$\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

$\large\frac{P^0-P}{P^0}=\frac{n}{N}$

$\large\frac{639.7-631.9}{639.7}=\frac{n\times 78}{1000}$

(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

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