At the exact azeotropic point for a binary mixture of highly volatile liquids A and B, the mole frac — Solutions 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 ($y_B$)?
💡 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, $y_B = x_B = 0.26$.