Question

The vapour density of a mixture containing $N{O}_2$ and $N_2{O}_4$ is 38.3 at ${27}^{0}C$ .The mole of $N{O}_2$ in 100 gm mixture is:

• A. 0.437
• B. 0.347
• C. 0.728
• D. cannnot be predicted

Answer 1

The correct option is C 0.728

Given,

Vapour Density of mixture having $N{O}_2$ & $N_2{O}_4$=$38.3$

$\Rightarrow$ Molar mass of mixture=$2\times V.D$

$=2\times 38.3$

$=76.6$

Now, Let us consider in a $100$% mixture

$x$% $N_2{O}_4$ is present $\Rightarrow N{O}_2$ is $(100-x)$%

$\Rightarrow N_2{O}_4$ is $x$ & $N{O}_2$ will be $1-x$

Now, $x($Molar mass of $N_2{O}_4)+(1-x\left)\right($Molar mass of $N{O}_2$)=$76.6$

$\Rightarrow x\left(92\right)+(1-x)\left(46\right)=76.6$

$\Rightarrow x=0.665$

$\therefore$ Percentage of $N_2{O}_4=66.5$%

$\Rightarrow$ Percentage of $N{O}_2$=$33.5$% $⟶(100-x)$

This shows that in $100g$ mixture $33.5g$ of $N{O}_2$ will be present

That means, No. of moles of $N{O}_2$ present= $\frac{33.5g}{46g/mol}$

$=0.728mol$