Question

The vapour pressure of pure benzene at ${50}^{o}C$ is 268 mm of Hg. How many moles of non-volatile sohite per mole of benzene are required to prepare a solution of benzene having a vapour pressure 167 mm of Hg at ${50}^{o}C$?

Answer 1

Applying Raoult's law in the following form:
$\frac{P_0-P_s}{P_s}=\frac{wM}{Wm}=\frac{w/m}{W/M}$
$=$ No. of moles of solute per mole of benzene
or $\frac{n}{N}=\frac{(268-167)}{167}=0.6047=0.605$
Alternative method: We know that, $P_s=$ Mole fraction of solvent $\times P_0$
or 167 = Mole fraction of solvent $\times 268$
So, Mole fraction of solvent $=\frac{167}{268}=0.623$
Mole fraction of solute $=1-0.623=0.377$
$\frac{n}{N}=\frac{Molefractionofsolute}{Molefractionofsolvent}=\frac{0.377}{0.623}=0.605$