Find the amount of ${\mathrm{CO}}_2\text{ gas }$in 250 mL of soda water when packed under 3.0 atm ${\mathrm{CO}}_2$ at 25⁰C. 

(K_H for CO₂  in water $\left.=1.67\times {10}^8\mathrm{Pa}\right)$

Henry's Law expression:

${\mathrm{p}}_{{\mathrm{CO}}_2}={\mathrm{K}}_{\mathrm{H}}{\mathrm{X}}_{{\mathrm{CO}}_2}$

Putting the given values in the above expression:

${\mathrm{X}}_{{\mathrm{CO}}_2}=\frac{3.0 \mathrm{atm}\times 1.01325\times {10}^5}{1.67\times {10}^8 \mathrm{Pa}}$

Mole fraction of CO₂:

$\mathrm{Mole} \mathrm{fraction}=\frac{\mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2}{\mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2+\mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{water}}$

Mass of water= volume x density=250 mL x 1 g/mL=250 g

Number of moles of water:

$\mathrm{Moles} \mathrm{of} \mathrm{water}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}=\frac{250 \mathrm{g}}{18 \mathrm{g}/\mathrm{mol}}=13.89 \mathrm{mol}$

Determining the moles of CO₂:

$$\begin{gathered} \frac{3.0 \mathrm{atm}\times 1.01325\times {10}^5}{1.67\times {10}^8}=\frac{\mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2}{\mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2+13.89} \\ \mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2=0.025 \mathrm{mol} \end{gathered}$$

Mass of CO₂ is calculated as:

$$\begin{gathered} 0.025 \mathrm{mol}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}} \\ 0.025 \mathrm{mol}=\frac{\mathrm{mass}}{44.01 \mathrm{g}/\mathrm{mol}} \\ \mathrm{mass}=1.1 \mathrm{g} \end{gathered}$$