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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3mc219

An equimolar mixture of gases X ( $M = 44$ ) and Y has a vapour density of $18$ . Select the correct statement(s).

AThe molar mass of Y is $28 g mo l^{- 1}$ .
BIn the equal-mass mixture the mole fraction of X is $0.389$ .
CIf X and Y are mixed in equal masses instead, the average molar mass is $36$ .
DIf X and Y are mixed in equal masses, the average molar mass is $34.2 g mo l^{- 1}$ .

Answer: A, B, D

Equimolar mixture:

(44 + M_{Y}) /2 = 36

gives

M_{Y} = 28

. For equal masses (

1

g each): moles

= 1/44 + 1/28 = 0.05844

, so

M_{a vg} = 2/0.05844 = 34.22

, not

36

. Mole fraction of X

= (1/44) /0.05844 = 0.389

.

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