Question #100685

2 g of benzoic acid is dissolved in 25 g of benzene,shows a depression in freezing point equal to 1.62 K. Kf for benzene is 4.9 K kg mol^-1. What is percentage association of benzoic acid if it forms dimer in benzene?

Expert's answer

W2 = 2g; Kf = 4.9K kg mol-1; W1 = 25g; deltaTf = 1.62K

Substituting 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-1

Thus experimental molar mass of benzoic acid in benzene is = 241.98 g mol-1

Now consider the following equilibrium for the acid : 2C6H5COOH -------> (C6H5COOH)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 x/2 as associated moles of benzoic acid at equilibrium is : 1 – x + x/2 = 1 – x/2

Thus, 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-1

Or x/2 = 1 – 122/ 241.98

= 1 – 0.504

= 0.496

Or x = 2 x 0.496

= 0.992

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

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