What is the mole percentage of $O_2$ in a mixture of $7 g$ of $N_2$ and $8{ g}_{ }$ of $O_2$ ?

 $\mathrm{moles}=\frac{\mathrm{given} \mathrm{mass}}{\mathrm{molecular} \mathrm{mass}}$

Moles of 7g of nitrogen is $\frac{7}{28}=0.25$

Moles of 8g of oxygen is $\frac{8}{32}=0.25$

Therefore, mole percentage of oxygen= $\frac{\mathrm{moles} \mathrm{of} \mathrm{oxygen}}{\mathrm{moles} \mathrm{of} \mathrm{oxygen}+\mathrm{moles} \mathrm{of} \mathrm{nitrogen}}$ x 100 = $\frac{0.25}{0.25+0.25}=0.5$ x 100 = 50%

Hence the correct option is (C).