Question

The vapour density of a mixture containing $N{O}_2$ and $N_2{O}_4$ is $38.3$ at $300K$. The number of moles of $N{O}_2$ in $100g$ of the mixture is approximate:

• A. $0.44$
• B. $4.4$
• C. $33.4$
• D. $3.34$

Answer 1

Answer: (A)

$0.44$

Let us asssume that $x$ moles of $N{O}_2$ in mixture.

Hence, there will be $(100-x)$ moles of $N_2{O}_4$ in mixture . (since total no. of moles is $100.$)

$1$ mole of $N{O}_2$ weight $46g=(14+2\times 16)gms$

$1$ mole of $N_2{O}_4$ weight $92g=(2\times 14+4\times 16)gms$

Total molecular mass of mixturee $=\left(x\right)\left(46\right)+(100-x)92$

We know that,

Molecular mass$=2\left(vapordensity\right)=2\times \left(38.3\right)=76.6g$

Molar mass of mixture$=\frac{massofmixture}{totalno.ofmoles}76.6=\frac{46x+(100-x)92}{100}(100-x)92+46x=7660x=33.48$

$33.48$ moles of $N{O}_2$ are in mixture.