What will be the percentage of dimerisation of 61g of benzoic acid in 1000g of a solvent and producing a depression in freezing point of 2 degree celsius considered kf to be 6?
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W2 = 2g; Kf = 4.9K kg mol-1; W1 = 25g; deltaTf = 1.62KSubstituting this in the equation deltaTf = (Kf x W2 x 1000) / (M2 x W1)Therefore, M2 = (4.9 x 2 x 1000) / (25 x 1.62) which is equal to 241.98 g mol-1Thus experimental molar mass of benzoic acid in benzene is = 241.98 g mol-1 Now consider the following equilibrium for the acid : 2C6H5COOH -------> (C6H5COOH)2If 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 x/2 as associated moles of benzoic acid at equilibrium is : 1 – x + x/2 = 1 – x/2Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i,But i = Normal molar mass / Abnormal molar mass = 122 g mol-1 / 241.98 g mol-1Or x/2 = 1 – 122/ 241.98 = 1 – 0.504 = 0.496Or x = 2 x 0.496 = 0.992
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