total
Mole fractio xA = nA / (n
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фA = 1/ηA
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log фA log фB
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= =
slope from calculations, log фA - log фB
Master your semester with Scribd slope from the graph, & The New York Times dy/dx Special offer for students: Only $4.99/month. Result:
=
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Where, d1 and d2 are the density of the liquids ; t1 and t2 are the time of flow of the liquids.
The coefficients of viscosity and composition of solution, we get a s line from which the composition of unknown mixture can be determined. Procedure:
Ostwald‟s viscometer is cleaned and dried. It is fitted vertically on a .A mixture of different compositions are prepared by mixing the two liqu different proposition. Time of flow for each of these is determined. Each time amount of liquid is taken. The time of flow for the given unknown ar determined. A graph is plotted between viscosity and composition of solutio with the help of it composition of given unknown mixture is obtained. Observations: Liquid composition
1
Flow time 2 Mean
Relative density
Coefficient of viscosity
Percenta compositio
Unknown
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Calculations: Relative density,
ρ1
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v1d1+v2d2
= Mass of empty bottle Mass of bottle and unknown Master your semester with Scribd Mass of unknown & The New York Times Relative density of unknown
= = =
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=
Useful Not useful weight of unknown Cancel anytime. Weight of water
Special offer for students: Only $4.99/month. Coefficient of viscosity of unknown =
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_______________________________________________________ Experiment 1: Determination of Molecular Weight
___________________________________________________________________ Aim:
To determine the molecular weight of an unknown solute using naph as the solvent. Principle:
When a non volatile solute is added the freezing point of the solv lowered. The difference in the freezing point of the pure solvent and the solu You're Reading a Preview referred to as the depression in freezing point. And it is related to the mo weight of the nonvolatile solute as given bel ow. Unlock full access with a free trial.
M = (kf X w X 1000) / ∆T X W
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Where w is the weight of the solute of molecular mass M and W weight of the solid solvent. Knowing the depression in freezing point ( ∆T) cryoscopic constant (k f ) of naphthalene, the molecular weight of unknown can be calculated. Advantage of the solid cryoscopy over liquid cryoscopy is t solid cryoscopy ordinary thermometer are used and the measurements a easy. Read Free Foron 30this Days Sign up to vote title Requirements: Not useful Useful Cancel anytime. Flat bottom test tube, Naphthalene, Biphenyl, water bath Special offer for students: Only $4.99/month. Procedure: A clean dry flat bottomed test tube is weighed and about 2/3 of it i
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Observation: Time
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Solvent
1st Addition
2nd Addition
3rd Addition
Result: The Molecular weight of given solute =
_______________________________________________________ Experiment 2: Determination of Cryoscopic Constant
___________________________________________________________________ Aim:
To determine the molecular weight of an unknown solute using naph as the solvent. Principle:
When a non volatile solute is added the freezing point of the solv lowered. The difference in the freezing point of the pure solvent and the solu referred to as the depression in freezing point. And it is related to the mo You're Reading a Preview weight of the nonvolatile solute as given bel ow. Unlock full access with a free trial.
Kf = (M X W X ∆T) / 1000 X w
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Where w is the weight of the solute of molecular mass M and W weight of the solid solvent. Knowing the depression in freezing point a molecular weight of solute, the Cryoscopic constant of unknown solvent c calculated. Advantage of the solid cryoscopy over liquid cryoscopy is that in cryoscopy ordinary thermometer are used and the measurements are too eas
Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Requirements: Flat bottom test tube, Naphthalene, Biphenyl, & The New York Times Useful water useful Notbath
Cancel anytime. Procedure: Special offer for students: Only $4.99/month. A clean dry flat bottomed test tube is weighed and about 2/3 of it i with naphthalene and again weighed. Then it is closed with a two hole
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Time
Solvent
1st Addition
2nd Addition
3rd Addition
Result: The Molecular weight of given solute =
___________________________________________________________________
HEAT OF SOLUTION BY SOLUBILITY DETERMINATION
___________________________________________________________________ Aim:
To determine the heat of a solution of a given substances by solubility metho Principle:
The dissolution of a solid into a liquid is usually accompanied with a effect, heat absorbed or evolved. The heat absorbed or evolved where 1 m the solid is dissolved into a solution, which is already practically saturated You're Reading a Preview from that of infinite dilution from saturation to infinite dilution. The influence of temperature on solubility is best expressed by Van‟t hoff eq Unlock full access with a free trial. ie. D ln S/dt = ΔH/RT2 Where, „S‟ is theDownload solubility With of the saltTrial in moles per 1000g of the solva Free ΔH is the heat of a solution per mole. On integration of the above expressio have log S = - ΔH +C 2.303 RT ΔH is assumed to be independent of temperature; A plot of log S (ordinate) against corresponding value 1/T (abscissa) will practicallybe a s Read Free For Days Sign up to vote this title Δ line or a smooth curve. The slope of the straight line is on –30 H/2.303R. Thus be calculated. Useful Not useful Cancel anytime. Alternatively, integrating equation between 2 temperatures T 1 and T Special offer for students: Only $4.99/month. logS2 – logS1 = ΔH |1- 1|
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