Question

Hexane and heptane are completely miscible. At 25°C, the vapor pressures of hexane and heptane are 0.198 atm and 0.06 atm, respectively. The mole fractions of hexane and heptane in the vapor phase for a solution containing 4 M hexane and 6 M heptane, respectively, are:

• A. 0.688 and 0.312
• B. 0.400 and 0.600
• C. 0.312 and 0.688
• D. 0.600 and 0.400

Hint:

Use Raoult's Law to calculate the mole fraction of components in the vapor phase, remembering to sum the partial pressures to find the total pressure and use it to find mole fractions.

Answer 1

The Correct Option is A

This is a Raoult's Law problem, where we are calculating the mole fraction of hexane and heptane in the vapor phase. According to Raoult's Law, the partial vapor pressure of each component in a mixture is proportional to its mole fraction in the liquid phase: \[ P_A = X_A P_A^0 \text{and} P_B = X_B P_B^0, \] where \( P_A \) and \( P_B \) are the partial pressures of hexane and heptane, \( P_A^0 \) and \( P_B^0 \) are their vapor pressures in pure form, and \( X_A \) and \( X_B \) are the mole fractions of hexane and heptane in the liquid phase. Step 1: Calculate the total pressure. The total pressure \( P_{\text{total}} \) is the sum of the partial pressures: \[ P_{\text{total}} = P_A + P_B = 0.198 + 0.06 = 0.258 \, \text{atm}. \] Step 2: Calculate the mole fractions in the vapor phase. The mole fraction of each component in the vapor phase is given by: \[ y_A = \frac{P_A}{P_{\text{total}}} = \frac{0.198}{0.258} \approx 0.688, \] \[ y_B = \frac{P_B}{P_{\text{total}}} = \frac{0.06}{0.258} \approx 0.312. \] Thus, the mole fraction of hexane in the vapor phase is 0.688, and the mole fraction of heptane is 0.312. Final Answer: (A) 0.688 and 0.312