Liquid A and B form an ideal solution at 30^oC. At this temperature vapour pressure of pure A and pure B is 200 mm Hg and 50 mm Hg respectively. Calculate the vapour pressure of the solution at same temperature if equal moles of A and B are added. (Given: Molecular mass of A and B are 40 and 60 respectively)

125 mm Hg

The correct answer is option 2.125 mm Hg.

To calculate the vapor pressure of an ideal solution formed by equal moles of liquids A and B, we use Raoult's Law. According to Raoult's Law, the partial vapor pressure of each component in the solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

Given:

Vapor pressure of pure A (\(P_A^0\)) = 200 mm Hg

Vapor pressure of pure B (\(P_B^0\)) = 50 mm Hg

Equal moles of A and B are added, hence their mole fractions are equal.

Since we have equal moles of A and B:

\( x_A = x_B = \frac{1}{2} \)

Using Raoult's Law, the total vapor pressure of the solution \(P_{solution}\) is:

\(P_{solution} = P_A + P_B \)

\(P_{solution} = x_A \cdot P_A^0 + x_B \cdot P_B^0 \)

\(P_{solution} = \left( \frac{1}{2} \right) \times 200 \, \text{mm Hg} + \left( \frac{1}{2} \right) \times 50 \, \text{mm Hg} \)

\(P_{solution} = 100 \, \text{mm Hg} + 25 \, \text{mm Hg} \)

\(P_{solution} = 125 \, \text{mm Hg} \)

Therefore, the vapor pressure of the solution at 30°C, when equal moles of A and B are added, is 125 mm Hg.