(a) Write two characteristics of non-ideal solution.(b) 2g of benzoic acid $(C_{6}H_{5}COOH)$ dissolved in 25g of benzene shows a dpression in freezing point equal to -1.62K. Molal depression constant for benzene is 4.9 K kg mol-1. What is the percentage association of acid if it forms dimer in solution?

(a) 1) The enthaply change of mixyure is not zero. The heat is either released or absorbed on mixing solute and solvent.

$\Delta H_{mix}\neq 0$ (-ve or +ve)

2) Change in the volume of solution an addition of solute to solute is not zero, $\Delta V\neq 0$. There is either expansion or contraction.

(b) Weight of solute (W2) = 2g

Kf = 4.9 K kg mol-1

Depression in freezing point = $\Delta T_{f}= 1.62K$

Weight of solvent (W1) = 25g

$\Delta T_{f}= K_{f}\times \frac{W_{2}}{M_{2}}\times \frac{1000}{W_{1}}$

$M_{2}= \frac{K_{f}\times W_{2}\times 1000}{\Delta T_{f}\times W_{1}}= \frac{4.92\times 2\times 1000}{1.62\times 25}= 242gmol^{-1}$

$2C_{6}H_{5}COOH \leftrightharpoons (C_{6}H_{5}COOH)_{2}$

Let x be the degree of association.

Undissociated benzoic acid = 1 - x for $\frac{n}{2}$ associated molecule.

Total no. of particles at equilibrium $= i = 1-x+\frac{x}{2}= 1-\frac{x}{2}$

i = $\frac{normal molar mass}{observed moar mass}= \frac{122}{242}=1-\frac{x}{2}$

$\frac{x}{2}= 1-\frac{122}{242}= \frac{120}{242}$

$x = \frac{120}{242}\times 2 = 0.992$

% association = 0.992 X 100 = 99.2%

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