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(B) The number of moles of $N_2$ is $n_1 = \frac{7 \ g}{28 \ g/mol} = 0.25 \ mol$.The number of moles of $O_2$ is $n_2 = \frac{8 \ g}{32 \ g/mol} = 0.

Vedclass · Accepted Answer

$X_{N_2} = X_{O_2}$

Vedclass · Answer

(B) The number of moles of $N_2$ is $n_1 = \frac{7 \ g}{28 \ g/mol} = 0.25 \ mol$.The number of moles of $O_2$ is $n_2 = \frac{8 \ g}{32 \ g/mol} = 0.25 \ mol$.The mole fraction of $N_2$ is $X_{N_2} = \frac{n_1}{n_1 + n_2} = \frac{0.25}{0.25 + 0.25} = 0.5$.The mole fraction of $O_2$ is $X_{O_2} = \frac{n_2}{n_1 + n_2} = \frac{0.25}{0.25 + 0.25} = 0.5$.Therefore,$X_{N_2} = X_{O_2}$.

Choose the correct option for the mole fraction of both components in a gaseous mixture containing $7 \ g$ of nitrogen and $8 \ g$ of oxygen.

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