The vapour pressure of a pure liquid 'A' is 70 torr at 27 ⁰C. It forms an ideal solution with another liquid B. The mole fraction of B in the solution is 0.2 and total vapour pressure of solution is 84 torr at 27 ⁰C. What is the vapour pressure of pure liquid B at 27 ⁰C?

140 torr

The correct answer is option 3. 140 torr.

To determine the vapor pressure of pure liquid B, we can use Raoult's Law for ideal solutions. Raoult's Law states that the partial vapor pressure of each component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

Given Data:

Vapor pressure of pure liquid A (\(P_A^0\)): 70 torr

Mole fraction of B (\(\chi _B\)): 0.2

Total vapor pressure of the solution (\(P_{\text{total}}\)): 84 torr

Steps to Calculate the Vapor Pressure of Pure Liquid B (\(P_B^0\)):

Calculate the mole fraction of A (\(\chi_A\)):

\(\chi _A = 1 - \chi _B = 1 - 0.2 = 0.8\)

Applying Raoult's Law:

\(P_A = \chi _A \cdot P_A^0\)

\(P_A = 0.8 \cdot 70 \, \text{torr} = 56 \, \text{torr}\)

Determine the partial vapor pressure of B (\(P_B\)) using the total vapor pressure:

\(P_{\text{total}} = P_A + P_B\)

\(84 \, \text{torr} = 56 \, \text{torr} + P_B\)

\(P_B = 84 \, \text{torr} - 56 \, \text{torr} = 28 \, \text{torr}\)

Calculate the vapor pressure of pure liquid B (\(P_B^0\)) using Raoult's Law:

\(P_B = \chi _B \cdot P_B^0\)

\(28 \, \text{torr} = 0.2 \cdot P_B^0\)

\(P_B^0 = \frac{28 \, \text{torr}}{0.2} = 140 \, \text{torr}\)

Therefore, the vapor pressure of pure liquid B at 27°C is 140 torr.