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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\\1.233\\1.333\\1.533\end{array} $

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Given :
We know $\large\frac{P^0-P_s}{P_s}=\frac{w\times M}{m\times W}$
$\Rightarrow \large\frac{w}{m\times W}$$\times 1000\times \large\frac{M}{1000}$
As molality =$\large\frac{\text{number of moles of the solute}}{\text{Mass of the solvent in kilograms}}$
$\Rightarrow \large\frac{w\times 1000}{m\times W}$
$\therefore \large\frac{P^0-(98/100)P^0}{(98/100)P^0}$$=Molality \times \large\frac{18}{1000}$
$\therefore$ Molality =$\large\frac{2P^0}{100\times \large\frac{98}{100}P^0}$$\times \large\frac{1000}{18}$
$\Rightarrow 1.133$
answered Aug 7, 2014 by sreemathi.v

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