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**[AI 2004 C]**

**Sol.**Mass of solvent does not change with temperature.

**Sol.**$\pi V=n R T$ where $n$ is the no. of moles of solute present in $V$ litres of solution, $\pi$ is osmotic pressure, $T$ is temperature and $R$ is gas constant or solution constant.

**[Delhi 2004; CBSE 2004]**

**Sol.**It is used for desalination of sea water.

**[Delhi 1999]**

**Sol.**Colligative properties are those properties which depends upon number of particles of solute and not on natures of solute.

**[Delhi 2002 C]**

**Sol.**In pressure cooker, high pressure is exerted by steam due to which boiling point becomes higher than in open pan.

**Sol.**$0.1 M N a C /$ has higher elevation in boiling point because it contains more number of particles than $0.1 M$ solution of glucose. $N a C l \rightarrow N a^{+}+C l^{-}$

**Sol.**Common salt is called de-icing agent because it lowers the freezing point of water to such an extent that it does not freeze to form ice. Hence, it is used to clear snow from roads.

**[A.I.S.B. 2006]**

**Sol.**At a number of sites, non-volatile glucose molecules will occupy the surface instead of water molecules. Hence, effective surface area for evaporation decreases and so the vapour pressure is lower.

**[H.P.S.B. 1998]**

**Sol.**Iodine vapour in air, aerated drinks.

**[Delhi 2003, 04; D.S.B. 2004 C]**

**Sol.**Due to osmosis, water molecules move into blood cells through the cells walls. As a result, blood cells swell and may even burst.

**Sol.**$\left(p^{\circ}-p_{s}\right) / p^{\circ}=n_{2} /\left(n_{1}+n_{2}\right)$ where $p^{\circ}=V . P .$ of pure solvent $p_{s}=V . P .$ of solution, $n_{2}=$ no. of moles of solute and $n_{1}=$ no. of moles of solvent.

**Sol.**Lowering of V.P. $=p^{\circ}-p_{s} .$ Relative lowering of V.P. $=\left(p^{\circ}-p_{s}\right) / p^{\circ}$

**[Delhi 1999; Foreign 1999]**

**Sol.**Ideal solutions

(i) Follow Raoult’s law

(ii) They can be separated by fractional distillation

**[Delhi 1999 C]**

**Sol.**It is because of $H$ -bonding between acetone and $C H C l_{3}$, the force of attraction increases, therefore energy is released.

**Sol.**Because they can not form

*H-*bonds with water.

**[AI 1999 C]**

**Sol.**Benzene is insoluble in water because benzene is non polar where as water is polar. Toluene is non polar solvent therefore, Benzene is soluble in toluene.

**Sol.**Acetic acid forms dimer due to

*H*-bonding.

*M*sodium chloride solution nearly twice that of 0.1

*M*glucose solution ?

**[A.I.S.B. 2006]**

**Sol.**$\mathrm{NaCl},$ being an electrolytes, dissociates almost completely to give $N a^{+}$ and $\quad C I^{-}$ ions whereas glucose, being nonelectrolyte, does not dissociate. Hence, the number of particles in 0.1 MNaCl solution is nearly double than in

$0.1 M$ glucose solution. Freezing point depression, being a colligative properly, is therefore, nearly twice for solution than for glucose solution of same molarity.

**[DSB2005]**

**Sol.**According to Henry’s law.

Partial pressure of gas above the solution

$=k_{H} \times$ mole fraction of the gas in the solution

or $\quad p_{A}=k_{H} \times x_{A^{\prime}}$ where $k_{H}=$ Henry’s constant.

**Sol.**When fruits and vegetables that have dried and are placed in water osmosis takes place,

*i.e.,*water molecules pass through semipermiable membranes present in cell walls, therefore, they swell. If temperature is increased, osmosis will be faster.

**Sol.**At higher pressure over the liquid (due to weight of the pressure cooker lid), the liquid boils at higher temperature. Therefore, cooking occurs faster.

**Sol.**Freezing point of a liquid depresses on the addition of a non-volatile solute and therefore, a solution of sodium chloride freezes at lower temperature than freezing point of water. On he other hand, there is elevation in boiling point on the addition of a non-volatile solute and consequently boiling point of sodium chloride solution is more than that of water.

**Sol.**HgI$_{2}$ forms a complex with $K I$ and therefore, the number of particles in solution increases.

$H g I_{2}+2 K I \rightarrow K_{2}\left[H g I_{4}\right] \Longrightarrow 2 K^{+}+\left[H g I_{4}\right]^{2-}$

Therefore, the osmotic pressure increases.

**[Delhi 2000; AI 2000]**

**Sol.**Ethanol and water, methanol and $H_{2} O$ show positive deviation from Raoult’s law because force of attraction between them decreases on mixing. and and, show negative deviation from Raoult’s law because force of attraction between them increases on mixing.

**[AI 2002 C]**

**Sol.**The diagram shows that solution containing non-volatile solute boils at temperature $T_{2}$ which is higher than boiling point of pure solvent, i.e., $T_{1}$

Variation of vapour pressure with Temperature showing boiling point of solution is higher than that of pure solvent.

**[AI 2003]**

**Sol.**Carbon tetrachloride is a non-polar compounds, it cannot form $H$ -bond with water, that is why $C C l_{4}$ and water do not mix with each other. Ethanol is a polar compound and $H_{2} \mathrm{O}$ is polar solvent, there is $H$ -bond between ethanol and water, therefore, ethanol and water are miscible in all proportions.

(Atomic mass $H=1, O=16, S=32 \text { a. } m . u .)$

**[Foreign 1999]**

**Sol.**$M=\frac{w_{B}}{M_{B}} \times \frac{1000}{w_{A}}=\frac{13 \times 1000}{98 \times \frac{100}{1.02}}=\frac{13 \times 10 \times 1.02}{98}=\frac{1326}{98}$

$=1.353 \mathrm{mol} L^{-1}$ or $1.353 \mathrm{M}$

$m=\frac{w_{B}}{M_{B}} \times \frac{100}{w_{A}}=\frac{13}{98} \times \frac{1000}{87}=1.524 \mathrm{mol} / \mathrm{kg}_{\mathrm{or}} 1.524 \mathrm{m}$

*g*of methyl alcohol to 120

*g*of water. Calculate the mole fraction of methanol and water.

**[H.P. 1995]**

**Sol.**Mass of methanol = 60

*g*

Moles of methanol $\left.=\frac{60}{32}=1.875 \text { (Molar mass }=32\right)$

Moles of water $=\frac{120}{18}=6.667$

Total No. of moles = 1.875 + 6.667 = 8.542

Mole fraction of methanol $=\frac{1.875}{8.542}=0.220$

Mole fraction of water $=\frac{6.667}{8.542}=0.780$

**Sol.**$\frac{P_{A}^{\circ}-P_{A}}{P_{A}^{\circ}}=X_{B} \Rightarrow \frac{31.82-30.95}{31.82}=\frac{27 / M}{100 / 18}$

$\Rightarrow \frac{0.87}{31.82}=\frac{27}{M_{B}} \times \frac{18}{100} \Rightarrow M_{B}=\frac{27 \times 18 \times 31.82}{0.87 \times 100}$

$\Rightarrow 177.75 g \mathrm{mol}^{-1}$

$\left[k_{f} \text { for water }=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}, \text { molar mass of urea }=60 \mathrm{g}\right.$ $\left.m o l^{-1}\right]$

**[Delhi 2002 C]**

**Sol.**$\Delta T_{f}=k_{f} \times m$

$\Delta T_{f}=1.86 \times \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}}=\frac{1.86 \times 3}{60} \times \frac{1000}{100}=0.93$

Freezing point of solution $=273 K-0.93=272.07 K$

**[CBSE 2004]**

**Sol.**Solution of $n$ hexane and ethanol shows positive deviations from Raoults law because $n$ -hexane molecular weaken the H-bonds between ethanol molecules which increases its vapour

Solution of acetone and chloroform shows negative deviations from Raoult’s law because of the formation of the $H$ -bonds between acetone and chloroform molecules.

**[AI 1999 C]**

**Sol.**The molecular mass obtained with the help of colligative property sometimes is different from normal molecular mass, it is called abnormal molecular mass. The factor which bring abnormality are

(i) *Association *: When solute particles undergo association, number of particles become less and molecular mass determined with the help of colligative property will be more.

(ii) *Dissociation *: It leads to increase in number of particles, therefore, increase in colligative property, therefore, therefore, decrease in molecular weight because colligative property is inversely proportional to molecular weight.

**Sol.**It is observed that the presence of a non-volatile solute in a solution reduces the escaping tendency of solvent molecules into vapour phase. It is because some of the solute particles occupy the position of the solvent molecules on the liquid surface and thus dower the vapour pressure of the solvent.

**[MP 2001, 02, 06]**

**Sol.**

**Sol.**According to Van’t Hoff equation

$\pi v=n R T$ …..(i)

$\pi=$ osmotic pressure; $T=$ Temperature for a solution

If $w$ gram of solute is dissolved in $v$ litres of he solution and $M$ is the molecular mass of the solute, then

$n=\frac{w}{M}$

Substituting this value in equation (v) we get,

$\pi v=\frac{w}{M} R T \Rightarrow M=\frac{w R T}{\pi v}$ …..(ii)

Thus, measuring the osmotic pressure of a solution containing grams of the solute in litres of the solution, at temperature the molecular man, of the solute can be calculated using equation (ii).

**[AI 1999 C]**

**Sol.**

**[CBSE Board 1998]**

**Sol.**

**[CBSE 2004]**

**Sol.**Here $W_{A}=25 g, W_{b}=2 g, \Delta T_{f}=1.62 K, K_{f}=4.9 K \mathrm{kg} \mathrm{mol}^{-1}$

Observed molecular mass of benzoic acid,

$M_{B}=\frac{1000 K_{f} \times W_{B}}{\Delta T_{f} \times W_{A}}=\frac{1000 \times 4.9 \times 2}{1.62 \times 25}=242$

Calculated molecular mass of benzoic acid

$=72+5+12+32+1=122$

Van’t Hoff factor

$i=\frac{\text { Calculated molecular mass }}{\text { Observed molecular mass }}=\frac{122}{242}=.504$

Let $\alpha$ be the degree of dissociation

\[ 2 C_{6} H_{5} C O O H \rightleftharpoons\left(C_{6} H_{5} C O O H\right)_{2} \]

Total number of moles after association

$=1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$

$\therefore \quad i=\frac{1-\frac{\alpha}{2}}{1}=.504$

or $\quad \alpha=(1-.504) \times 2=.496 \times 2=.992$

percentage of association $=99.2 \%$

**[MP 2002, 04, 06]**

**Sol.**Experimental Measurement of Osmotic Pressure. A number of methods are available for measurement of osmotic pressure. The best out of these is Berkeley and Hartley’s Method.

This method is based upon applying pressure on the solution which is just sufficient to prevent the entry of the solvent into the solution through the semi-permeable membrane. The apparatus used by Berkeley and nartley is shown in Figure.

It consists of a porous tube (open at both ends) containing the semi-permeable membrance of copper ferrocyanide. The porous tube is fitted with a reservoir R on one side and a tube T on the other.

The porous tube is filled with the pure solvent so that the level in the tube T stands at the mark M. The porous tube is fitted into an outer vessel made of gun metal. This vessel has a wide tube at the top which is fitted with a frictionless piston and a pressure gauge as shown in figure. The solution under study is taken in the outer gun metal vessel. As a result of osmosis, the level in the tube tends to fall. The pressure applied on the solution by means of the piston which keeps the level in the tube at is a measure of the osmotic pressure and can be read directly on the pressure gauge.

This method is superior to the other methods because of the following reasons :

(a) In this method, the osmotic pressure is balanced by the external pressure so that there is no strain on the membrane.

(b) The concentration of the solution does not change because the entry of the solvent into the solution is prevented by the external pressure.

(c) The time taken for the measurement of osmotic pressure is much less in this method as compared to the other methods.

*A*and

*B*between vapour pressure and mole fractions of components at constant temperature.

**[CBSE 2002; Delhi 2002]**

**Sol.**

Relationships between V.P. and mole fraction for ideal solution

**Sol.**The given quantities are : $w_{2}=2 g ; K_{f}=4.9 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$ ;

$w_{1}=25 g, \Delta T_{f}=1.62 K$

Substituting these values in equation

$M_{2}=\frac{k_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}}$

$M_{2}=\frac{4.9 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} \times 2 \mathrm{g} \times 1000 \mathrm{g} \mathrm{kg}^{-1}}{25 \mathrm{g} \times 1.62 \mathrm{K}}=241.98 \mathrm{gmol}^{-1}$

Thus, experimental molar mass of benzoic acid in benzene is $=241.98 \mathrm{g} \mathrm{mol}^{-1}$

Now consider the following equilibrium for the acid:

$2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$ BHA $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)_{2}$

If $x$ represents the degree of association of the solute then we would have $(1-x)$ mol of benzoic acid left in

unassociated form and correspondingly $\frac{x}{2}$ as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is:

$1-x+\frac{x}{2}=1-\frac{x}{2}$

Thus, total number of moles of particles at equilibrium equals Van’t Hoff factor $i$

But $i=\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}=\frac{122 g \mathrm{mol}^{-1}}{241.98 \mathrm{gmol}^{-1}}$

$\Rightarrow \frac{x}{2}=1-\frac{122}{241.98}=1-0.504=0.496$

$\Rightarrow x=2 \times 0.496=0.992$

Therefore, degree of association of benzoic acid in benzene is $99.2 \%$