Molecules of benzoic acid ( $C_{6} H_{5} C O O H$ ) dimerise in benzene. $' w'$ g of the acid dissolved in $30 g$ of benzene shows a depression in freezing point equal to $2 K$ . If the percentage association of the acid to form dimer in the solution is 80, then w is
(Given that $K_{f} = 5 K kg {mol}^{- 1}$ , Molar mass of benzoic $a c i d = 122 g m o l^{- 1}$ )

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$2.4 g$

$1.5 g$

$1.8 g$

$1.0 g$

answer is B.

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Detailed Solution

$\begin{array}{l} \underset{1 - 0.8}{2 (C_{6} H_{5} COOH)} \to_{C_{6} H_{6}}^{(30 g)} \underset{0.4}{C_{6} H_{5} COOH}; x = 80 % = 0.8 \\ Δ_{f} T = i k_{f} m \\ 2 = 0.6 \times 5 \times \frac{w \times 1000}{122 \times 30} (i = 1 - 0.8 + 0.4 = 0.6) \\ w = 2.44 g . \end{array}$

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