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A very small amount of a non-volatile solute (that does not dissociate) is 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 of benzene is 100 mm 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}.S\\0.1252,3.027Kmolality^{-1}.S\\0.1352,4.027Kmolality^{-1}.S\\1.1452,6.027Kmolality^{-1}.S\end{array} $

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1 Answer

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 $\times \large\frac{78}{1000}$
Molecular weight of benzene =78
Or,Molality $\big(\large\frac{w\times 1000}{m\times W}\big)=\frac{1.12\times 1000}{78\times 98.88}$
$\Rightarrow 0.1452$
We know that
$\Delta T=K_f\times$ molality
$\therefore K_f=\large\frac{\Delta T}{\text{molality}}$
$\Rightarrow \large\frac{0.73}{0.1452}$
$\Rightarrow 5.027K molality^{-1}.S$
answered Aug 6, 2014 by sreemathi.v
 

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