Question

Two liquids $A$ and $B$ are mixed. The partial vapour pressures of $A$ and $B$ in pure state are $100$ and $200mm$ respectively. If they are mixed in $1:4$ mole ratio, then at equilibrium pressure of the mixture, assuming that mixture obeys Raoult's law, the mole fractions of $A$ and $B$ present in gaseous state. are:

• A. $\frac{1}{5},\frac{4}{5}$
• B. $\frac{1}{3},\frac{2}{3}$
• C. $\frac{1}{7},\frac{6}{7}$
• D. $\frac{1}{9},\frac{8}{9}$

Answer 1

The correct option is A $\frac{1}{5},\frac{4}{5}$

The ratio of moles of A and B is $1:4$.

Let no. of moles be m

Moles of A $=1\times m=m$

Moles of B $=4\times m=4m$

Mole fraction of A = $\frac{m}{m+4m}=\frac{1}{5}$

Mole fraction of B = $\frac{4m}{m+4m}=\frac{4}{5}$

The correct option is A.