Question

3 moles of liquid A and 1 mole of liquid B are mixed to form an ideal solution. The vapour pressure of solution becomes $500 \text{ mm Hg}$. If 1 mole of A is further added then vapour pressure of solution increases by $20 \text{ mm Hg}$. Find vapour pressure of pure B ($P_B^o$) in $\text{mm Hg}$ ?

• A. 200
• B. 400
• C. 600
• D. 800

Hint:

When solving systems of equations derived from Raoult's Law for multiple compositions, algebraic subtraction is the simplest way to find the vapor pressure of one pure component.

Answer 1

The Correct Option is A

To solve this problem, we will use Raoult's Law, which states that the vapour pressure of an ideal solution is the sum of the partial pressures of the components. The partial pressure of each component is equal to the mole fraction of the component multiplied by the vapour pressure of the pure component.

The formula for Raoult's Law is given by: 

\(P_{\text{solution}} = x_A \cdot P_A^o + x_B \cdot P_B^o\)

where:

• \(P_{\text{solution}}\) is the vapour pressure of the solution.
• \(x_A\) and \(x_B\) are the mole fractions of A and B, respectively.
• \(P_A^o\) and \(P_B^o\) are the vapour pressures of pure A and B, respectively.

First, we calculate the initial mole fractions when 3 moles of A and 1 mole of B are mixed:

\(x_A = \frac{3}{3+1} = \frac{3}{4}\)

\(x_B = \frac{1}{3+1} = \frac{1}{4}\)

The initial vapour pressure of the solution is 500 mm Hg, so:

\(500 = \frac{3}{4} \cdot P_A^o + \frac{1}{4} \cdot P_B^o \quad \quad (1)\)

When 1 mole of A is added, the new mole fractions are:

\(x_A = \frac{4}{4+1} = \frac{4}{5}\)

\(x_B = \frac{1}{4+1} = \frac{1}{5}\)

The new vapour pressure is 520 mm Hg, so:

\(520 = \frac{4}{5} \cdot P_A^o + \frac{1}{5} \cdot P_B^o \quad \quad (2)\)

We now have two equations with two unknowns:

1. \(500 = \frac{3}{4} \cdot P_A^o + \frac{1}{4} \cdot P_B^o\)
2. \(520 = \frac{4}{5} \cdot P_A^o + \frac{1}{5} \cdot P_B^o\)

To solve these equations, we first solve equation (1) for \(P_A^o\):

\(500 = \frac{3}{4} \cdot P_A^o + \frac{1}{4} \cdot P_B^o\)

Multiplying through by 4 to clear fractions:

\(2000 = 3P_A^o + P_B^o \quad \quad (3)\)

Next, multiply equation (2) by 5:

\(2600 = 4P_A^o + P_B^o \quad \quad (4)\)

Now, subtract equation (3) from equation (4):

\(2600 - 2000 = 4P_A^o + P_B^o - (3P_A^o + P_B^o)\)

\(600 = P_A^o\)

Substitute \(P_A^o = 600\) into equation (3):

\(2000 = 3 \times 600 + P_B^o\)

\(2000 = 1800 + P_B^o\)

\(P_B^o = 200\)

Thus, the vapour pressure of pure B is 200 mm Hg. The correct answer is 200.