A mixture contains 16 g oxygen, 28 g of nitrogen and 8 g of methane. Total pressure of the mixture is 740 mm. What is the partial pressure $\left(\mathrm{mm}\right)$ of nitrogen In mm?

Number of moles of O₂:

$$\begin{gathered} \mathrm{Moles}=\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{mass}} \\ \mathrm{Moles}=\frac{16 \mathrm{g}}{32 \mathrm{g}/\mathrm{mol}}=0.5 \mathrm{mol} \end{gathered}$$

Number of moles of N₂:

$$\begin{gathered} \mathrm{Moles}=\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{mass}} \\ \mathrm{Moles}=\frac{28 \mathrm{g}}{28 \mathrm{g}/\mathrm{mol}}=1 \mathrm{mol} \end{gathered}$$

Number of moles of CH₄:

$$\begin{gathered} \mathrm{Moles}=\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{mass}} \\ \mathrm{Moles}=\frac{8 \mathrm{g}}{16 \mathrm{g}/\mathrm{mol}}=0.5 \mathrm{mol} \end{gathered}$$

$$\begin{gathered} \mathrm{Partial} \mathrm{pressure}=\mathrm{total} \mathrm{pressure}\times \mathrm{mole} \mathrm{fraction} \mathrm{of} {\mathrm{N}}_2 \\ \mathrm{Mole} \mathrm{fraction} \mathrm{of} {\mathrm{N}}_2=\frac{\mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{N}}_2}{\mathrm{total} \mathrm{number} \mathrm{of} \mathrm{moles}} \\ \mathrm{Mole} \mathrm{fraction} \mathrm{of} {\mathrm{N}}_2=\frac{1}{0.5+0.5+1}=\frac{1}{2} \\ \mathrm{Partial} \mathrm{pressure}=\frac{1}{2}\times 740=370 \mathrm{mm} \end{gathered}$$