Question

Calculate the percentage by mole of ${\mathrm{NO}}_2$ in a mixture of ${\mathrm{NO}}_2\left(\mathrm{g}\right)$ and $\mathrm{NO}\left(\mathrm{g}\right)$ having average molecular mass equal to $34$?

• A. $20\%$
• B. $25\%$
• C. $40\%$
• D. $75\%$

Answer 1

The correct option is B

$25\%$

Explanation for correct option:

Option (b) $25\%$

Step 1: Create algebraic equations by using average mass of ${\mathrm{NO}}_2$ and $\mathrm{NO}$

Let's assume that the mole fraction of ${\mathrm{NO}}_2$ is ‘$\mathrm{x}$’ and $\mathrm{NO}$ is ‘$\mathrm{y}$’.

Now let's assume that the container contains only these two gases. So we can say that,

$\mathrm{x}+\mathrm{y}=1$

Now we know that molecular weight of ${\mathrm{NO}}_2$ is $46\mathrm{g}{\mathrm{mol}}^{-1}$ and that of $\mathrm{NO}$ is $30\mathrm{g}{\mathrm{mol}}^{-1}$.

Now, the average molecular mass is given by

$\frac{46\times \mathrm{x}+30\times \mathrm{y}}{2}=34$

Step 2: Find The percentage by mole of ${\mathrm{NO}}_2$

Now solving the two equations we find that ,

$\mathrm{x}=0.25\mathrm{and}\mathrm{y}=0.75$

Percentage by mole of

$$\begin{gathered} {\mathrm{NO}}_2=\mathrm{x}\times 100=0.25\times 100=25\% \\ {\mathrm{NO}}_2=25\% \end{gathered}$$

So, the percentage by mole of ${\mathrm{NO}}_2$ is $25\%$