Question

At 300 K, the vapour pressure of an ideal solution containing 1 mole of A and 3 moles of B is 500 mm Hg. At the same temperature, 2 moles of B are added to this solution. The vapour pressure of the solution increases by 10% of the original vapour pressure. The correct statement(s) about the vapour pressure are :

• A. The vapour pressure of A in the pure state is 1300 mm Hg
• B. The vapour pressure of B in the pure state is 650 mm Hg
• C. The ratio of the final vapour pressure to the initial vapour pressure is 2 : 1
• D. The ratio of the vapour pressure of pure B to the vapour pressure of pure A is 13 : 1

Answer 1

Answer: (B), (D)

The correct options are
B The vapour pressure of B in the pure state is 650 mm Hg
D The ratio of the vapour pressure of pure B to the vapour pressure of pure A is 13 : 1

$n_A$ = 1; $n_B$ = 3; $P_T$ = 500 mm Hg
When 2 moles of B is added
$n_A$ = 1; $n_B$ = 5
As it is given that vapour pressure increases by 10%,
$P_T^{\prime }=500+\frac{500\times 10}{100}=550$
$500=P_A^{\circ }\left(\frac{1}{1+3}\right)+P_B^{\circ }\left(\frac{3}{1+3}\right)$
$2000=P_A^{\circ }+3P_B^{\circ }-----\left(1\right)$
$550=P_A^{\circ }\left(\frac{1}{6}\right)+P_B^{\circ }\left(\frac{5}{6}\right)$
$3300=P_A^{\circ }+5P_B^{\circ }-----\left(2\right)$
Solving (1) and (2) we get,
$P_A^{\circ }=50\mathrm{mm}\mathrm{Hg}{P}_B^{\circ }=650\mathrm{mm}\mathrm{Hg}$
$\frac{P_A^{\circ }}{P_B^{\circ }}=\frac{650}{50}=13:1$