Question

Vapour pressures of pure liquids 'A' and 'D' are 500 mm Hg and 800 mm Hg, respectively.The binary solution of ’A’ and ’D’ boils at 50◦C and 700 mm Hg pressure.The mole percentage of D in the solution is:

• A. 33.33 mole percent
• B. 66.67 mole percent
• C. 25.75 mole percent
• D. 75.25 mole percent

Answer 1

The Correct Option is B

To determine the mole percentage of component 'D' in a solution, Raoult's Law is applied. This law states that the total vapor pressure of a solution is the sum of the partial pressures of its individual components. The partial pressure of a component is calculated by multiplying its mole fraction by its pure vapor pressure. For a binary solution comprising 'A' and 'D', the relationship is:

P_solution = P_A + P_D

Where:

• P_A represents the partial pressure of component 'A'.
• P_D represents the partial pressure of component 'D'.
• P_A = x_A × P⁰_A
• P_D = x_D × P⁰_D
• x_A is the mole fraction of 'A'.
• x_D is the mole fraction of 'D'.
• P⁰_A is the vapor pressure of pure 'A'.
• P⁰_D is the vapor pressure of pure 'D'.

The provided data is:

• P⁰_A = 500 mm Hg
• P⁰_D = 800 mm Hg
• P_solution = 700 mm Hg

Given that the sum of mole fractions in a binary solution is 1 (x_A + x_D = 1), we can substitute these values into the Raoult's Law equation:

700 = x_A × 500 + (1 - x_A) × 800

Simplifying the equation yields:

700 = 500x_A + 800 - 800x_A

700 = 800 - 300x_A

Rearranging to solve for x_A:

300x_A = 800 - 700

300x_A = 100

x_A = 100/300 = 1/3

Subsequently, the mole fraction of 'D' is found using x_D = 1 - x_A:

x_D = 1 - 1/3 = 2/3

The mole percentage of 'D' is calculated as:

Mole percentage of 'D' = x_D × 100 = (2/3) × 100 = 66.67%

Therefore, the mole percentage of 'D' in the solution is 66.67 mole percent.