At $300\ K$, two pure liquids $A$ and $B$ have vapour pressures $150\ mm\ Hg$ and $100\ mm\ Hg$ respectively. In an equimolar liquid mixture of $A$ and $B$, the mole fraction of $B$ in the vapour mixture at this temperature is:

In an equimolar mixture of liquids A and B, the mole fraction of B is 0.5. Using Raoult's law, the partial pressures are calculated as B equals 0.5 multiplied by 100 and A equals 0.5 multiplied by 150, giving 50 mm Hg and 75 mm Hg respectively. The mole fraction of B in the vapour phase equals its partial pressure divided by the total pressure, which is 50 divided by 125, resulting in 0.4.

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