Question

A mixture of $NO_2$ and $N_2O_4$ has a vapour density of 38.3 at 300 K. What is the number of moles of $NO_2$ in 100 g of the mixture ?

• A. 0.043
• B. 4.4
• C. 3.4
• D. 0.437

Answer 1

Answer: (D)

0.437

Answer 2

Average mol. mass of mixture $=2 \times V.D$ $= 2 \times 38.3 = 76.6$ Let mixture contains $a$ moles of $NO_{2}$ and $b$ moles of $N_{2}O_{4}$ $\therefore \frac{a\times46+b\times92}{a+b}=76.6$ (Molar mass of $NO_{2}=46\,g\,mol^{-1}$ molar mass of $N_{2}O_{4}=92\,g\,mol^{-1})$ Let mass of $NO_{2}=x$ and mass of $N_{2}O_{4}=100-x$ $\therefore a=\frac{x}{46}$ and $b=\frac{100-x}{92}$ $\therefore\frac{x+100-x}{\frac{x}{46}+\frac{100-x}{92}}=76.6$ $\frac{x}{46}+\frac{100-x}{92}=\frac{100}{76.6}$ $x=\frac{100\times15.4}{76.6}$ No. of moles of $NO_{2} a=\frac{x}{46}$ $=\frac{100\times15.4}{46\times76.6}=0.437$