Thank you for reporting, we will resolve it shortly
Q.
A mixture of $NO_{2}$ and $N_{2}O_{4}$ 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?
States of Matter
Solution:
No. of molecules $= 2 \times V.d$
$2 \times 38.3 = 76.6$
Weight of $NO_{2} =x$
So that weight of $N_{2}O_{4}=100-x$
Hence, $\frac{x}{46}+\frac{100-x}{92}=\frac{100}{76.6}$
$=\frac{2x+100-x}{92}=\frac{100}{76.6}$
$x \,=\, 20.10$, the number of moles of $NO_{2}$
$=\frac{20.10}{46}=0.437$