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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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A very small amount of a non-volatile solute (that does not dissociate) $P_s$ dissolved in $56.8cm^3$ of benzene (density =$0.889gcm^{-3}$).At room temperature,vapour pressure of this solution is 98.88mm Hg while that benzene is 100mm Hg.Find the molality of this solution.If the freezing temperature of this solution is 0.73 degree lower than that of benzene,what is the value of $K_f$ for benzene .

$\begin{array}{1 1}0.1452,5.027Kmolality ^{-1}\\0.1352,6.027Kmolality ^{-1}\\0.2452,2.027Kmolality ^{-1}\\1.1452,1.027Kmolality ^{-1}\end{array} $

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1 Answer

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We know that $\large\frac{P^0-P_s}{P_s}=\frac{w\times M}{m\times w}$
Substituting values,
$\large\frac{100-98.88}{98.88}=\frac{w}{m}\times \frac{78}{W}\times \frac{1000}{1000}$
$\Rightarrow $ Molality $\times \large\frac{78}{1000}$
(or) Molality $(\large\frac{w\times 1000}{m\times W})=\frac{1.12\times 1000}{78\times 98.88}$
$\Rightarrow 0.1452$
We know that
$\Delta T_f=K_f\times$ molality
$\therefore K_f=\large\frac{\Delta T_f}{\text{Molality}}=\frac{0.73}{0.1452}$
$\Rightarrow 5.027Kmolality ^{-1}$
answered Aug 7, 2014 by sreemathi.v
 
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