The vapour pressure of pure benzene at `25^@C` is 639.7 mm of Hg and the vapour pressure of a solution of a solute in `C_6H_6` at the same tempeature is 631.9 mm of Hg. Calculate molality of solution.

Options

A. 0.372mol/kg
B. 0.869 mol/kg
C. 0.635mol/kg
D. 0.158 mol/kg

Answer

D

Explanation

`because (P^(@)-P_(S))/(P_(S))= (w xx M)/(m xx W)`
Molality `=(w)/(m xx W)xx 1000 = (P^(@)- P_(S))/(P_(S)xx M) xx 1000= (639.7-631.9)/(631.9 xx 78) xx 1000= 0.158` mol/kg of solvent