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The vapour pressure of pure benzene at $25^{\large\circ}C$ is 639.7mm of mercury and the vapour pressure of a solution of a solute in benzene at the same temperature is $631.9$mm of mercury.Calculate the molality of the solution.

$\begin{array}{1 1}0.156mole/kg\\0.146mole/kg\\0.126mole/kg\\0.176mole/kg\end{array} $

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Let the molality of the solution =m
Now the solution contains m moles of solute per 1000gm of benzene.
Vapour pressure of pure benzene,$P^0=639.7mm$
Vapour pressure of solution ,P=631.9mm
Moles of benzene (mol wt. 78)$N=\large\frac{1000}{78}$
Moles of solute,n=?
Substituting these values in the Raoult's equations
$\large\frac{639.7-631.9}{639.7}=\frac{n\times 78}{1000}$
(Or) $\large\frac{7.8}{639.7}=\frac{78n}{1000}$
$\therefore n=\large\frac{1000\times 7.8}{78\times 639.7}$
$\Rightarrow 0.156$
Hence molality of solution =0.156mole/kg
answered Aug 7, 2014 by sreemathi.v

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