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

Let's solve this problem step by step:

Given:

• Vapour pressure of pure liquid A (P°A) = 500 mm Hg
• Vapour pressure of pure liquid D (P°D) = 800 mm Hg
• Total pressure of the solution (P_total) = 700 mm Hg

We need to find:

• Mole percentage of D in the solution.

Applying Raoult's Law:

Raoult's Law states that the partial pressure of a component in a solution is equal to the product of its mole fraction and its vapour pressure in the pure state.

P_total = P_A + P_D

P_A = x_A * P°A

P_D = x_D * P°D

Where:

• x_A is the mole fraction of A
• x_D is the mole fraction of D

Also, x_A + x_D = 1, so x_A = 1 - x_D

Substituting the values into the equation for P_total:

700 = (1 - x_D) * 500 + x_D * 800

700 = 500 - 500x_D + 800x_D

700 - 500 = 300x_D

200 = 300x_D

x_D = 200 / 300 = 2/3

x_D = 0.6667

Now, to find the mole percentage of D:

Mole percentage of D = x_D * 100

Mole percentage of D = 0.6667 * 100

Mole percentage of D = 66.67%

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

The correct answer is:

Option 2: 66.67 mole percent