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Find the molality of a solution containing a non-volatile solute if the vapour pressure is 2% below the vapour pressure of pure water.

$\begin{array}{1 1}1.133\\1.123\\1.433\\0.133\end{array} $

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We know that
$\large\frac{P^0-P_s}{P_s}=\frac{w\times M}{m\times W}$
$\Rightarrow \large\frac{w}{m\times W}$$\times 1000 \times \large\frac{M}{1000}$
Molality =$\large\frac{w\times 1000}{m\times W}$
$\therefore \large\frac{P^0-P_s}{P_s}=$molality $\times \large\frac{M}{1000}$
$\therefore \large\frac{P^0-(98/100)P^0}{(98/100)P^0}=$molality $\times \large\frac{1000}{18}$
$\therefore$ Molality =$\large\frac{2P^0/100}{\Large\frac{98}{100}\normalsize P^0}\times \large\frac{1000}{18}$
answered Aug 6, 2014 by sreemathi.v

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