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12.2gm of benzonic acid (M=122) in 100gm benzene has depression in freezing point $2.6^{\large\circ}C$; $K_f = 5.2 ^{\circ} mol^{-1}kg$. If there is 100% polymerization, the number of molecules of benzonic acid in associated state is

$\begin{array}{1 1}(a)\;4\\(b)\;3\\(c)\;2\\(d)\;1\end{array}$

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Answer: 2
The depression of freezing point $\Delta T_f$ for a dilute solution (ideal solution) can be expressed as$\Delta T_f = \large\frac{1000 K_f w_2 }{M_2 w_1}$
Given $\Delta T_f = 2.6^{\circ}$, $K_f = 5.2 ^{\circ} mol^{-1}kg$, $w_2 = 12.2\; gm$, and $M_2 = 122$ and 100% polymerization,
$n_1 = \large\frac{1000 \times 5.2 \times 12.2}{122 \times 2.6} $$= 2$
$\Rightarrow$ Two molecules of benzoic acid associates to form one molecule.
answered Feb 14, 2014 by sreemathi.v
edited Jul 15, 2014 by balaji.thirumalai
 

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