Question

Henry's law constant for \(\mathrm{CO_2}\) in water is \(1.67 \times 10^{8}\) Pa at 298 K. Calculate the number of moles of \(\mathrm{CO_2}\) in 540 g of soda water when packed under \(3.34 \times 10^{5}\) Pa at the same temperature.

Hint:

\(x = p/K_H = 2.0\times10^{-3}\); 540 g water is 30 mol.
\(n_{\mathrm{CO_2}} = x \times n_{\mathrm{H_2O}} \approx 0.06\) mol.

Answer 1

Step 1: Understanding the Concept:
Henry's law says the partial pressure of a gas over a solution is proportional to the mole fraction of the gas in the solution: \(p = K_H \, x\).
We use it to find the mole fraction of \(\mathrm{CO_2}\), and then convert the mole fraction to moles.

Step 2: Find the mole fraction of CO2:
\[ x_{\mathrm{CO_2}} = \frac{p}{K_H} = \frac{3.34 \times 10^{5}}{1.67 \times 10^{8}} = 2.0 \times 10^{-3} \]

Step 3: Moles of water in 540 g:
Soda water is mostly water, so 540 g of it is taken as 540 g of water.
\[ n_{\mathrm{H_2O}} = \frac{540}{18} = 30 \ \mathrm{mol} \]

Step 4: Moles of CO2:
\[ x_{\mathrm{CO_2}} = \frac{n_{\mathrm{CO_2}}}{n_{\mathrm{CO_2}} + n_{\mathrm{H_2O}}} \approx \frac{n_{\mathrm{CO_2}}}{n_{\mathrm{H_2O}}} \]
since \(n_{\mathrm{CO_2}}\) is very small compared with \(n_{\mathrm{H_2O}}\).
\[ n_{\mathrm{CO_2}} = 2.0 \times 10^{-3} \times 30 = 0.06 \ \mathrm{mol} \]

Final Answer:
The soda water contains about \(0.06\) mol of \(\mathrm{CO_2}\).