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When $20\;g$ of naphthoic acid $(C_{11}H_gO_2)$ is dissolved in $50 \;g$ of benzene $(K_t =1.72\;K\;kg\;mol^{-1}).$, a freezing point depression of $2 K$ is observe. The van't Hoff factor (i) is

$\begin{array}{1 1} 0.5 \\ 1 \\ 2 \\ 3 \end{array} $

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Answer :0.5
Amount of naphthoic acid $=20\;g/172\;g\;mol^{-1}$
Molality of naphthoic acid $=(20/172)\;mol/0.05kg=(400/172)\;mol \;kg^{-1}$
Using the expression $(- \Delta T_t) =i K_t m$ we get,
$i= \large\frac {(- \Delta T_t)}{K_t m}=\frac{(2K)}{(1.72 K \;kg mol^{-1} ) ((400 /172) mol \;kg^{-1})}$$=0.5$
$\quad= 0.5$
answered Aug 6, 2014 by meena.p

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