A mixture of A and B in the molar ratio 1 : 2 forms a maximum boiling azeotrope. | TestHub

ChemistryMediummultiple choice

A mixture of A and B in the molar ratio $1 : 2$ forms a maximum boiling azeotrope. Identify the correct statements, if A is more volatile.

[Molar mass of A = 100, Molar mass of B = 50]

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Answer:

A, D

Solution:

(A) Mass % of A in liquid solution, $50 = \frac{100 \times n _{A} + 50 \times n _{B}}{n _{A} + n _{B}}$ , then $\frac{n _{A}}{n _{B}} = \frac{1}{2}$ . This is an azeotropic mixture, so $Y_{A} = 50%$ .

(B) Mass % of A in liquid solution, $50 < \frac{100 \times n _{A} + 50 \times n _{B}}{n _{A} + n _{B}}$ . The vapor must be richer in A, so $A > 50% (P_{A}^{0} > P_{B}^{0})$ .

(C) If $X_{A} = \frac{1}{4} < \frac{1}{3}$ , pure A cannot be obtained.

(D) If $X_{A} = \frac{2}{5} < \frac{1}{3}$ , pure A can be obtained.

Stream:JEESubject:ChemistryTopic:Liquid SolutionSubtopic:Azeotropic Mixtures/Non-ideal solution.

⏱ 2mℹ️ Source: QB

Chemistry - Liquid Solution Question with Solution | TestHub

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