Question

The mole percentage of oxygen in a mixture of $7g{N}_2$ and $8g{O}_2$ is :

• A. 25
• B. 75
• C. 50
• D. 40

Answer 1

The correct option is C 50

7 g of $N_2\Rightarrow$ Number of moles of

$N_2=n_{N_2}=\frac{7}{28}=\frac{1}{4}$

8 g of $O_2\Rightarrow$ Number of moles of

$O_2=n_{O_2}=\frac{8}{32}=\frac{1}{4}$

Mole of percentage of $O_2=$ mole fraction of

$O_2\times 100=\frac{n_{O_2}}{n_{O_2}+n_{N_2}}\times 100.....\left(1\right)$

Substituting the values from the data in the equation (1), we get

Mole percentage of oxygen

$=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{4}}\times 100=\frac{\frac{1}{4}}{\frac{1}{2}}\times 100=50$