A mixture of $CH X_{4}$ and $C X_{3} H X_{8}$ has a density of $1.50 g L^{- 1}$ at STP (molar volume $22.4 L mo l^{- 1}$ ). The mole percentage of $CH X_{4}$ is (C = 12, H = 1)

Answer: C

M(mix) = 1.50 x 22.4 = 33.6 g/mol. With x = mole fraction of CH4: 16x + 44(1 - x) = 33.6, so 28x = 10.4 and x = 0.3714, i.e. 37.1%.

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