Two liquids A and B form an Ideal solution. At 300K, the V.P of solution containing one mole of 'A' and 4 mole 'B' is 560mm Hg. At the same temperature. If one mole of 'B' is taken out from the solution the V.P of the solution has decreased by 10mm Hg, the V.P, of pure A & B are (in mm)

Given data:It is given to identify pure vapour pressure of A and B.

Explanation:

Let the pure A vapour pressure be $=p_A^0$ and the pure B vapour pressure be ${\mathrm{p}}_{\mathrm{B}}^0.$
Solution total vapour pressure (1 mole A + 3 mole B)
$=X_A·p_A^0+X_B·p_B^0$ [$X_A$ is the mole fraction of A and $X_B$ is the mole fraction of B]

$$\begin{gathered} \begin{array}{l}550=\frac{1}{4}{p}_A^0+\frac{3}{4}{p}_B^0\\ 2200=p_A^0+3p_B^0\end{array} \\ or Total solution vapour pressure (1 mole A + 4 mole B) \\ \begin{array}{l}=\frac{1}{5}{p}_A^0+\frac{4}{5}{p}_B^0\\ 560=\frac{1}{5}{p}_A^0+\frac{4}{5}{p}_B^0\\ 2800=p_A^0+4p_B^0 \dots \text{ (ii) }\end{array} \\ or Solving equations I and \left(ii\right) \left(ii\right), \\ {\mathrm{p}}_{\mathrm{B}}^0=400 \mathrm{mm} Hg = pure A vapour pressure \\ {\mathrm{p}}_{\mathrm{B}}^0=600 \mathrm{mm} Hg = pure B vapour pressure \end{gathered}$$

Hence, the correct option is: (A)  400,600