A solution consists of 0.695 moles of C7H8 and 0.600 moles of C6H12. What is the mole fraction of C7H8 in the vapor above this solution at 25oC?

Question

A solution consists of 0.695 moles of toluene (C₇H₈) and 0.600 moles of cyclohexane (C₆H₁₂). The vapor pressures of toluene and cyclohexane at 25^oC are 28.4 mmHg and 107.5 mmHg, respectively. What is the mole fraction of toluene in the vapor above this solution at 25^oC?

Answer 1

Your answer is correct and I have performed the calculations below to verify that answer.

Raoult's Law

Find total moles of material present:

0.695 moles C₇H₈ + 0.600 mole C₆H₁₂ = 1.295 total moles

mole fraction of C₇H₈ = 0.695 / 1.295 = 0.537 = X_tol

mole fraction of C₆H₁₂ = 0.600 / 1.295 = 0.463 = X_cyclo

Total vapor pressure of the mixture = X_tol*P_tol + X_cycl*P_cyclo = 0.537*28.4 + 0.463*107.5 = 15.3 + 49.8 = 65.1 mm Hg

Mole fraction toluene in the vapor phase = 15.3 / 65.1 = 0.235