If two substances A and B have \(P^o_A : P^o_B = 1 : 2\) and have mole fraction in the ratio \(1 : 2\), then mole fraction of \(A\) in vapours is

0.2

The correct answer is option (2) → 0.2

• Vapour pressure ratio: $P_A^o : P_B^o = 1 : 2$. Let $P_A^o = p$ and $P_B^o = 2p$.
• Mole fraction ratio in liquid: $x_A : x_B = 1 : 2$.
• Since $x_A + x_B = 1$, we can solve for the specific values:
• $x_A = \frac{1}{1+2} = \frac{1}{3}$
• $x_B = \frac{2}{1+2} = \frac{2}{3}$

Partial Pressures ($P_A$ and $P_B$):

Using Raoult's Law ($P_i = P_i^o \cdot x_i$):

• $P_A = P_A^o \cdot x_A = p \cdot \left( \frac{1}{3} \right) = \frac{p}{3}$
• $P_B = P_B^o \cdot x_B = 2p \cdot \left( \frac{2}{3} \right) = \frac{4p}{3}$

Total Vapour Pressure ($P_{total}$):

$P_{total} = P_A + P_B = \frac{p}{3} + \frac{4p}{3} = \frac{5p}{3}$

 Mole Fraction in Vapour Phase ($y_A$):

According to Dalton's Law, the mole fraction of a component in the vapour phase is the ratio of its partial pressure to the total pressure:

$y_A = \frac{P_A}{P_{total}}$

$y_A = \frac{p/3}{5p/3} = \frac{1}{5}$

$y_A = 0.2$