A mixture of $CH X_{4}$ (16) and $C X_{2} H X_{6}$ (30) has a density of $1.050 g L^{- 1}$ at $300 K$ and $1.00 atm$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the mole percent of $C X_{2} H X_{6}$ (one decimal) is

Numerical answer type, answer in %, tolerance 0.2.

Answer: 70.4 %

\overset{ˉ}{M} = d RT / P = 1.050 \times 0.0821 \times 300 = 25.86 g mo l^{- 1}

16 (1 - x) + 30 x = 25.86

x = \frac{25.86 - 16}{14} = 0.7044

, i.e.

70.4

mole percent.