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Pure Benzene freezes at $5.4 ^{\circ}\;C$. A solution of $0.223 \;g$ of phenyl acetic acid $(C_6 H_5 CH_2 COOH)$ in $4.4\;g$ benzene $(K_f =5.12 K\; Kg\;mol^{-1}) $ freezes at $4.47 ^{\circ}$ . From this observation , one can conclude that

$\begin{array}{1 1} \text{phenyl acetic acid exists as such in benzene} \\ \text {phenyl acetic acid undergoes partial ionization in benzene} \\ \text {phenyl acetic acid undergoes complete ionization in benzene} \\ \text{phenyl acetic acid dimerizes in benzene} \end{array} $

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Answer :phenyl acetic acid dimerizes in benzene
$m = \large\frac{( - \Delta T_t}{K_t} = \frac{(5.4 -4.47 )}{5.12}$$ \;mol\;kg^{-1}$
$\quad= \large\frac{0.97}{5.12 }$$ mol\;kg^{-1}$
Hence $\large\frac{0.223 g/M}{(4.4/1000)\;kg}$
$\quad= \large\frac{0.97}{5.12 }$$ mol\;kg^{-1}$
Solving for M, we get $M= 279 \;g mol^{-1}$.
Number of molecules of phenyl acetic acid $= \large\frac{276}{136}$$=2$
answered Aug 5, 2014 by meena.p

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