Question

One mole of a mixture of $N_2,N{O}_2$ and $N_2{O}_4$ has a mean molar mass of $\mathrm{55.4.}$ On heating to a temperature at which $N_2{O}_4$ is completely dissociated: $N_2{O}_4\to 2N{O}_2$, the mean molar mass tends to a lower value of $39.57$. What is the mole ratio of $N_2:N{O}_2:N_2{O}_4$ in the original mixture?

• A. 0.4 : 0.2 : 0.4
• B. 0.5 : 0.1 : 0.4
• C. 0.6 : 0.2 : 0.2
• D. 0.4 : 0.4 : 0.2

Answer 1

The correct option is B 0.5 : 0.1 : 0.4

Let
$n_{N_2}=x$
$n_{N{O}_2}=y$
$n_{N_2{O}_4}=z$
Total moles $=1$
$x+y+z=1\left(i\right)$
mean molar mass before the reaction:
$\frac{28x+46y+92z}{x+y+z}=55.4\left(ii\right)$
Once the reaction happens, all the `z` moles of `N_2O_4` get converted into `2y` moles of `NO_2`
mean molar mass after the reaction:
$\frac{28x+46(y+2z)}{x+y+2z}=39.57\left(iii\right)$
Using `(ii)` and `(iii)`, we get `z=0.4`
Also, we get
$x=0.5$
$y=0.1$