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2.0g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant ($K_f$) of benzene is 4.9 K $kg mol^{−1}$. What is the percentage association of the acid if it forms dimer in the solution?

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Mass of solute (benzoic acid), $w_2 = 2.0$g
Mass of solvent (benzene), $w_1$ = 25.0g
Observed molar mass of benzoic acid $M_2=\large\frac{1000\times K_f\times W_2}{\Delta T_f\times W_1}$
$\Rightarrow 242gmol^{-1}$
Calculated molar mass of benzoic acid ($C_6H_5COOH)$ = 72 + 5 + 12 + 32 + 1 = 122 g$ mol^{−1}$
van’t Hoff factor,i=calculated mol.mass/Observed mol.mass=$\large\frac{122}{242}$$=0.504$
Total number of moles after association =1-$\alpha+\large\frac{\alpha}{2}$
$\alpha= (1 – 0.504) \times 2 = 0.496 \times 2 = 0.992$ or % are association = 99.2%
Hence (B) is the correct answer.
answered Jun 13, 2014 by sreemathi.v

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