(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?

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

(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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