A quantity of 2 g of C6H5COOH dissolved in 25 g of benzene shows a depression in freezing point equa

Question

A quantity of 2 g of C6H5COOH dissolved in 25 g of benzene shows a depression in freezing point equal to 1.96 K. Molar depression constant for benzene is 4.9 K-kg mol−1. What is the percentage association of acid if it forms double molecules (dimer) in solution?

Accepted Answer

$\Delta T_{f} = K_{f}.m\left\lbrack 1 + \alpha\left( \frac{1}{n} - 1 \right) \right\rbrack$

$\therefore 1.96 = 4.9 \times \frac{2/122}{25/1000}\left\lbrack 1 + \alpha\left( \frac{1}{2} - 1 \right) \right\rbrack \Rightarrow \alpha = 0.78$

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