Find the amount of ${\mathrm{CO}}_2$ gas in 250 mL of soda water when packed under $3.0 \mathrm{atm}$ ${\mathrm{CO}}_2$ at $25°\mathrm{C}$ (${\mathrm{K}}_{\mathrm{H}}$ for ${\mathrm{CO}}_2$ in  ${\mathrm{H}}_2\mathrm{O}=1.67\times {10}^8$ Pa)

Henry's formula:

$\mathrm{p}={\mathrm{K}}_{\mathrm{H}}\times \mathrm{x}$

First determine the mole fraction using the above expression:

$\mathrm{x}=\frac{3\times 1.013\times {10}^5}{1.6\times {10}^8}$

$$\begin{gathered} {\mathrm{x}}_{{\mathrm{CO}}_2}=\frac{ \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2}{\mathrm{moles} \mathrm{of} {\mathrm{CO}}_2+\mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{O}}\cong \frac{\mathrm{moles} \mathrm{of} {\mathrm{CO}}_2}{\mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{O}} \\ \frac{{\mathrm{n}}_{{\mathrm{CO}}_2}}{{\displaystyle \raisebox{1ex}{$1000$}\!\left/ \!\raisebox{-1ex}{$18$}\right.}}=\frac{3\times 1.013}{1.6\times 1000} \\ {\mathrm{n}}_{{\mathrm{CO}}_2}=\frac{30}{16}\times \frac{1.013}{18} \\ \mathrm{mass} \mathrm{of} {\mathrm{CO}}_2 \mathrm{dissolved}=\frac{30}{16}\times \frac{44\times 1.013}{18}=4.64 \mathrm{g} \end{gathered}$$