As given, vapour pressure of pure benzene, p⁰ = 639.7 mm Hg and that of solution, p = 631.9 mm Hg.

If, n₁ and n₂ be the number of moles of benzene and solute respectively, then according to Raoults' law,

$${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$$ or $${{639.7 - 631.9} \over {639.7}} = {{{n_2}} \over {{n_1}}}$$

$$\therefore$$ $${n_2} = 0.0122 \times {n_1}$$

Amount of benzene in solution = n₁ mol = 78 $$\times$$ n₁g [$$\because$$ Molar mass of benzene = 78 g mol^$$-$$1]

$$\therefore$$ Molality of the solution $$ = {{{n_2} \times 1000} \over {78 \times {n_1}}} = {{0.0122 \times {n_1} \times 1000} \over {78 \times {n_1}}}$$ = 0.156 mol kg^$$-$$1