Question

If \(P_A^{0}\) and \(P_B^{0}\) are the vapour pressures of pure liquids A and B, what is the total pressure of an ideal solution where the mole fraction of A is \(0.4\)?

• A. \(0.4P_A^{0} + 0.6P_B^{0}\)
• B. \(0.6P_A^{0} + 0.4P_B^{0}\)
• C. \(P_A^{0} + P_B^{0}\)
• D. \(0.5(P_A^{0} + P_B^{0})\)
• E.

Hint:

For ideal solutions, always apply Raoult's law: total vapour pressure equals the sum of mole fraction multiplied by vapour pressure of each pure component. Remember that \(x_A + x_B = 1\).

Answer 1

The Correct Option is A

Concept: According to Raoult's Law, the partial vapour pressure of each component in an ideal solution is proportional to its mole fraction. The total vapour pressure of the solution is: \[ P_{\text{total}} = P_A + P_B \] where: \[ P_A = x_A P_A^{0}, \quad P_B = x_B P_B^{0} \] Thus, \[ P_{\text{total}} = x_A P_A^{0} + x_B P_B^{0} \]

Step 1: Use the given mole fraction of component A. The problem states: \[ x_A = 0.4 \]

Step 2: Find the mole fraction of component B. Since the sum of mole fractions in a binary solution is always 1: \[ x_A + x_B = 1 \implies x_B = 1 - 0.4 = 0.6 \]

Step 3: Substitute into Raoult's law. Substituting the values into the total pressure formula: \[ P_{\text{total}} = (0.4)P_A^{0} + (0.6)P_B^{0} \] Hence, the total vapour pressure is: \[ \boxed{0.4P_A^{0} + 0.6P_B^{0}} \]