Question

The vapour density of a mixture containing $NO _{2}$ and $N _{2} O _{4}$ is $38.3$ at $300\, K$. The number of moles of $NO _{2}$ in $100 \,g$ of the mixture is approximately

• A. 0.44
• B. 4.4
• C. 33.4
• D. 3.34

Answer 1

Answer: (A)

Molecular weight of mixture $=38.3 \times 2=76.6$

Let mass of $NO _{2}$ in mixture $=x\,g$

Then mass of $N _{2} O _{4}=(100 -x) g$

Number of moles of $NO _{2}=x / 46$

Number of moles of $N _{2} O _{4}=\frac{100-x}{92}$

$\frac{\text { Weight }}{\text { Number of moles }}=$ Molecular weight

$\frac{x+(100-x)}{\frac{x}{46}+\frac{(100-x)}{92}}=76.6$

$ \Rightarrow \frac{x}{46}+\frac{(100-x)}{92}=\frac{100}{76.6}$

$ \Rightarrow x=20.1$

Number of moles of $NO _{2}=\frac{20.1}{46}=0.437 \approx 0.44$