A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be :

To find the mole fraction of pentane in the vapor phase, we can use Raoult's Law, which states that the partial vapor pressure of a component in a solution is equal to the mole fraction of that component times the vapor pressure of the pure component. Given the mole ratio of pentane to hexane is 1:4, the mole fraction of pentane in the liquid phase is 1/(1+4) = 1/5 = 0.2. The mole fraction of hexane is 4/5 = 0.8. Using Raoult's Law, the partial vapor pressure of pentane (P_pentane) is the mole fraction of pentane times the vapor pressure of pure pentane, and similarly for hexane. The total vapor pressure (P_total) is the sum of the partial vapor pressures of pentane and hexane. We can then find the mole fraction of pentane in the vapor phase by dividing the partial vapor pressure of pentane by the total vapor pressure. So, P_pentane = 0.2 440 mmHg and P_hexane = 0.8 120 mmHg. Then, P_total = P_pentane + P_hexane. The mole fraction of pentane in the vapor phase = P_pentane / P_total. Calculating these values will give us the mole fraction of pentane in the vapor phase.