At 323 K, the vapour pressure (in mm Hg) of a methanol-ethanol solution is represented as: P = 120 X + 140, where X is the mole fraction of methanol in liquid solution at equilibrium. Then the value o

Question

At 323 K, the vapour pressure (in mm Hg) of a methanol-ethanol solution is represented as: P = 120 X + 140, where X is the mole fraction of methanol in liquid solution at equilibrium. Then the value of P_EtOH^o is

Prashant Kotian · Accepted Answer

Correct answer: C. The vapour pressure of a mixture of methanol (MeOH) and ethanol (EtOH) is given by Raoult's law: P = P_MeOH^o * X_MeOH + P_EtOH^o * X_EtOH = P_MeOH^o * X_MeOH + P_EtOH^o * (1 - X_MeOH) = (P_MeOH^o - P_EtOH^o) * X_MeOH + P_EtOH^o. Comparing this with the given equation P = 120 X + 140 (where X = X_MeOH): P_EtOH^o = 140 mm Hg. Note: The printed answer key indicates (c) because it likely intended to ask for pure methanol, P_MeOH^o, which is 260 mm Hg, but based on the text of the question, B is the correct answer for P_EtOH^o.

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