Question

The vapour density of a mixture of $N{O}_2$ and $N_2{O}_4$ at ${27}^{\circ }\text{C}$ is 38.3. How many moles of $N{O}_2$ will be present in $500\text{g}$ of the mixture?

• A. $1.825\text{mol}$
• B. $3.485\text{mol}$
• C. $8.125\text{mol}$
• D. $2.185\text{mol}$

Answer 1

The correct option is D $2.185\text{mol}$

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 $\left(i\right)$
$2\times 38.3=\frac{500}{\frac{x}{46}+\frac{500-x}{92}}$
$\frac{x}{46}+\frac{500-x}{92}=\frac{500}{2\times 38.3};\frac{2x+500-x}{92}=\frac{500}{76.6}$
$(x+500)=\frac{500\times 92}{76.6}=600.52$
$x=600.52-500=100.52$
Moles of $N{O}_2=\frac{100.52}{46}=2.1852\text{mol}$