IIT JEE Syllabus - KEEEL
INTRODUCTION
Solutions are homogenous mixtures of two or more substances in a single phase. Most of solutions can be considered as having a majority ingredient called a solvent and one more minority ingredients called solutes. For the sake of simplicity in this unit, we shall consider only binary solutions. Here each component may be in solid, liquid or in gaseous state and, therefore, several types of possible solutions are summarized in the following table:
DIFFERENT TYPES OF SOLUTIONS
|Type of Solution |Common Example |
|Gaseous Solutions | |
|Gas in gas |a mixture of oxygen and nitrogen gases. |
|Liquid in gas |chloroform vapours mixed with nitrogen gas. |
|Solid in gas |camphor |
|Liquid Solutions | |
|Gas in liquid |Oxygen gas dissolved in water |
|Liquid in liquid |ethanol dissolved in water |
|Solid in liquid |sucrose dissolved in water |
|Solid Solutions | |
|Gas in solid |solutions of hydrogen in palladium |
|Liquid in solid |Amalgam of mercury with sodium |
|Solid in solid |copper dissolved in gold |
CONCENTRATION UNITS
The concentration of a solute is the amount of solute dissolved in a given quantity of solvent or solution. The quantity of solvent or solution can be expressed in terms of volume or in terms of mass or molar mass. Thus there are several ways of expressing the concentration of a solution.
(a) Molarity (M): Moles of solute present in one litre solution.
[pic]
(b) Molality (m): Moles of solute present in one kilogram of solvent.
[pic]
(c) Normality (N): No. of equivalents present in one litre solution.
[pic]
(d) Mole fraction: The mole fraction of a component substance A(XA) in a solution is defined as the moles of component substance divided by the total moles of solution.
[pic]
Mass percent: The mass percent of a component A in solution is defined as
Mass % of A = [pic]
Part per million (PPM): It is defined as the parts of given component in one million parts of solution. Mathematically
[pic]
Illustration 1. 5 g of NaCl is dissolved in 1000 g of water. If the density of the resulting solution is 0.997 g per cc, calculate molality, molarity, normality and mole fraction of the solute.
Solution: Mole of NaCl = [pic] = 0.0854 (Mol. wt. of NaCl = 58.5)
Molality = [pic]
= [pic] = 0.0854 m
Volume of the solution= [pic]= [pic]
Again by definition
Molarity = [pic]
= [pic] = 0.085 M
( Normality = 0.085 N (for NaCl, eq. wt. = mol. wt)
Further, Mole of H2O = [pic] = 55.55
(1000 gram of water = 1000 ml of water, because density = 1 g/cc)
Total mole = Mole of NaCl + Mole of H2O
= .0854 + 55.55 = 55.6354
Mole fraction of NaCl = [pic] = [pic] = 1.535 ( 10–3
Illustration 2. If 20 ml of ethanol (density = 0.7893 gm / ml) is mixed with 40 ml of water
(density = 0.9971 gm/ml) at 25°C, the final solution has density of 0.9571 gm / ml. Calculate the percentage change in total volume of mixing. Also calculate the molality of alcohol in the final solution.
Solution: Mass of ethanol = v ( d
W = 20 ( 0.7893 gm = 15.786 gm
Mass of water = v ( d
= 40 ( 0.9971 gm = 39.884 gm
Total volume = 60 ml
Total mass = 15.786 + 39.884 = 55.67 gm
Let the volume of solution = x ml
X = [pic] = [pic] = 58.165 ml
Change in volume = 60 – 58.165 = 1.835 ml
% change in volume = [pic]
= 3.05%
Molality, m = [pic] = [pic]
= 8.604 m
Exercise 1.
A solution of ethanol in water is 10% by volume. If the solution and pure ethanol have densities of 0.9866 g/cc and 0.785 g/cc respectively, find the per cent by weight of ethanol.
Exercise 2.
The density of a 3M sodium thiosulphate solution is 1.25 gm cm–3. Calculate the molalities of Na+ and S2O3– – ions; and mole fraction of sodium thiosulphate.
SOLUBILITY OF GASES
We are familiar that gases are completely miscible with each other. Gases also dissolve in liquids and solids. For example, soda-water contains carbon dioxide dissolved in water under high pressure. Oxygen is sufficiently soluble in water to allow survival of aquatic life in lakes, rivers and oceans. An example of dissolution of gas in a solid is the solubility of hydrogen gas in palladium.
The solubility of a gas in a liquid is determined by several factors. In addition to the nature of the gas and the liquid, solubility of the gas depends on the temperature and pressure of the system. The solubility of a gas in a liquid is governed by Henry’s law which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas. Dalton, a contemporary of Henry, also concluded independently that the solubility of a gas in a liquid solution is a function of the partial pressure of the gas. If we use the mole fraction of the gas in the solution as a measure of its solubility, then:
Mole fraction of the gas in a solution is proportional to the partial pressure of the gas.
Or, partial pressure of the gas in solution = KH ( mole fraction of the gas in solution
Here KH is Henry’s law constant
or, p = KHx
If we draw a graph between partial pressure of the gas versus mole fraction of the gas in solution, then we should get a plot of the type as shown in figure.
[pic]
Different gases have different KH values at the same temperature. This suggests that KH is a function of the nature of the gas. Table gives KH values of some common gases at specified temperature
Values of Henry’s law constant (KH) for some selected gases in water
|Gas |Temp/K |KH/kbar |
|He |293 |144.97 |
|H2 |293 |69.16 |
|N2 |293 |76.48 |
|N2 |303 |88.84 |
|O2 |293 |34.86 |
|O2 |393 |46.82 |
It is obvious from figure that the higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid. It can be seen from table that KH value for both N2 and O2 increases with increase in temperature indicating that solubility of gases decreases with increase of temperature. It is due to this reason that aquatic species are more comfortable in cold waters rather than warm waters.
Illustration 3. If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 litre of water. Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N2 at 293 K is 76.84 kbar.
Solution: The solubility of gas is related to its mole fraction in the aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry’s law. Thus,
[pic]
As 1 litre water contains 55.5 mol of it, therefore, if n represents number of moles of N2 in solution,
[pic]
Thus, n = 1.29 ( 10(5 ( 55.5 mol = 7.16 ( 10(4 mol
[pic]
= 0.716 m mol
Henry’s law finds several applications in industry and explains some biological phenomenon. Notable among these are:
(i) To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.
(ii) To minimize the painful effects accompaynig the decompression of deep sea divers, oxygen diluted with less soluble helium gas is used as breathing gas.
(iii) In lungs, where oxygen is present in air with high partial pressure, haemoglobin combines with oxygen to form oxyhaemoglobin. In tissues where partial pressure of oxygen is low, oxyhaemoglobin releases oxygen for utilization in cellular activities.
SOLID SOLUTIONS
Solid solutions are formed by mixing two solid components in the molten state in appropriate proportion and then cooling the molten mass. Solid solutions are of two types: substitutional solid solutions and interstitial solid solutions. In a substitutional solid solution, atoms, molecules or ions of one substance take the place of similar species of other substance in its crystal lattice as shown in figure (a). Brass, bronze, monel metal and steel are familiar examples of this type of solid solution.
Interstitial solid solutions constitute the other type and are formed by placing atoms of one kind into voids or interstices, that exist between atoms in the host lattice. This is illustrated in figure (b).
[pic]
Tungsten carbide, WC, an extremely hard substance, is an example of interstitial solid solution. Here tungsten atoms are arranged in a face-centred cubic pattern with carbon atoms in the octahedral holes, where these are surrounded by six tungsten atoms placed at the vertices of the octahedron. Tungsten carbide has many industrial uses in making of cutting and grinding tools.
VAPOUR PRESSURE OF A SOLUTION
Relative lowering of vapour pressure of a solvent is a colligative property equal to the vapour pressure of the pure solvent minus the vapour pressure of the solution. For example, water at 20(C has a vapour pressure of 17.54 mmHg. Ethylene glycol is a liquid whose vapour pressure at 20(C is relatively low, an aqueous solution containing 0.010 mole fraction of ethylene glycol has a vapour pressure of 17.36 mmHg. Thus the vapour pressure lowering, (P = 17.54 mmHg ( 17.36 mmHg = 0.18 mmHg.
RAOULT’S LAW
Vapour pressure of a number of binary solution of volatile liquids such as benzene and toluene at constant temperature gave the following generalization which is known as the Raoult’s law.
The partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution.
Suppose a binary solution contains nA moles of a volatile liquid A and nB moles of a volatile liquid B, if PA and PB are partial pressure of the two liquid components, the according to Raoult’s law
[pic]
Where xA is the mole fraction of the component A given by nA/nA+ nB; xB is the mole fraction of the component B, given by nB/nA+nB and [pic] are the vapour pressures of pure components A and B respectively.
If the vapour behaves like an ideal gas, then according to Dalton’s law of partial pressures, the total pressure P is given by
[pic]
= [pic]
[pic]
[pic]
The relationship between vapour pressure and mole fraction of an ideal solution at constant temperature is shown. The dashed lines I and II represent the partial pressure of the components. (It can be seen from the plot that pA and pB are directly proportional to xA and xB respectively). The total vapour pressure is given by line marked III in the figure.
Illustration 4. The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40g of methanol. Calculate total vapour pressure of the solution.
Solution: Number of moles of ethanol = [pic] = 1.5
Number of moles of methanol = [pic] = 1.25
XA = [pic] = 0.4545 and XB = 1–0.4545 = 0.545
Let A = CH3OH, B = C2H5OH
Total pressure of the solution
PT = XA[pic] + XB[pic]
= 0.4545 ( 88.7 + 0.545 ( 44.5 = 40.31 + 24.27 = 64.58 mm 4g
Illustration 5. The composition of vapour over a binary ideal solution is determined by the composition of the liquid. If XA and YA are the mole-fraction of A in the liquid and vapour, respectively find the value of XA for which YA(XA has a minimum. What is the value of the pressure at this composition.
Solution: [pic]
Subtracting xA from both the sides, we get
[pic]
Now differentiating w.r.t. xA, we get
[pic]
The value of xA at which yA ( xA has a minimum value can be obtained by putting the above derivative equal to zero. Thus we have
[pic]
Solving for xA, we get xA = [pic] and hence P = [pic]
Illustration 6. A very small amount of non-volatile solute (that does not dissociate) is dissolved in 56.8 cm3 of benzene (density 0.889gcm-3). At room temperature, the vapour pressure of this solution is 99.88 mm Hg while that of benzene is 100mm Hg. Find the molality of this solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene. What is the value of molal freezing point depression constant of benzene?
Solution: As [pic]
[pic]
[pic]
[pic]
[pic]
Also, [pic]
[pic]
[pic]
Illustration 7. What is the composition of the vapour which is in equilibrium at 30°C with a benzene-toluene solution with a mole fraction of benzene of 0.400?
([pic]= 119 torr and [pic] = 37.0 torr)
Solution: Total pressure of the solution is given by
PT = XB[pic] + XT[pic]
= 0.4 ( 119 + 0.6 ( 37
= 47.6 + 22.2
= 69.8 torr
Applying Dalton’s law for mole fraction in vapour phase.
YB = [pic]
= 0.763
YT = 1– 0.763 = 0.237
Exercise 3.
If PA is the vapour pressure of a pure liquid A and the mole fraction of A in the mixture of two liquids A and B is x, then, what is the partial vapour pressure of A?
Exercise 4.
Vapour pressure of C6H6 and C7H8 mixture at 50°C is given by P (mm Hg) = 180 XB + 90, where XB is the mole fraction of C6H6. A solution is prepared by mixing 936 g benzene and 736 g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of 50°C, what would be mole fraction of C6H6 in the vapour state?
Exercise 5.
When a solute is added to a pure solvent, which statement is not true?
(A) Vapour pressure of the solution becomes lower than the vapour pressure of the pure solvent.
(B) Rate of evaporation of the pure solvent is reduced.
(C) Solute does not affect the rate of condensation.
(D) None of these.
Exercise 6.
According to Raoult's law the relative decrease in solvent vapour pressure over the solution is equal to
(A) The mole fraction of the solvent.
(B) The number of moles of the solute.
(C) 'i' times the mole fraction of the solute which undergoes dissociation or association in the solvent ( i = Vant Hoff factor).
(D) None.
Measurement of Vapour Pressure Lowering (Ostwald and Walker’s Apparatus)
In this method a stream of dry air is bubbled successively through (i) the solution (ii) the pure solvent and (iii) a reagent which can absorb the vapour of the solvent. As the solvent is usually water the reagent is generally anhydrous Calcium Chloride. The complete assembly is shown in the figure.
[pic]
The first three bulbs contain a weighed amount of the solution under examination and the next three bulbs contain a weighed amount of the pure solvent. A weighed amount of anhydrous calcium chloride is taken in the set of U-tubes at the end.
All the bulbs must be kept at the same temperature and air must be bubbled gradually to ensure that it gets saturated with the vapours in each bulb.
The dry air, as it passes through the solution, takes up an amount of vapour which is proportional to the vapour pressure of the solution at the prevailing temperature. This moist air passes through water (solvent), it takes up a further amount of vapour which is proportional to the difference in vapour pressure of the pure solvent and the solution.
It is evident that,
Loss in mass of solution ( P
Loss in mass of solvent ( Po ( P
Loss in mass of solution + loss in mass of solvent
( Po (( P + P ( Po
Where P° = Vapour pressure of pure solvent, P = Vapour pressure of solution
The calcium chloride tubes are weighed at the end of the experiment. The gain in mass should be equal to the total loss in mass of the solution and solvent which, in turn, is proportional to P0, as shown above.
In other words, [pic]
Thus, knowing the loss in mass of the solvent and gain in the mass of the calcium chloride tubes, it is possible to calculate the lowering of vapour pressure.
Illustration 8. Dry air is passed through a solution containing 20g. of an organic non-volatile solute in 250 ml of water. Then the air was passed through pure water and then through a U-tube containing anhydrous CaCl2. The mass lost in solution is 26g and the mass gained in the U-tube is 26.48 g. Calculate the molecular mass of the organic solute.
Solution: Loss in mass in solution = 26g
Gain in mass in U-tube = 26.48 g.
Therefore, Loss in mass in solvent = 0.48g
We know that
[pic]
[pic]= mole fraction of solute in the solution
[pic]
[pic]
[pic]
[pic]
( Molecular mass of the organic solute = 78.
Exercise 7.
In Ostwald and Walker's Apparatus, if the loss in wt. of solution is 6 units and loss in wt of solvent in 3 units, then calculate the mole fraction of the solute.
Exercise 8.
A current of dry air was bubbled through a bulb containing 26.66 g of an organic substance in 200 gms of water, then through a bulb at the same temperature containing pure water and finally through a tube containing fused calcium chloride. The loss in weight of water bulb is 0.0870 gms and gains in weight of CaCl2 tube is 2.036 gm. Calculate the molecular weight of the organic substance in the solution.
IDEAL SOLUTION
An ideal solution of substance A and B is one in which both substances follow Raoult’s law for all values of mole fractions. Such solutions occurs when the substances are chemically similar so that the intermolecular forces between A and B molecules are similar to those between two A molecules or between two B molecules. The total vapour pressure over an ideal solution equals the sum of the partial vapour pressures, each of which is given by Raoult’s law.
[pic]
For example: Solution of benzene C6H6 and toluene C6H5CH3 are ideal. Note the similarity in their structural formula.
|[pic] |
Suppose a solution is 0.70 mole fraction in benzene and 0.30 mole fraction in toluene. The vapour pressure of pure benzene and pure toluene are 75 mmHg and 22 mmHg respectively. Hence the total vapour pressure is
[pic]
Ethylene bromide and ethylene chloride, n-Hexane and n-heptane, n-butyl chloride and
n – butyl bromide, carbon tetrachloride and silicon tetra chloride are few other examples of ideal solutions.
Condition for a solution to be ideal.
(a) It should obey Raoult’s law i.e. [pic]
(b) [pic]= 0 i.e. no heat should be absorbed or evolved during mixing.
(c) [pic]= 0 i.e. no expansion or contraction on mixing
REAL OR NON - IDEAL SOLUTIONS
Those solution which do not obey Raoult’s law over entire range of composition and deviate from ideal behaviour, are real or non – ideal solution. They are divided into two types
(a) The solution showing positive deviations from ideal behaviour for those type of solutions,
(i) [pic]
(ii) [pic]
(iii) [pic]
Positive deviation
Solution of ethyl alcohol in water shows positive deviation. Attractive forces between
A – A or B ( B > A ( B.
|[pic] |
Positive deviations occur when the interaction between unlike molecules is weaker than the interaction between like molecules. The solutions of this type are characterized by positive enthalpy of mixing, (Hmix and positive volume of mixing, (Vmix. Example are (acetaldehyde (carbon disulphide, water (propyl alcohol, ethyl alcohol (chloroform and cyclohexane (carbon tetrachloride.
(b) The solution showing large negative deviations from ideal behaviour and the vapour pressure of each component is considerably less than that predicted by Raoult’s law, for these type of solutions.
(i) [pic]
(ii) [pic]
(iii) [pic]
Negative deviation
The solution of HNO3 in water shows negative deviation. Attractive forces between A ( A or B ( B < A ( B
|[pic] |
Negative deviations occurs when the interaction between unlike molecules is stronger than the interaction between like molecules. The solutions of this type are characterized by negative enthalpy of mixing and negative volume of mixing.
For example - acetone - chloroform, water - sulphuric acid and water - nitric acid.
Illustration 9. Liquids A and B form an ideal solution. The vapour pressure of A and B at 100°C are 300 and 100 mm Hg respectively. Suppose that vapour above solution is composed of 1 mole of A and 1 mole of B is collected and condensed. This condensate is then heated at 100°C and vapour are again condensed to form a liquid L. What is the mole fraction of A in the vapours of L?
Solution: Vapour pressure due to vapours above solution
PT = XA[pic] + XB[pic]
PT = 300XA + 100 XB
It is given that in vapour phase each of A and B are one mole each hence each of them have mole fraction 0.5 in vapour phase
After condensation of vapours
In condensate (1)
X(A = 0.5, X(B = 0.5
P(T = 0.5 ( 300 + 0.5 ( 100
= 150 + 50 = 200 mm
Mole fraction of A & B in vapour phase of condensate
Y(A = [pic] = [pic] = 0.75
Y(B = 1 – 0.75 = 0.25
When the vapours of the condensate (1) will again be vapourised in condensate (2) liquid L
XA(( = 0.75
XB(( = 0.25, where XA(( and XB(( are mole fractions of A and B in liquid L
PT(( = 300 ( 0.75 + 100 ( 0.25 = 225 + 25 = 250 mm
and mole fraction of A in vapour phase of the condensate (2) is
given by
YA(( = [pic] = 0.9
Exercise 9.
Which pair will show positive and negative deviation from Raoult's Law.
(i) Methanal and Chloroform
(ii) Methanol and Cyclopentane
Exercise 10.
Cyclohexane and ethanol at a particular temperature have vapour pressure of 280 mm and 168 mm of Hg respectively. If these two solutions having mole fraction value of cyclohexane equal to 0.32 are mixed and the mixture has a total vapour pressure of 376 nm, will the mixture be an ideal solution?
AZEOTROPIC MIXTURE
Some liquids on mixing form azeotropes which are binary mixtures having same composition in liquid and vapours phase and boil at a constant temperature. In such cases, it is not possible to separate the components by fractional distillation. There are two types of azeotropes called as minimum boiling azeotrope and maximum boiling azeotrope, respectively. Solutions of ethanol and water show such a large deviation from Raoult’s law that there is a maximum in the vapour pressure curve and hence a minimum in the boiling point.
Ethanol-water mixtures (obtained by fermentation of sugars) are rich in water. Fractional distillation is able to concentrate the alcohol to, at best, the azeotropic composition of approximately 95% by volume of ethanol. Once this composition has been achieved, the liquid and vapour have the constant composition, and no additional concentration occurs. Other methods of separation have to be used for preparing 100% C2H5OH.
There are also solutions that show large negative deviation from ideality and, therefore, have a minimum in their vapour pressure curves. This leads to a maximum on the boiling point diagram. HNO3 and H2O form examples of this class of the azeotrope. This azeotrope has the approximate composition, 68% nitric acid and 32% water by mass, with a boiling point of 393K.
COLLIGATIVE PROPERTIES
These are the properties that depend on the concentration of solute molecule or ions in solution but not on the chemical identity of the solute. For example, addition of ethylene glycol to water lowers the freezing point of water below 0(C.
The magnitude of freezing point lowering is directly proportional to the number of solute molecules added to a particular quantity of solvent.
If 0.01 mole of ethylene glycol is added to 1 kg of water, the freezing point is lowered to (0.019(C while on adding 0.020 mole of ethylene glycol to 1 kg of water, the freezing point is lowered to (0.038(C.
The various colligative properties are
(i) Relative lowering of vapour pressure
(ii) osmotic pressure
(iii) elevation of boiling point
(iv) depression of freezing point
RELATIVE LOWERING OF VAPOUR PRESSURE
The addition of a non – volatile solute to a solvent at a given temperature and pressure results in the lowering of the vapour pressure of the solvent.
If a solution is formed by dissolving n moles of a non volatile solute in N moles of a volatile solvent. Then
Mole fraction of solvent, X1 = [pic]
Mole fraction of solute X2 = [pic]
Since the solute is non – volatile, so it would have negligible vapour pressure and thus the vapour pressure of the solution, is approximately same as the vapour pressure of the solvent.
According to Raoult’s law
[pic]
Since X1 + X2 = 1 so X1 = 1 ( X2
Now P1 = [pic]
Or
[pic]
in equation (ii) [pic]is the vapour pressure lowering and [pic] is the relative vapour pressure lowering. From equation (ii) it is clear that the relative vapour pressure lowering is proportional to the mole fraction of the non – volatile solute in solution. It is independent of the nature of the solute. The relative vapour pressure lowering is, therefore a colligative property.
The mole fraction, X2 = [pic]
For a dilute solution, the number of moles of the solute (n) can be neglected as compared to the number of moles of the solvent (N). Hence
[pic]
Where W1 is the amount of the solute dissolved in W1 amount of solvent, M1 is the molar mass of the solvent and M2 is the molar mass of the solute.
[pic]
Vapour pressure in terms of molality of the solution:
[pic]
Illustration 10. The vapour pressure of a solvent decreased by 10 mm Hg when a non volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg.
(A) 0.8 (B) 0.6
(C) 0.4 (D) 0.2
Solution: Po – Ps = Po ( mole fraction of solute; 10 = Po ( 0.2
Again, 20 = Po ( mole fraction solute
Hence, n = 0.4, so, mole fraction of solvent = 1 – 0.4 = 0.6
Hence, (B) is correct.
Illustration 11. The density of a 0.438 M solution of potassium chromate at 298 K is 1.063 g cm-3. Calculate the vapour pressure of water above this solution.
Given: P0 (water) = 23.79 mm Hg.
Solution: A solution of 0.438 M means 0.438 mol of K2 CrO4 is present in 1L of the solution. Now,
Mass of K2CrO4 dissolved per litre of the solution = 0.438 ( 194 = 84.972 g
Mass of 1L of solution = 1000 (1.063 = 1063 g
Amount of water in 1L of solution = 978.028[pic]
Assuming K2CrO4 to be completely dissociated in the solution, we will have;
Amount of total solute species in the solution = 3 ( 0.438 = 1.314 mol.
Mole fraction of water solution =[pic]
Finally, Vapour pressure of water above solution = 0.976 ( 23.79 = 23.22 mm Hg
Exercise 11.
Ethylene dibromide (C2H4Br2) and 1, 2–dibromopropane (C3H6Br2) forms a series of ideal solution over the whole range of composition. At 85(C, the vapour pressure of these pure liquids are 173 mmHg and 127mm Hg respectively.
10 gm of ethylene dibromide is dissolved in 80 gm of 1, 2 – dibromo–propane. Calculate the partial pressures of each components and the total pressure of the solution at 85(C. Calculate the composition of vapour in equilibrium with the above solution and express as mole fraction of ethylene dibromide.
What would be the mole fraction of 1, 2 dibromopropane in solution with 50:50 mole mixture in the vapour.
Exercise 12.
A solution comprising 0.1 mol of naphthalene and 0.9 mol of benzene is cooled until some benzene freezes out. The solution is then decanted off from the solid and warmed to 353 K, where its vapour pressure is found to be 670 torr. The freezing and normal boiling points of benzene are 278.5 K and 353 K, respectively and its enthalpy of fusion is 10.67 kJ mol–1. Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume conditions of ideal solution. Kf of benzene = 5K kg mole–1.
Exercise 13.
The pressure of the water vapour of a solution containing a non volatile solute is 2% below than that of the vapour pressure of pure water. The molality of the solution in mol/g is:
(A) 1.134 (B) 0.1134
(C) 1.051 (D) 0.0975
OSMOSIS AND OSMOTIC PRESSURE
There is a natural tendency of solutes in a solution to diffuse from a higher concentration to a lower concentration so as to bring about a uniform distribution throughout.
Certain membranes allow solvent molecules to pass through them but not solute molecules, particularly not those of large molecular weight. Such a membrane is called semipermeable and might be an animal bladder, a vegetable tissues or a piece of cellophone.
Osmosis is the phenomenon of solvent flow thorough a semi permeable membrane to equalize the solute concentrations on both sides of the membrane.
Osmotic pressure is a colligative property of a solution equal to the pressure that, when applied to the solution just stops osmosis. It is denoted by (. The osmotic pressure ( of a solution is related to the molar concentration of solute M.
[pic], Where R is the gas constant and T is absolute temperature.
Van’t Hoff equation for dilute solution is similar to ideal gas equation,
[pic]
Where ( = osmotic pressure
R = gas constant
Illustration 12. Calculate the molecular weight of cellulose acetate if its 0.5% (wt./vol) solution in acetone (sp. gr. = 0.9) shows an osmotic rise of 23 mm against pure acetone at 27°C.
Solution: 0.5% (wt. / vol) solution means 0.5 gm of cellulose acetate is dissolved in 100 ml solution.
Osmotic pressure = 23 mm of pure acetone
( = 2.3 cm of pure acetone = [pic]cm of Hg = 0.1522 cm of Hg
( = [pic] atm = 0.002 atm
Let the molecular weight of the cellulose acetate be M
Here, number of mole of cellulose acetate (n) [pic]
Volume = V = 100 ml = 0.1 lit
R = 0.082 Lit atm mol–1 K–1, T = (27 + 273) = 300 K
Osmotic pressure (() = [pic] (RT
( 0.002 = [pic]( 0.0821 ( 300
( m = 61575
Isotonic solutions
A pair of solutions having same osmotic pressure is called isotonic solution. When two solutions having the same osmotic pressure are put into communication with each other through a semi permeable membrane, there will be net transference of solvent from one solution to the other, but there is a dynamic equilibrium between two solutions. According to Van’t Hoff equation, it is evident that isotonic solutions must have the same molar concentration, if both are at same temperature.
For isotonic solution, [pic]
Or [pic]
Or [pic]
Illustration 13. Calculate the osmotic pressure of sucrose solution containing 1.75 gms in 150 ml of solution at 17°C.
Solution: We know that
(V = nRT = [pic]
( ( 0.15 = [pic]
( = 0.812 atm
Illustration 14. Osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose that will be isotonic with blood is ----- wt/ vol.
(A) 5.41% (B) 3.54%
(C) 4.53% (D) 53.4%
Solution: We know (V = nRT for glucose and blood;
If isotonic, (glucose = (blood
Thus, 7.65 ( V = [pic]( 0.0821 ( 310
( [pic]= 54.1 g/ lit = 5.41 % (wt/vol)
Hence, (A) is correct.
Exercise 14.
A 5% solution (wt./vol) of cane sugar is isotonic with 0.877% (wt /vol) of urea solution. Find m. wt. of urea if the m. wt. of sugar is 342.
Exercise 15.
Which has maximum osmotic pressure at temperature T?
(A) 100 ml of 1M urea solution
(B) 300 ml of 1M glucose solution
(C) Mixture of 100 ml of 1M urea solution and 300 ml of 1M glucose solution
(D) All are isotonic
Reverse osmosis and water purification
The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solutions side. That is, now the pure solvent flows out of the solution through the semipermeable membrane (SPM). This phenomenon is called reverse osmosis and is of great practical utility. Reverse osmosis is used in the desalination of sea water. A schematic set up for the process is shown in figure. When pressure more than osmotic pressure is applied pure water is squeezed out of the sea water through the membrane. A variety of polymer membrane are available for this purpose.
[pic]
The pressure required for reverse osmosis are quite high and a workable porous membrane is a film of cellulose acetate placed over a suitable support. Cellulose acetate is permeable to water but impermeable to impurities and ions present in sea water. These days many countries use desalination plants to meet their water requirement.
BOILING POINT ELEVATION BY NON VOLATILE SOLUTE
The boiling point Tb of a liquid is the temperature at which its pressure becomes equal to the atmospheric pressure. When a non – volatile solute is added to a liquid, the vapour pressure of the liquid is decreased. Hence it must be heated to a higher temperature in order that its vapour pressure becomes equal to that of the atmospheric pressure. This means that the addition of a non – volatile solute to a liquid raises its boiling point.
If [pic] is the boiling point of the solvent and Tb is the boiling point of the solution, then
[pic](Elevation of boiling point)
(Tb ( molality (m)
[pic]
(Tb = Kb ( m, kb: molal elevation constant or Ebullioscopic constant
m: molality of the solution
Elevation in boiling point can be calculated as
[pic] [M1 is the molar mass of solvent, [pic] is enthalpy of vaporization per gram of solvent.]
For a given solvent, [pic]is constant, it is denoted by Kb and called molal elevation constant of the solvent thus, [pic]
Hence [pic]
If molality m = 1, i.e. 1 mole of the solute is dissolved in 1 kg of the of the solvent. Then Kb = (Tb. Thus molal boiling point elevation constant or ebullioscopic constant of a solvent is defined as the elevation in boiling point of the solution which may be theoretically be produced when 1 mole of the non volatile, non electrolyte solute is dissolved in 1 kg of the solvent.
The elevation in boiling point depends only on the molality of the solute and is independent of the nature of the solute, it is therefore a colligative property.
[pic]
Determination of molar mass using boiling point elevation method
If W2 grams of the solute of molar mass M2 is dissolved in W1 kg of the solvent, then the number of moles of the solute dissolved in 1 kg of the solvent would be given by:
[pic]
[pic]
Illustration 15. Y g of non - volatile organic substance of molecular mass M is dissolved in 250 g benzene. Molal elevation constant of benzene is Kb. Elevation in its boiling point is given by
(A)[pic] (B)[pic]
(C)[pic] (D)[pic]
Solution: (T = [pic] = [pic]
Hence, (B) is correct.
Illustration 16. A certain non-volatile, non-electrolyte contains 40.0% carbon, 6.71% hydrogen and 53.3% oxygen. An aqueous solution containing 5% by mass of that solute which boils at 100.15°C. Kb(water) = 0.512 K kg mol-1. Calculate the molecular formula of that solute.
Solution: First of all, we compute the empirical formula of the compound as shown:
|Element |% |Mass taken |No. of moles of element |Simple ratio |
|C |40 |40 g |[pic] |[pic] |
|H |6.7 |6.7 g |[pic] |[pic] |
|O |53.3 |53.3 g |[pic] |[pic] |
Hence the empirical formula is CH2O. Empirical formula mass = 30g mol-1.
Mass of solute = 5 g, mass of solvent = 100 – 5
= 95 g
[pic]
= 0.180 kg mol-1
= 180 g mol-1
Let the molecular formula is (CH2O)x
[pic]
Hence molecular formula = C6H12O6.
Illustration 17. Molal elevation constant [pic] values of following alcohols are in the order:
[pic]
Explain in brief
Solution: Moving from [pic] to [pic] via [pic] the branching increases, the surface area decreases resulting into decrease in Vander Waals' force and hence boiling point.
[pic]
Molecular weight (M) remains the same for all the three isomeric alcohols. According to Trouton’s rule
[pic]is almost constant for all the liquids of almost similar kind of association.
Thus, [pic]is constant and [pic]
Exercise 16.
An aqueous solution of glucose boils at 100.01°C. The molal elevation constant for water is 0.5 K Kg/mol–1 Kg. What is the number of glucose molecule in the solution containing 100 gm of water? (Mglucose = 180).
Exercise 17.
An aqueous solution containing 288 gm of a non– volatile compound having the stochiometric composition CxH2xOx in 90 gm water boils at 101.24°C at 1.00 atmospheric pressure. What is the molecular formula?
Kb(H2O) = 0.512 K mol–1 kg, Tb (H2O) = 100°C.
DEPRESSION OF FREEZING POINT BY A NON – VOLATILE SOLUTE
|The temperature at which solid and liquid states of a substance |[pic] |
|have the same vapour pressure is called freezing point. If a non| |
|volatile solute is mixed with a pure solvent, the freezing point| |
|of pure solvent is always greater than the impure one. | |
|[pic], where Tf0 is the freezing point of the pure solvent and | |
|Tf is the freezing point of a non – volatile solute containing | |
|solvent and (Tf is the depression in freezing point. | |
|(Tf ( molality | |
|or (Tf = Kf ( m | |
Kf = molal freezing point depression constant of the solvent (cryoscopic constant)
m = molality of the solution
Molal freezing point depression constant of the solvent or cryoscopic constant is defined as the depression in freezing point which may be theoretically produced by dissolving 1 mole of any Non-volatile, non-electrolyte solute in 1000 gm of solvent.
[pic]
Where m1 = molecular weight of solute
w = weight of solute
W = weight of solvent
Illustration 18. The amount of ice that will separate on cooling a solution containing 50g of ethylene glycol in 200g water to –9.3°C is : [Kf = 1.86 K molality–1]
(A) 38.71 g (B) 38.71 mg
(C) 42 g (D) 42 mg
Solution: (T = [pic]
or 9.3 = [pic]
( W = 161.29
( Ice separated = 200 – 161.29 = 38.71 g
Hence, (A) is correct.
Illustration 19. The freezing point of aqueous solution contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is: ([pic]= 1.86 K molality–1)
(A) 290.2 K (B) 285.5 K
(C) 269.93 K (D) 250 K
Solution: (Tf = (Tf for glucose + (Tf for KCl + (Tf for urea
= [pic] + [pic] + [pic] = 3.069°
( Freezing point = 273 – 3.069 = 269 .93 K
Hence, (C) is correct.
Illustration 20. The molal freezing point constant for water is 1.86 K.molarity–1. If 34.2 g of cane sugar (C12H22O11) are dissolved in 1000g of water, the solution will freeze at
(A) –1.86°C (B) 1.86°C
(C) –3.92°C (D) 2.42°C
Solution: (Tf = [pic] = 1.86°
( Tf = 0 – 1.86 = –1.86° C
Hence, (A) is correct.
Illustration 21. An aqueous solution containing 0.25 g of a solute dissolved in 20g of water freezes at 0.4°C. Calculate the molar mass of the solute. Kf = 1.86 K kg mol-1.
Solution: (Tf = Kf ( m
[pic]= 58. 13g/mol
Illustration 22. Two solutions of non-volatile solutes A and B are prepared. The molar mass ratio,[pic]. Both are prepared as 5% solutions by weight in water. Calculate the ratio of the freezing point depressions, [pic]of the solutions. If the two solutions are mixed to prepared two new solutions S1 and S2, the mixing ratio being 2 : 3 and 3 : 2 by volume for S1 and S2 respectively what would be the ratio [pic]
Solution: [pic]
( [pic]
Further for solution S1, molality = [pic]
For solution S2, molality = [pic]
[pic]
[pic]
Illustration 23. A very small amount of a non volatile solute (that does not dissociate) is dissolved in 56.8 cc of benzene (density 0.889 g/cc). At room temperature the vapour pressure of this solution is 98.88 mm Hg while that of benzene is 100 mmHg. Find the molality of this solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene, what is the value of molal freezing point depression constant of benzene.
Solution: From Raoult’s law
P1 = x1P10
98.88 = x1 (100
x1= 0.988
x2 = mole fraction of solute = 0.0112
weight of solvent = 0.988 ( 78
= 77.116 g
molality of solution = [pic]
= 0.1452 mol/kg
We know that Kf = [pic]= 5.028 K kg/mol
Illustration 24. The freezing point of 0.02 mole fraction of acetic acid in benzene is 277.4K. Acetic acid exists partly as dimmer. Calculate the equilibrium constant for dimerisation. Freezing point of benzene is 278.4 K and heat of fusion of benzene is 10.042 Kj/mol. Assume molarity and molality same.
Solution: [pic]
For acetic acid :
[pic]
C O
(1-() ((/2)
i = 1-( + (/2 = (1-(/2)
[pic]i = (1-(/2)……………(ii)
[pic]
[pic]
[pic]
Given, mol fraction of acitic acid [pic]
Mole fraction of benzene =[pic]= 0.98
[pic]
Molality [pic]
=[pic]
= 0.262 m.
From equation (i) & (ii)
1= 50 × 0.0262 × [pic]
(( = 0.48
Kc =[pic]
[pic]
Illustration 25. The freezing point of a 0.08 molal aqueous solution of NaHSO4 is (0.372(C. Calculate the dissociation constant for the reaction.
HSO4( [pic] H+ + SO4(2
Assuming no hydrolysis (Kf of water = 1.86 k kg/mole)
Solution: NaHSO4 (( Na+ + HSO4(
0.08 0 0
(0.08) (0.08)
Suppose the degree of dissociation of HSO4( is (
HSO4( [pic] H+ + SO4(2
Eqm. C(1-() C( C(
( [H+] = 0.08 (
[SO4-2] = 0.08 (
[HSO4(] = 0.08 (1-()
Total ions present = 0.08 + 0.08(1(() + 0.08 ( + 0.08 (
= 0.16 + 0.08 (
= [pic] = 2 + (
(Tf = kf ( m ( i
0.372 = ((2 + () ( 1.86 ( 0.08
( ( = 0.5.
Exercise 18.
If for a particular solution, (Tf = 0.73°, Kf = 5.028 K Kg/mole, what is the molality of the solution.
Exercise 19.
1 kg of an aqueous solution of Sucrose is cooled and maintained at –4°C. How much ice will be separated out if the molality of the solution is 0.75?
Kf (H2O) = 1.86 Kg mol–1K.
Exercise 20.
A complex is represented as CoCl3.xNH3. Its 0.1 m solution in aqueous solution shows (Tf = 0.558°, Kf(H2O) = 1.86 mol–1 K and assume 100% ionization and
co–ordination number of Co(III) is six . What is the complex?
Exercise 21.
30 ml of CH3OH (d = 0.7980 gm cm–3) and 70 ml of H2O (d = 0.9984 gm cm–3) are mixed at 25°C to form a solution of density 0.9575 gm cm–3. Calculate the freezing point of the solution. Kf (H2O) is 1.86 Kg mol–1 K. Also calculate its molarity.
Exercise 22.
In a cold climate, water gets frozen causing damage to the radiator of a car. Ethylene glycol is used as an anti-freezing agent. Calculate the amount of ethylene glycol to be added to 4 kg of water to prevent it from freezing at – 6°C.
Kf for water = 1.86 kg mol K–1.
Exercise 23.
A solution contains 0.1 mol of acetamide in 1 Lit of glacial acetic acid. When the solution is cooled, the first crystal that appeared at the freezing point contains the molecules of
(A) acetamide only (B) Acetic acid only
(C) both acetamide and acetic acid (D) None
ABNORMAL MOLECULAR WEIGHT AND VAN'T HOFF FACTOR
Since colligative properties depend upon the number of particles of the solute, in some cases where the solute associates or dissociates in solution, abnormal results for molecular masses are obtained.
• Van't Hoff Factor: Van't Hoff, in order to account for all abnormal cases introduced a factor i known as the Van't Hoff factor, such that
[pic]
[pic]
i > 1 for dissociation of solute in solution
i < 1 for association of solute.
i = 1 when solute neither associate non dissociate.
Illustration 26. The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is
(A) 60% (B) 83.5%
(C) 46.7% (D) 60.23%
Solution: i for AgNO3 = [pic] = 1 + (
( = [pic] – 1 = 0.835 = 83.5 %
Hence, (B) is correct.
Illustration 27. 20 g of a binary electrolyte (mol.wt. = 100) are dissolved in 500 g of water. The freezing of the solution is –0.74°C, Kf = 1.86 K. molality–1. The degree of ionisation of the electrolyte is
(A) 50% (B) 75%
(C) 100% (D) 0
Solution: (Tf = [pic]
( 0.74 = [pic] ( M = 100
Now, [pic]= 1 + ( = [pic]
So, ( = 0
Hence, (D) is correct.
Illustration 28. Acetic acid (CH3COOH) associates in benzene to form double molecules. 1.65 g of acetic acid when dissolved in 100g of benzene raised the boiling point by 0.36°C. Calculate the Van't Hoff Factor and the degree of association of acetic acid in benzene (Molal elevation constant of benzene is 2.57).
Solution: Normal molar mass of acetic acid = 60
Observed molar mass of acetic acid
[pic]=[pic]
Van't Hoff Factor = [pic]
= 0.508
0.508 = [pic] = 1 – (/2
(/2 = 1– 0.508 = 0.492
( ( = 2 ( 0.492 = 0.984
Thus acetic acid is 98.3% associated in benzene.
Illustration 29. River water is found to contain 11.7% NaCl, 9.5% MgCl2, and 8.4%. NaHCO3 by weight of solution. Calculate its normal boiling point assuming 90% ionization of NaCl,70% ionization of MgCl2 and 50% ionization of NaHCO3
(Kb for water = 0.52 k kg mol(1)
Solution: nNaCl = [pic] = 0.2
[pic]= [pic] = 0.1
[pic]= [pic] = 0.1
iNaCl = 1+ ( = 1+ 0.9 = 1.9
[pic]= 1 + 2( = 1+ 0.7 ( 2 = 2.4
[pic]= 1+ 2( = 1+ 0.5 ( 2 = 2.0
Weight of solvent = 100 – (11.7 + 9.5 + 8.4) = 70.4 g
(Tb = [pic]
= [pic]
= 6.0657°C
( Boiling point of solution = 100 + 5.94 = 106.057°C
Illustration 30. Calculate the Van’t Hoff factor when 0.1 mole [pic] is dissolved in 1 L of water. Degree of dissociation of [pic] is 0.8 and its degree of hydrolysis is 0.1.
Solution: [pic]
[pic]
[pic]
i = 1.8 + 0.8[pic]
= 1.8 + 0.08
i = 1.88
Illustration 31. Calculate the molecular weight of cellulose acetate if its 0.5% (wt./vol) solution in acetone (sp. gr. = 0.9) shows an osmotic rise of 23 mm against pure acetone
at 27°C.
Solution: 0.5% (wt. / vol) solution means 0.5 gm of cellulose acetate is dissolved in 100 ml solution.
Osmotic pressure = 23 mm of pure acetone
( = 2.3 Cm of pure acetone
= [pic] cm of Hg (13.6 is specific gravity of Hg)
= 0.1522 cm of Hg
( = [pic] atm = 0.002 atm
Let the molecular weight of the cellulose acetate be M Here, number of mole of cellulose acetate (n) [pic]
Volume = v = 100 ml = 0.1 lit
R = 0.082 Lit atm mol–1 K–1, T = (27 + 273) = 300 K
Osmotic pressure (() = [pic] (RT
( 0.002 = [pic]( 0.0821 ( 300
( m = 61575
Illustration 32. A milimolar solution of potassium ferricyanide is 70% dissociated at 27°C. Find out the osmotic pressure of the solution.
Solution: Given, concentration of solution = C = [pic]M = 10–3M
K3[Fe(CN)6] (( 3K+ + [Fe(CN)6]–3
Initially moles 1 0 0
Moles after dissociation 1-( 3( (
Where ( = degree of dissociation = 70% =[pic] = 0.7
Total moles of after dissociation = (1–() + 3( + ( = 1+ 3(
( i = [pic]
i = 1+3( = 1+(3( 0.7) = 3.1
Since osmotic pressure
( = iCRT
( ( = 3.1 ( 10–3 ( 0.0821 ( 300
= 0.31 ( 3 ( 0.0821 atm = 0.07635 atm
Illustration 33. A storage battery contains a solution of H2SO4 38% by weight. At this concentration, Van’t Hoff factor is 2.50. At what temperature will the battery contents freeze? (Kf = 1.86° mol–1 kg)
Solution: (Tf = [pic] = 11.633
i = [pic]
([pic] = 2.50 ( 11.633
= 29.05
Thus F.P. = 273 – 29.08 = 243.92° K = – 29.08°C
Illustration 34. The molar heat of vaporization of water at 100(C is 40.585 kJ/mol. At what temperature will a solution containing 5.60 g of Al2(SO4)3 per 1000 g of water boil? [Assuming the degree of ionization of salt to be 1]
Solution: Kb = [pic]
= [pic]
= 0.513 K kg/mol
Now mol weight of Al2(SO4)3 = 342
Al2(SO4)3 ( 2Al+3 + 3SO4(2
1-( 2( 3(
i = 1 ( ( + 2( + 3(
= 1 + 4(
i = 5 ([pic])
(Tb = i ( Kb ( m
= 5 ( 5.13 ( [pic]
= 0.042
( Boiling point of solution = 100 + 0.042 = 100.042(C
Illustration 35. Acetic acid (CH3COOH) associates in benzene to form double molecules. 1.65 g of acetic acid when dissolved in 100g of benzene raised the boiling point by 0.36°C. Calculate the Van't Hoff Factor and the degree of association of acetic acid in benzene (Molal elevation constant of benzene is 2.57).
Solution: Normal molar mass of acetic acid = 60
Observed molar mass of acetic acid.
[pic]
[pic]
Van't Hoff Factor = [pic] = 0.508
0.508 = 1–( (1–1/n) = (1–(/2)
(/2 = 1– 0.508 = 0.492 ( ( = 2 ( 0.492 = 0.984
Thus acetic acid is 98.4% associated in benzene.
Illustration 36. A 0.5 percent aqueous solution of Potassium Chloride was found to freeze at 0.24°C. Calculate the Van't Hoff factor and degree of dissociation of the solute at this concentration (Kf for water is 1.86).
Solution: (Tf = [pic]
0.24 = [pic]
i = [pic] = 1.923; i = 1+(n–1)(
1.923 = 1+ (2–1)( ( 1.923 = 1+ (
( = 1.923 – 1 = 0.923
or ( = 0.923 or 92.3%
Illustration 37. If the solution of mercuric cyanide of strength 3 g/l has an osmotic pressure 0.3092 ( 105 Nm–2 at 298 K, what is the apparent molecular weight and degree of dissociation of Hg(CN)2? (Hg = 200.61, C = 12, N = 14)
Solution: Let the apparent molecular weight = mo
Osmotic pressure = ( = [pic]( RT =[pic]
or, 0.3092 ( 105 = [pic]( 8.314 ( 298
( mo = 240.2 ( 10–3 kg
Since, [pic]
( Hg (CN)2 [pic] Hg++ + 2CN–
1–( ( 2(
( Number of particles after dissociation = (1–() + ( + 2( = 1+2(
i.e., [pic]=[pic]
( ( = 0.02457 or 2.4567 %
Illustration 38. If the apparent degree of ionization of KCl (KCl = 74.5 gm mol–1) in water at 290K is 0.86. Calculate the mass of KCl which must be made up to 1 dm3 of aqueous solution so as to produce the same osmotic pressure as the 4.0% solution of glucose at that temperature.,
Solution: Due to ionization of KCl
KCl ((( K+ + Cl–
Initial moles 1 0 0
Moles after dissociation 1–( ( (
( i = [pic] = 1+ (
From question, degree of ionization = ( = 0.86
( i = 1+ 0.86 = 1.86
For Osmotic pressure of glucose,
For 4% glucose solution, Weight of glucose = 40 gm,
Volume of solution = V = 1L = 1dm3
Molecular weight of glucose C6H12O6= m = 180
( (glucose = n ( [pic]( RT, where (glucose = osmotic pressure of glucose
= [pic]( [pic]( R ( T = [pic]( [pic]( RT
Similarly, (KCl = [pic]= 1.86 ( n ( [pic](RT
= 1.86 ( [pic]( [pic]( RT = 1.86 ( [pic]( [pic](RT
As both solutions are isotonic,
( (KCl = (glucose
1.86 ( [pic]( RT = [pic] ( RT
( W1 = [pic]
= 8.9 gm
Illustration 39. The ratio of the value of any colligative property for KCl solution to that for sugar is nearly ----- times.
(A) 1 (B) 0.5
(C) 2 (D) 2.5
Solution: iKCl= [pic]
= [pic] = 2
isugar = 1
Hence, (C) is correct.
Exercise 24.
An aqueous 0.01m solution of K3[Fe(CN)6] freezes at – 0.062°C. What is the apparent percentage of dissociation? (Kf of H2O = 1.86oC/m).
Exercise 25.
2.0 gm of benzoic acid dissolved in 25 gm of benzene shows a depression of freezing point equal to 1.62 K. Molal depression constant (Kf) of benzene is
4.9 kg mol–1 K. What is the percentage association of the acid?
Exercise 26.
A weak electrolyte, AB is 5% dissociated in aqueous solution of AB having molality 0.1 m. Freezing point of the solution will be (Kf for water 1.86 (C /molal)
Exercise 27.
The freezing point of an aqueous solution of KCN containing 0.1892 mol kg–1 of solvent was found to be – 0.704°C. On adding 0.095 mol of Hg(CN)2, the freezing point of the solution was found to be – 0.530°C. If the complex – formation takes place according to the following equation.
Hg(CN)2 + nKCN [pic] Kn [Hg(CN)n+2]
What is the formula of the complex? Kf(H2O) is 1.86 kg mol–1 K.
Exercise 28.
The freezing point of a solution of acetic acid (mole fraction is 0.02) in benzene 277.4 K. Acetic acid exists partly as a dimer 2A [pic] A2. Calculate equilibrium constant for dimerisation. Freezing point of benzene is 278.4 k and (Kf for benzene is 5).
Exercise 29.
The freezing point depression of a 0.109 M aq. Solution of formic acid is –0.21°C. Calculate the equilibrium constant for the reaction,
HCOOH (aq) [pic] H+ (aq) + HCOO– (aq)
Kf for water = 1.86 kg mol–1 K.
Exercise 30.
Which of the following aqueous solutions has the highest boiling point?
(A) 0.1 – M KNO3 (B) 0.1– M Na3PO4
(C) 0.1 –M BaCl2 (D) 0.1 – M K2SO4
Exercise 31.
Among the following, the solution which shows the highest osmotic pressure is
(A) 0.05 – M NaCl (B) 0.10 – M BaCl2
(C) 0.05 – M FeCl3 (D) 0.05 – M Na2SO4
Exercise 32.
When mercuric iodide is added to an aqueous solution of KI, the
(A) Freezing point is raised (B) Freezing point is lowered
(C) Boiling point does not change (D) Freezing point does not change
.
Exercise 33.
In a solute undergoes dimerisation and trimerisation, the minimum values of the Vant Hoff factors are
(A) 0.5 and 1.50 (B) 1.5 and 1.33
(C) 0.5 and 0.33 (D) 0.25 and 0.67
Exercise 34.
Which of the following pairs of solutions can be expected to be isotonic at the same temperature?
(A) 0.1 - M urea and 0.1 - M NaCl (B) 0.1 –M urea and 0.2 - M MgCl2
(C) 0.1 - M NaCl and 0.1 M Na2SO4 (D) 0.1-M Ca(NO3)2 and 0.1-M Na2SO4
Exercise 35.
Study the following figure, and choose the correct option
[pic]
(A) There will be no movement of any solution across the membrane
(B) BaCl2 will flow towards the NaCl solution
(C) The osmotic pressure of 0.1 - M NaCl is higher than the osmotic pressure of
0.05 - M BaCl2, assuming complete dissociation of electrolyte.
(D) NaCl will flow towards the BaCl2 solution.
`
Exercise 36.
In which of the following pairs of solution will the values of the Van't Hoff factor be the same?
(A) 0.05 - M K4[Fe(CN)6] and 0.1 M FeSO4
(B) 0.05 - M K4[Fe(CN)6] and 0.05 - M FeSO4.(NH4)2SO4.6H2O
(C) 0.2 M NaCl and 0.1 M BaCl2
(D) 0.05 -M FeSO4(NH4)2SO4.6H2O and 0.05 - M Na3PO4
Exercise 37.
The Van't Hoff factor for an electrolyte which undergoes dissociation and association in solvent are respectively
(A) Greater than one and greater than one
(B) Less than one and greater than one
(C) Less than one and less than one
(D) Greater than one and less than one
Some other properties of liquids
Many familiar and observable properties of liquids can be explained by the intermolecular forces. Some liquids such as water or gasoline flow easily when poured, whereas other such as motor oil or maple syrup, flow sluggishly.
The measure of a liquids resistance to flow is called its viscosity. Viscosity is related to ease with which individual molecules move around in the liquid and thus to the intermolecular forces present. Substance with small nonpolar molecules, such as pentane and benzene, experience only weak intermolecular forces and have relatively low viscosities whereas more polar substances such as glycerol (C3H5(OH)3) experience stronger intermolecular forces and so have higher viscosities.
Another familiar property of liquids is surface tension, the resistance of a liquid to spread out and increase its surface area. Surface tension is caused by the difference in intermolecular forces experienced by molecules at the surface of a liquid and those experienced by molecules in the interior. Molecules at the surface feel attractive forces on single side only and are thus drawn in toward the liquid, white molecules in the interior are surrounded and are drawn equally in all directions. Surface tension, like viscosity is generally higher in liquids that have stronger intermolecular forces. Both properties are also temperature dependent because molecules have more kinetic energy to counteract the attractive forces holding them together.
Phase changes
1. Fusion (melting) Solid ((((( Liquid
2. Freezing Liquid ((((( Solid
3. Vaporization Liquid ((((( Gas/vapour
4. Condensation Gas ((((( liquid
5. Sublimation Solid ((((( Gas
6. Deposition[Sublimation] Gas ((((( Solid
Every phase change has associated with it a free energy change, (G. This free energy change (G is made up of two contributions on enthalpy part (H and a temperature – dependent entropy part T(S, according to equation [pic]
The enthalpy part is the heat flow associated with making or breaking the intermolecular attractions that hold liquids and solids together, while the entropy part is associated with the change in disorder or randomness between the various phases.
|[pic] |
The Clausius-clayperon equation
The molecules in a liquid are in constant motion but at a variety of speeds depending on the amount of energy (K.E.) they have. The higher the temperature and lower the boiling point of the substance, the greater the fraction of molecules in the sample that have sufficient kinetic energy to break free from the surface of the liquid and escape into the vapour.
|[pic] |
Only the faster moving molecules have sufficient kinetic energy to escape from the liquid and enter the vapour.
The vapour pressure of a liquid rises with temperature in a non linear way. A linear relationship is found, however, when the logarithm of the vapour pressure, in vapour is plotted against the inverse of the Kelvin temperature, 1/T.
[pic]
y = mx + c
Where (Hvap is the heat of vaporization of liquid, R is the gas constant and c is a constant characteristic of each specific substance.
|[pic] |[pic] |
The clauses clausuis equation makes it possible to calculate the heat of vaporization of a liquid by measuring its vapour pressure at several temperature and than plotting to obtain the slope of the line
It can be written in this way too.
[pic]
ANSWER TO EXERCISES
Exercise 1:
8.02%
Exercise 2:
(i) 7.73m, 3.865 m
(ii) [pic]= 0.065
Exercise 3:
[pic]
Exercise 4:
0.93
Exercise 5:
(D)
Exercise 6:
(C)
Exercise 7:
0.33
Exercise 8.
53.75
Exercise 9:
(i) Negative deviation
(ii) Positive deviation
Exercise 10:
Non ideal and shows positive deviation.
Exercise 11:
0.882, 0.118, 20.42 mm Hg, 112.014 mm Hg, 0.154, YA = 0.5, YB = 0.5
Exercise 12:
12.1 g, 269.9 K
Exercise 13:
(A)
Exercise 14:
59.99
Exercise 15:
(D)
Exercise 16 :
1.2 ( 1021 molecule
Exercise 17:
C44H88O44.
Exercise 18:
0.1452
Exercise 19:
518.31 gm
Exercise 20:
[Co(NH3)5Cl]Cl2.
Exercise 21:
(19.91°C, 7.63 M
Exercise 22.
800 g
Exercise 23:
(B)
Exercise 24:
78%
Exercise 25.
98.3%
Exercise 26:
(0.1953(C
Exercise 27:
K2[Hg(CN)4]
Exercise 28:
3.39 kg mol–1
Exercise 29:
Ka = 1.46 ( 10–4
Exercise 30:
(B)
Exercise 31:
(B)
Exercise 32:
(A)
Exercise 33:
(C)
Exercise 34:
(D)
Exercise 35:
(C)
Exercise 36:
(B)
Exercise 37:
(D)
MISCELLANEOUS EXERCISES
Exercise 1: Why a cook cries less on cutting onion if cooled in fridge rather than cutting onion at room temperature?
Exercise 2: A peeled egg when dipped in water swells, while in saturated solution it shrinks.
Exercise 3: Semipermeable membrane of Cu2Fe(CN)6 is not used for studying osmosis in non aqueous solutions.
Exercise 4: Evaporation of liquid from palm shows cooling, why?
Exercise 5: Liquid ammonia bottle is cooled before opening the seal, Why?
Exercise 6: Explain the difference between osmotic pressure and vapour pressure of a solution.
Exercise 7: What type of deviation (positive or negative) from ideal behaviour will be shown by the solution of cyclohexane and ethanol? Give suitable reason.
Exercise 8: Differentiate between molarity (M and molality (m) of a solution.
Exercise 9: When is the value of Van’t Hoff factor more then one?
Exercise 10: Which of the electrolytes is most effective for the coagulation of Fe(OH)3 sol and why?
NaCl, Na2SO4, Na3PO4
ANSWERS TO MISCELLANEOUS EXERCISE
Exercise 1: Onion kept in fridge has low vapour pressure of its contents at low temperature
Exercise 2: In 1st case endo osmosis occurs because solution inside egg is concentrated one. In case 2nd exo osmosis occurs because solution outside egg is concentrated one.
Exercise 3: Cu2Fe(CN)6 is soluble in non aqueous solutions.
Exercise 4: Liquid takes up energy of evaporation by palm itself.
Exercise 5: Liquid NH3 has high vaporization tendency and thus, bottle filled with liquid NH3 develops high vapour pressure. If seal is opened, liquid NH3 will bump out of the bottle. To avoid bumping of liquid NH3, bottle is cooled to lower down the vapour pressure inside it.
Exercise 6: Vapour pressure of a liquid at a given temperature is the normal force for the liquid acting on unit area of that liquid.
While osmotic pressure of a liquid is the pressure of the solution which prevents the entry of solvent molecules to solution side when the two are separated by a semipermeable membrane.
Exercise 7: A mixture of cyclohexane and ethanol shows positive deviation. It is because in pure ethanol a very high fraction of the molecules are hydrogen bonded. On addition of cyclohexane, these molecules get in among the molecules of ethanol. This intermolecular insertion breaks the hydrogen bonding between ethanol molecules. This reduces intermolecular attraction between ethanol-ethanol molecules. Hence escaping tendency of the solution is more. Thus the mixture in question shows positive deviation.
Exercise 8: Molarity is defined as no of moles of soluble present in one litre solution while molality is defined as the no. of moles of solute present in one kg of solvent
[pic], m = [pic]
Exercise 9: When the solute molecules undergo dissociation in the solution, Van’t Hoff factor is more then one.
Exercise 10: Na3PO4, because coagulating power decrease in the order
[pic]
SOLVED PROBLEMS
Subjective:
Board type Questions
Prob 1. A solution is prepared by dissolving 43 gm of naphthalene in 117 gm of benzene. Calculate the mole fractions of the two components of the solution.
Sol. Number of moles, [pic]
Moles of benzene, [pic]
Moles of naphthalene, [pic]
Mole fraction of naphthalene, [pic]
Mole fraction of benzene = 1 ( X2 = 1 ( 0.185 = 0.815
Prob 2. 1.2 gm of a non-volatile substance was dissolved in 100 gm of acetone at 20(C. The vapour pressure of the solution was found to be 182.5 torr. Calculate the molar mass of the substance (vapour pressure of acetone at 20(C is 185.0 torr)
Sol. From equation [pic]
[pic] or M1 = 42.92 gm/mol.
Prob 3. A solution containing 8.6 gm per dm3 of urea (molar mass = 60 gm/mole) was found to be isotonic with a 5 percent solution of an organic non – volatile solute. Calculate the molar mass of the latter.
Sol. According to Van’t Hoff theory, isotonic solutions have the same osmotic pressure at the same temperature and the same molar concentration.
Molar concentration of urea solution = [pic]
Let M2 be the molar mass of unknown solute.
Molar concentration of unknown solution = [pic]
Since both are isotonic,
[pic]or M2 = 348.8 gm/mol
Prob 4. Acetic acid associate in benzene to form dimer 1.65 gm of acetic acid when dissolved in 100 gm of benzene raised the boiling point by 0.36(C. Calculate the Van’t Hoff factor and the degree of association of acetic acid in benzene. (Kb = 2.57 K Kg mol(1)
Sol. Normal molar mass of acetic acid = 60 gm/mol
Observed molar mass of acetic acid, M2 = [pic]
= [pic]
[pic]
Since acetic acid associated to form double molecules. Hence
[pic]
The number of effective moles = 1 ( ( + (/2
= 1 + (/2
[pic]
0.508=1+ (/2 or ( = 0.982
Prob 5. At 25(C, the osmotic pressure of human blood due to the presence of various solutes in the blood is 7.65 atm. Assuming that molarity equals molality, calculate the freezing point of blood. Kf = 1.86 K Kg mol(1.
Sol. According to the Van’t Hoff eqation, ( = CRT
[pic]
i.e. the concentration of various solutes present in blood = 0.313 mol dm(3.
In the present concentration, as given, molality of solution = molarity of solution
m = C = 0.313 mol kg(1
[pic]
[pic]
[pic] = (0.582 (C
IIT level Questions
Prob 6. The density of 2.0 M solution of acetic acid in water is 1.02 g ml(1. Calculate the mole fraction of acetic acid.
Sol. Mass of acetic acid in 2.0 M acetic acid solution = [pic]
Mass of the solution = V ( D = 1 ( 1.02 = 1.02 kg
Mass of solvent = 1020 ( 120 = 900 kg
Mole fraction of acetic acid = [pic]
= 0.038
Prob 7. Liquid A and B form an ideal solution obeying Raoult’s law. At 50(C, the total vapour pressure of a solution containing 1 mole of A and 2 mole of B is 300 torr. When 1 more mole of A is added to the solution, the vapour pressure increases to 400 torr. Calculate the vapour peessure of pure components.
Sol. [pic]
[pic]
given, [pic]
[pic]
[pic]
Now from these equations
[pic], [pic]
Prob 8. A solution of A and B with 30 mole percent of A is in equilibrium with its vapour which contains 60 mole percent of A. Assuming ideality of the solution and the vapour. Calculate the ratio of the vapour pressure of pure A to that of pure B.
Sol. Given, xA = 0.3 xB = 0.7 in liquid or in solution.
In the vapour phase, XA vapour = 0.6 and xB, vapour = 0.4
Using Dalton’s law of partial pressures and Raoult’s law,
[pic]
[pic]
[pic]
[pic]
Prob 9. What would be the vapour pressure of 0.5 molal solution of a non volatile solute in benzene at 30(C? The vapour pressure of pure benzene at 30(C is 119.6 torr.
Sol. [pic]
Given that, m = 0.5 mol, M1 = 87 gm/mol
[pic]
Putting these values in above equation
[pic]
P1 = 114.94 torr
Prob 10. The molar heat of vaporization of water at 100(C is 40.585 KJ/mol. At what temperature will a solution containing 5.6 gm of glucose per 1000 gm of water boil?
Sol. [pic]
= [pic]
[pic]
Boiling point of solution = 373 K + 0.16 K = 373.16 K
= 100.16 (C
Prob 11. An aqueous solution of non – volatile solute boils at 100.17(C. At what temperature would it freeze? For water, Kb = 0.52 K Kg mol(1 and Kf = 1.86 K Kg mol(1
Sol. [pic]
[pic]
[pic]
[pic]
1.86 ( 0.327 = 0.608 K
Freezing point, Tf = [pic] = (0.608 (C
Prob 12. The complex compound K4[Fe(CN)6] is 45% dissociated in 0.1M aqueous solution of the complex at 27(C. What would be the osmotic pressure of the solution?
Sol. If the complex K4[Fe(CN)6] had not been dissociated, its normal osmotic pressure would have been given by that Van’t Hoff equation,
( = CRT = 0.1 ( 0.0821 ( 300
= 2.462 atm
The complex, however is dissociated in equation solution
[pic]
No. of moles after dissociation = 1 ( ( + 4( + ( = 1 + 4(
[pic]
( = 0.45 hence
[pic]
Prob 13. The formula for low molecular weight starch is (C6H10O5)n where n averages
2.00 ( 102. When 0.798 gm of starch is dissolved in 100 ml of water solution, what is the osmotic pressure at 25(C?
Sol. No. of mole of starch = [pic]
[As M of starch = 162 ( 2 ( 102 = 32400]
The molarity of the solution is = [pic]
Osmotic pressure ( = CRT
= [pic]
= 59.69 ( 10(4 atm
Prob 14. Phenol associates in benzene to a certain extent to form a dimer. A solution containing 20 ( 10(3 kg of phenol in 1.0 kg of benzene has its f. pt depressed by 0.69 K. Calculate the fraction of phenol dimerised. Kf for C6H6 = 5.120 mol(1kg.
Sol. We know that, (T = [pic]
[pic]
mexp = 148.41
mnormal = 94
[pic]
[pic]
or ( = 0.734 or 73.4 %
Prob 15. Calculate the amount of ice that will separate out on cooling a solution containing
50 gm of ethylene glycol in 200 gm water to (9.3(C.
Tf = (9.3(C. (kf for water = 1.86 K mol(1 kg)
Sol. Given (T = 9.3, W2 = 50 gm, kf = 1.86 K mol(1 kg
Mglycol = 62
[pic]
[pic]
( W = 161.29 grams
Wwater = 161.29 gm
Thus weight of ice separated = 200 ( 161.29
= 38.71 gm
Objective:
Prob 1. The freezing point of aqueous solution containing 5% by mass urea, 10% by mass
KCl and 10% by mass of glucose, is
(A) 290.2 K (B) 285.5 K
(C) 269.93 K (D) 264.67 K
Sol. [pic]
[pic]
[pic]
= 8.327 K
( Freezing point = 264.67 Ks
( (D)
Prob 2. The Van’t’ Hoff factor for 0.1 M La(NO3)3 solution is found to be 2.74 the percentage dissociation of the salt is
(A) 85% (B) 58%
(C) 65.8% (D) 56.8%
Sol. [pic]
i = 1 + 3(
2.74 = 1 + 3(
so ( = 0.58
% dissociation = 58%
( (B)
Prob 3. The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in contact with equimolar solution of benzene and toluene is
(A) 0.5 (B) 0.6
(C) 0.27 (D) 0.73
Sol. XB = XT = 0.5
PB = 0.5 ( 160 = 80 Torr
PT = 0.5 ( 60 = 30 Torr
Ptotal = 80 + 30 = 110 Torr
Hence mole fraction of toluene = [pic]
( (C)
Prob 4. 1.0 molal aqueous solution of an electrolyte X3Y2 is 25% ionized. The boiling point of the solution is (Kb for H2O = 0.52 K kg/mol)
(A) 375.5 K (B) 374.04 K
(C) 377.12 K (D) 373.25 K
Sol. [pic]
i = 1 + 4( = 1 + 4 ( 0.25 = 2
[pic]
Boiling point of the solution = 373 + 1.04 = 374.04 K
( (B)
Prob 5. A 0.2 molal aqueous solution of a weak acid is 20% ionized. The freezing point of this solution is (Kf = 1.86 K. kg mol(1 for water)
(A) (0.45 (C (B) ( 0.9 (C
(C) ( 0.31 (C (D) (0.53 (C
Sol. [pic]
i = 1 + ( = 1 + 0.2
(Tf = iKf.m
= 1.2 ( 1.86 ( 0.2 = 0.45
Hence freezing point of this solution = 0 ( 0.45 = ( 0.45
Prob 6. The fundamental cause of all colligative properties is
(A) higher entropy of the solution relative to that of pure solvent
(B) lower entropy of the solution relative to that of pure solvent
(C) higher enthalpy of the solution relative to that of pure solvent
(D) lower enthalpy of the solution relative to that of pure solvent
Sol. ( (A)
Prob 7. A mixture of two immiscible liquids nitrobenzene & water boiling at 99(C has a partial vapour pressure of water 733 mm & that of nitrobenzene 27 mm. The ratio of the weights of nitrobenzene to the water in the distillate is
(A) 2:1 (B) 4:1
(C) 3:1 (D) 1:4
Sol. [pic]
Where XA is mole fraction in gaseous phase.
[pic]
& [pic]
[pic]
( (B)
Prob 8. The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is
(A) 60% (B) 83.5%
(C)46.7 % (D) 60.23%
Sol. i = [pic]
Hence ( = [pic]
Hence % dissociation = 83.5%
( (B)
Prob 9. Dry air was passed successively through a solution of 5 gm of a solute in 180 gm of water and then through pure water. The loss in weight of solution was 2.50 gm and that of pure solvent is 0.04 gm. The molecular weight of the solute is
(A) 31.25 (B) 3.125
(C) 312.5 (D) none of these
Sol. P0 ( Ps ( loss in weight of water chamber
Ps ( loss in weight of solution chamber
[pic]
[pic]
hence m = 31.25
Prob 10. The boiling point of an aqueous solution of a non – volatile solute is 100.15(C. What is the freezing point of an aqueous obtained by diluting the above solution with an equal volume of water? The values of Kb and Kf for water are 0.512(C kg/m(1 and 1.86 (C kg/mol respectively
(A) (0.544 (C (B) (0.512 (C
(C) (0.272 (C (D) (1.86 (C
Sol. For same solution, [pic]
Or [pic]
[pic]
Now on diluting the solution to double the volume
[pic]
( [pic]
( Freezing point = (0.2725 (C
True/False
Prob 11. Sensitivity of molecular weight determination decreases with increase in molecular weight by ebullioscopy method.
Sol. True
Prob 12. Relative lowering of vapour pressure is equal to the mole fraction of the solute in solution.
Sol. True
Prob 13. The freezing point of the solution is always lower than that of pure solvent.
Sol. True
Prob 14. Every solution behaves as ideal solution.
Sol. False
Prob 15. R.B.C. on keeping in hypertonic solutions shrinks down.
Sol. True
Fill in the Blanks
Prob 16. Molal elevation constant is a characteristic constant for a given ………….
Sol. Solvent
Prob 17. Osmosis occurs from ………. osmotic pressure solution to ……….. osmotic pressure solution.
Sol. Lower, higher
Prob 18. When an azeotropic mixture is distilled, its composition remains ………….
Sol. Unchanged
Prob 19. Higher is the amount of solute in a solution ………….. is its vapour pressure.
Sol. Lower
Prob 20. Depression in f.pt. is …………. pronounced if camphor is used as a solvent in place of water for same amount of solute and solvent.
Sol. More
ASSIGNMENT PROBLEMS
Subjective:
Level – O
1. The Henry law constant for oxygen dissolved in water is [pic] at 25(C. The partial pressure of oxygen in air is 0.2 atm. Under ordinary atmospheric conditions, calculate the concentration (in moles/litre) of dissolved oxygen in water at equilibrium with air at 25(C.
2. Explain with suitable examples in each case why the molar masses of some substances determined with the help of colligative properties are (i) higher (ii) lower then actual values.
3. One litre aqueous solution of sucrose (molar mass = 342) weighing 1015 gm is found to record an osmotic pressure of 4.82 atm at 293 K. What is the molality of the sucrose solution? (R = 0.0821 atm mol(1 k(1)
4. With the help of suitable diagrams, illustrate the two types of non – ideal solutions.
5. An aqueous solution freezes at 272.4 K, while pure water freezes at 273.0 K. Determine
(i) molality of the solution
(ii) boiling point of the solution
Given Kf = 1.86 k Kg mol(1, Kb = 0.512 K kg mol(1
6. Illustrate elevation in boiling point with the help of vapour pressure temperature curve of a solution. Show that elevation in boiling point is a colligative property.
7. What is osmotic pressure and how it is related to the molecular mass of a non – volatile substance.
8. (a) Carbon tetrachloride and water are immiscible while ethyl alcohol and water are miscible in all proportions. Correlate the above behaviour with structural nature of molecules of these compounds.
(b) The vapour pressure of pure water at 373K is 760 mm Hg while that of a dilute aqueous solution of glucose, C6H12O6 is 750 mm Hg at the same temperature. Calculate the mole fraction of the solute and molality of the solution.
9. What is deliquescence? What type of substances generally show deliquescence?
10. 2 gm of C6H5COOH dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K Kg mol(1. What is the percentage association of acid if it exists as dimer in solution?
11. Find the molality of a solution containing a non – volatile solute if the vapour pressure is 2% below the vapour pressure of pure water.
12. An aqueous solution of 2% (wt/wt) non – volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute?
13. An aqueous solution of glucose containing 12 gm in 100 gm of water was found to boil at 100.34(C. Calculate Kb for water in[pic].
14. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The freezing point of pure water is
273.15 K.
15. What is the depression in freezing point of a solution of non – electrolyte if elevation in boiling point is 0.13 K, Kb = 0.52[pic]; Kf = 1.86[pic].
Level – I
1. At 10(C, the osmotic pressure of urea solution is 500 mm of Hg. The solution is diluted and the temperature is raised to 25(C, when the osmotic pressure is found to be 105.3 mm. Determine the extent of dilution.
2. The vapour pressure of an aqueous solution of glucose is 750 mm of Hg at 373 K. Calculate the molality and mole fraction of solute.
3. How much ethyl alcohol must be added to 1.00 litre of water so that the solution will freeze at 14(F?
(Kf for water = 1.86 (C kg/mol)
4. 1000 gm of 1 m sucrose solution in water is cooled to (3.534(C. What weight of ice would be separated out at this temperature?
[pic]
5. A 1% (wt/vol) KCl solution is ionized to the extent of 82%. What would be its osmotic pressure at 18(C?
6. Calculate the osmotic pressure of 20% (wt/vol) anhydrous Ca(NO3)2 solution at 0(C assuming 100% ionization.
7. 17.4% (wt/vol) K2SO4 solution at 27(C is isotonic to 5.85% (wt/vol) NaCl solution at 27(C. If NaCl is 100% ionized, what is % ionization of K2SO4 in aqueous solution.
8. An aqueous solution of liquid ‘X’ (mol. Wt. 56) 28% by weight has a vapour pressure 150 mm of Hg. Find the vapour pressure of ‘X’ if vapour pressure of water is 155 mm of Hg.
9. Calculate the osmotic pressure at 17(C of an aqueous solution containing 1.75 gm of sucrose per 150 ml solution.
10. A mixture of two immiscible liquids nitrobenzene and water boiling at 99(C has a partial vapour pressure of water 733 mm and that of nitrobenzene 27 mm. Calculate the ratio of weights of nitrobenzene to the water in distillate.
Level – II
1. Calculate O.P of a solution obtained by mixing 100 ml of 3.4% solution (wt/vol) of urea and 100 ml of 1.6% solution (wt/vol) of cane sugar at 20(C.
2. A beaker containing 20 gm sugar in 100 gm water and another containing 10 gm sugar in 10 gm water are placed under a bell – jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to other?
3. What will be osmotic pressure of 0.1 M monobasic acid? Its pH is 2 at 25(C.
4. Vapour pressure of a saturated solution of a sparingly soluble salt, A2B2, is 31.8 mm of
Hg at 40(C. If vapour pressure of pure water is 31.9 mm of Hg at 40(C. Calculate Ksp of A2B2 at 40(C.
5. A complex is represented as[pic]. It is 0.1 molal solution in aqueous solution shows [pic] Kf for H2O is 1.86 K molality(1. Assuming 100% ionization of complex and
co-ordination number of Co as six, calculate formula of complex.
6. A storage battery contains a solution of H2SO4 38% by weight. At this concentration,
Van’t Hoff factor is 2.50. At what temperature will the battery contents freeze?
[pic]
7. 1 gm of mono basic acid in 100 gm of water lowers the freezing point by 0.168 (C. If 0.2 gm of same acid require 15.1 ml of N/10 alkali for complete neutralization, calculate degree of dissociation of acid[pic].
8. 0.025 M solution of monobasic acid had a freezing point of (0.06(C. Calculate Ka for the acid. Kf [pic] Assume molality equal to molarity?
9. Phenol associate in C6H6 to form dimer. A solution of 2 gm of phenol in 100 gm C6H6 has its f.pt lowered by 0.72 K. Kf for C6H6 is 5.12 K mol(1 kg. Calculate degree of association of phenol.
10. What approximate proportion by volume of water (d = 1 gm/mL) and ethylene glycol
(d = 1.2 gm/mL) must be mixed to ensure protection of an automobile radiator to cooling (10(C. [pic]
Objective:
Level – I
1. If in a solvent, n simple molecules of solute combine to form an associated molecule, x is the degree of association, the Van’t Hoff factor ‘i’ is equal to
(A) [pic] (B) [pic]
(C) [pic] (D) [pic]
2. At a constant temperature with increase in concentration of solute, the osmotic pressure of the solution
(A) increases (B) decreases
(C) remains constant (D) none of these
3. Consider following cases:
I: 2 M CH3COOH solution in benzene at 27(C where there is dimer formation to the extent of 100%.
II: 0.5 M KCl aq. solution at 27(C, which ionizes 100%, which is/are true statement (s)?
(A) both are isotonic (B) I is hypertonic
(C) II is hypertonic (D) none is correct
4. Which of the following statements is correct, if the intermolecular forces in liquids A, B and C are in the order A < B < C?
(A) B evaporates more readily than A (B) B evaporates less readily than C
(C) A and B evaporate at the same rate (D) A evaporates more readily than C
5. The osmotic pressure of a 5% solution of cane sugar at 150(C is
(A) 4 atm (B) 3.4 atm
(C) 3.55 atm (D) approx 5 atm
6. 100 ml of liquid A was mixed with 25 ml of liquid B to give a non – ideal solution of
A – B mixture. The volume of this mixture would be
(A) 75 ml (B) 125 ml
(C) close to 125 ml but not exactly 125 ml (D) just more than 125 ml
7. The Van’t Hoff factors i for an electrolyte which undergoes dissociation and association in solvents are respectively
(A) greater than one and less than one (B) less than one and greater than one
(C) less than one in both cases (D) more then one in both cases
8. For the given electrolyte AxBy. The degree of dissociation ‘(’ can be given by
(A) [pic] (B) [pic]
(C) [pic] (D) either of these
9. Glucose is added to 1 litre water to such an extent that [pic]/kf becomes equal to 1/1000. The weight of glucose added is
(A) 180 gm (B) 18 gm
(C) 1.8 gm (D) 0.18 gm
10. When mercuric iodide is added to the aqueous solution of potassium iodide, the
(A) freezing point is raised (B) freezing point is lowered
(C) freezing point does not change (D) can not predict
11. According to Raoult’s law the relative decrease in the solvent vapour pressure over the solution is equal to
(A) the mole fraction of the solvent (B) the mole fraction of solute
(C) the number of moles of solute (D) all of these
12. The freezing point of equimolal aqueous solution will be highest for
(A) [pic] (B) Ca(NO3)2
(C) La(NO3)3 (D) C6H12O6
13. Molal depression constant is given by the expression
(A) [pic] (B) [pic]
(C) [pic] (D) [pic]
14. 0.01 M solution each of urea, common salt and Na2SO4 are taken, the ratio of depression of freezing point
(A) 1 : 1 : 1 (B) 1 : 2 : 1
(C) 1 : 2 : 3 (D) 2 : 2 : 3
15. A X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. Thus value of X is
(A) 14 (B) 3.2
(C) 1.4 (D) 2
Level – II
1. The azeotropic mixture of water (b.p. = 100(C) and HCl (b.p = 85(C) is distilled, it is possible to obtain:
(A) pure HCl (B) pure water
(C) pure HCl as well as water (D) neither HCl nor H2O in pure form
2. Mole fraction of C3H5(OH)3 in a solution of 36 gm of water and 46 gm of glycerine is
(A) 0.46 (B) 0.36
(C) 0.20 (D) 0.40
3. Colligative properties of the solution depend upon
(A) nature of the solution (B) nature of the solvent
(C) number of solute particles (D) number of moles of solvent
4. The vapour pressure of a dilute aqueous solution of glucose is 750 mm of mercury at 373K. The mole fraction of solute is
(A) [pic] (B) [pic]
(C) [pic] (D) [pic]
5. The ratio of the value of colligative property for KCl solution to that of sugar solution at the same concentration is nearly
(A) 1 (B) 2
(C) 0.5 (D) 2.5
6. The Van’t Hoff factor of a 0.005 M aqueous solution of KCl is 1.95. The degree of ionization of KCl is
(A) 0.95 (B) 0.97
(C) 0.94 (D) 0.96
7. Equimolal solution of A and B show depression in freezing point in the ratio of 2 : 1. A remains in normal state in solution, B will be in………………..state in solution.
(A) normal (B) associated
(C) hydrolysed (D) dissociated
8. 2.56 gm of sulphur in 100 gm CS2 has depression in f.p of 0.010(C;[pic]. Hence atomicity of sulphur in CS2 is
(A) 2 (B) 4
(C) 6 (D) 8
9. Which of the following azeotropic solutions has the boiling point less than boiling point of the constituents A and B?
(A) CHCl3 and CH3COCH3 (B) CS2 and CH3COCH3
(C) CH3CH2OH and CHCl3 (D) CH3CHO and CS2
10. If a solute undergoes dimerisation and trimerisation the minimum values of the Van’t Hoff factors are
(A) 0.5 and 1.5 (B) 1.5 and 1.33
(C) 0.5 and 0.33 (D) 0.25 and 0.67
11. Osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose then will be isotonic with blood is
(A) 5.41% (B) 3.54%
(C) 4.53% (D) 5.34%
12. 1 mole benzene [pic] and 2 moles toluene [pic] will have
(A) total vapour pressure 38 mm
(B) mole fraction of vapours of benzene above liquid mixture is 7/19
(C) ideal behaviour
(D) all of the above
13. The value of observed and calculated molecular weight of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is
(A) 60% (B) 83.5%
(C) 46.7% (D) 60.23%
14. Lowering of vapour pressure due to a solute in 1 molal aqueous solution at 100(C is
(A) 13.43 torr (B) 14.12 torr
(C) 312 torr (D) 352 torr
15. The vapour pressure of pure benzene at 50(C is 268 torr. How many mol of non volatile solute per mol of benzene is required to prepare a solution of benzene having a vapour pressure of 167 torr at 50(C
(A) 0.377 (B) 0.605
(C) 0.623 (D) 0.395
ANSWERS TO ASSIGNMENT PROBLEMS
Subjective:
Level – O
1. By Henry’s law, the mole fraction of oxygen in the solution is [pic]
As M for water = 55.5
Hence [pic]
Or n =[pic] (n ................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
Related searches
- english composition syllabus sample
- school syllabus in sinhala
- english 101 syllabus community college
- school syllabus sri lanka
- advanced level syllabus sri lanka
- school syllabus sri lanka 2016
- new syllabus biology 2019 pdf
- nied biology syllabus 2019
- al physics syllabus sinhala medium
- english 101 syllabus course outline
- acca f5 syllabus 2019
- a l syllabus teacher s guide