Vapor pressures of two volatile species A and B are 55 mm Hg and 120 mm Hg respectively. If mole fraction of ‘A’ in liquid state is 0.8, then mole fraction of ‘B’ in vapor state is

To find the mole fraction of species B in the vapor phase, we can apply Raoult's law and Dalton's law of partial pressures. Let us break down the solution step-by-step.

Step 1: Identify the given data
- Vapor pressure of pure species A, $P_A^{\circ }=55\text{mm Hg}$
- Vapor pressure of pure species B, $P_B^{\circ }=120\text{mm Hg}$
- Mole fraction of A in the liquid phase, $x_A=0.8$

Step 2: Calculate the mole fraction of B in the liquid phase
Since the liquid mixture consists of only species A and B, the sum of their mole fractions in the liquid state must equal 1:
$x_A+x_B=1$
Substituting $x_A=0.8$:
$x_B=1-0.8=0.2$

Step 3: Calculate the partial vapor pressures using Raoult's law
According to Raoult's law, the partial vapor pressure of each component is equal to the product of its mole fraction in the liquid state and its vapor pressure in the pure state:
- For species A:
$P_A=x_A\times P_A^{\circ }=0.8\times 55\text{mm Hg}=44\text{mm Hg}$
- For species B:
$P_B=x_B\times P_B^{\circ }=0.2\times 120\text{mm Hg}=24\text{mm Hg}$

Step 4: Calculate the total vapor pressure of the solution
The total vapor pressure ($P_{\text{total}}$) is the sum of the partial vapor pressures of A and B:
$P_{\text{total}}=P_A+P_B=44+24=68\text{mm Hg}$

Step 5: Calculate the mole fraction of B in the vapor phase
According to Dalton's law of partial pressures, the mole fraction of a component in the vapor state ($y_B$) is the ratio of its partial pressure to the total pressure:
$y_B=\frac{P_B}{P_{\text{total}}}$
Substituting the calculated values:
$y_B=\frac{24}{68}\approx 0.353$

Rounding to two decimal places, we get 0.35. Thus, the mole fraction of B in the vapor state is 0.35.