A mixture of $N X_{2}$ (molar mass 28) and $CO X_{2}$ (molar mass 44) has a density of $1.600 g L^{- 1}$ at $300 K$ and $1.00 atm$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the mole percent of $CO X_{2}$ in the mixture (one decimal) is

Numerical answer type, answer in %, tolerance 0.2.

Answer: 71.3 %

Average molar mass

\overset{ˉ}{M} = \frac{d RT}{P} = 1.600 \times 0.0821 \times 300 = 39.408 g mo l^{- 1}

. With

x

the mole fraction of

CO X_{2}

28 (1 - x) + 44 x = 39.408

x = \frac{39.408 - 28}{16} = 0.7130

, i.e.

71.3

mole percent.