Question

The vapour density of a mixture of $N{O}_2$ and $N_2{O}_4$ at ${27}^{\circ }\text{C}$ is 36.8. How many moles of $N{O}_2$ will be present in $92\text{g}$ of the mixture?

• A. $0.4\text{mol}$
• B. $0.8\text{mol}$
• C. $0.6\text{mol}$
• D. $0.5\text{mol}$

Answer 1

The correct option is D $0.5\text{mol}$

We know that:
$Vapourdensityofthemixture=\frac{averagemolarmass}{2}\left(i\right)$
Let the mass of $N{O}_2=x$
$\text{Moles of}N{O}_2+\text{moles of}{N}_2{O}_4=\text{total moles of the mixture}$
$\text{Average molar mass}=\frac{\text{Total mass}}{\text{Total moles}}$
Substituting in equation $\left(i\right)$
$2\times 36.8=\frac{92}{\frac{x}{46}+\frac{92-x}{92}}$
$\frac{x}{46}+\frac{92-x}{92}=\frac{92}{2\times 36.8};\frac{2x+92-x}{92}=\frac{92}{73.6}$
$(x+92)=\frac{92\times 92}{73.6}=115$
$x=115-92=23$
Moles of $N{O}_2=\frac{23}{46}=0.5\text{mol}$