Solutions and Colligative PropertieshardNUMERICAL

At the exact azeotropic point for a binary mixture of highly volatile liquids A and B, the mole fracSolutions and Colligative Properties Chemistry Question

Question

At the exact azeotropic point for a binary mixture of highly volatile liquids A and B, the mole fraction of component A in the bulk liquid phase is measured to be exactly $0.74$. Under steady boiling equilibrium at this state, what must be the precise mole fraction of component B in the escaping vapour phase ($y_B$)?

Answer: 0.26

💡 Solution & Explanation

In any binary mixture, the sum of mole fractions is 1, so the mole fraction of B in the liquid is $x_B = 1 - 0.74 = 0.26$. By definition, an azeotropic mixture vaporizes without a change in composition, meaning the vapour phase composition is identical to the liquid phase. Therefore, $y_B = x_B = 0.26$.

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