Solutions

Concept:Use Raoult’s law to relate vapour pressures and mole fractions in the liquid, then find the vapour-phase mole fraction of B.Explanation:The solution is 5 molal: 5 moles of B in 1 kg (1000 g) of A. Moles of A = $1000 / 50 = 20$. Total moles in liquid = $20 + 5 = 25$. Mole fraction of B in liquid, $x_B = 5 / 25 = 0.2$. Mole fraction of A, $x_A = 20 / 25 = 0.8$. Total vapour pressure $P_T = x_A P_A^o + x_B P_B^o$ (Raoult’s law). $P_T = 100$ mm Hg, $P_A^o = 105$ mm Hg. So, $100 = (0.8)(105) + (0.2) P_B^o$. $100 = 84 + 0.2 P_B^o$ → $P_B^o = (100 - 84) / 0.2 = 80$ mm Hg. Partial pressure of B in vapour: $P_B = x_B P_B^o = 0.2 \times 80 = 16$ mm Hg. Mole fraction of B in vapour: $y_B = P_B / P_T = 16 / 100 = 0.16$.Answer:$0.16$