Two liquids $$X$$ and $$Y$$ form an ideal solution. At $$300 \,K,$$  vapour pressure of the solution containing 1 mol of $$X$$ and 3 mol of $$Y$$ is $$550 \,mmHg.$$   At the same temperature, if 1 mol of $$Y$$ is further added to this solution, vapour pressure of the solution increases by $$10 \,mmHg.$$  Vapour pressure ( in $$mmHg$$  ) of $$X$$ and $$Y$$ in their pure states will be, respectively :

Solution :
$${P_{{\text{total}}}} = P_A^o{X_A} + P_B^o{X_B}\,;$$     $$550 = P_A^o \times \frac{1}{4} + P_B^o \times \frac{3}{4}$$
$$\eqalign{ & P_A^o + 3P_B^o = 550 \times 4\,...\left( {\text{i}} \right) \cr & {\text{In second case}} \cr & {P_{{\text{total}}}} = P_A^o \times \frac{1}{5} + P_B^o \times \frac{4}{5} \cr & P_A^o + 4P_B^o = 560'5\,\,\,\,\,\,...\left( {{\text{ii}}} \right) \cr & {\text{Subtract (i) from (ii)}} \cr} $$
$$\therefore \,\,P_B^o = 560 \times 5 - 550 \times 4 = 600$$         $$\because P_A^o = 400$$