A $5.0 L$ flask contains $N X_{2}$ and $O X_{2}$ at a total pressure of $3.0 atm$ and $300 K$ . The total mass of gas is $18.4 g$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the mole fraction of $O X_{2}$ is nearest to

Answer: A

n = \frac{3.0 \times 5.0}{0.0821 \times 300} = 0.6090 mol

\overset{ˉ}{M} = 18.4/0.6090 = 30.21

28 (1 - x) + 32 x = 30.21

x = 0.553 \approx 0.55