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 . Under steady boiling equili

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 ($y_B$)?

Prashant Kotian · Accepted Answer

Correct answer: 0.26. In any binary mixture, the sum of mole fractions is 1, so the mole fraction of B in the liquid is . By definition, an azeotropic mixture vaporizes without a change in composition, meaning the vapour phase composition is identical to the liquid phase. Therefore, .

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