Question

Two liquids A and B form an ideal solution. At 300 K, the vapour pressure of a solution containing 1 mole of A and 3 mole of B is 550 mm of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of A and B in their pure states.

Answer 1

Let the vapour pressure of pure A be $=p_A^0$; and the vapour pressure of pure B be $=p_B^0$.

Total vapour pressure of solution (1 mole A + 3 mole B)
$=X_A\cdot p_A^0+X_B\cdot p_B^0$ [$X_A$ is mole fraction of A and $X_B$ is mole fraction of B]

$550=\frac{1}{4}{p}_A^0+\frac{3}{4}{p}_B^0$

$2200=p_A^0+3p_B^0$ ...(i)
or
Total vapour pressure of solution (1 mole A + 4 mole B)

$=\frac{1}{5}{p}_A^0+\frac{4}{5}{p}_B^0$

$560=\frac{1}{5}{p}_A^0+\frac{4}{5}{p}_B^0$

$2800=p_A^0+4p_B^0$ ...(ii)
or
Solving eqs. (i) and (ii),

$p_B^0=600$ mm of Hg = vapour pressure of pure B

$p_B^0=400$ mm of Hg = vapour pressure of pure A