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)

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 $NO_{2} a=\frac{x}{46}$
$=\frac{100\times15.4}{46\times76.6}=0.437$