Question

Water in a container at 290 K is exposed to air containing \(3\ %\) \(\mathrm{CO_2}\) by volume. Air behaves like an ideal gas and is maintained at \(100\ \text{kPa}\) pressure. The liquid phase comprising of dissolved \(\mathrm{CO_2}\) in water behaves like an ideal solution. Use Henry’s constant of \(\mathrm{CO_2}\) dissolved in water at 290 K as \(12\ \text{MPa}\). Under equilibrium conditions, which one of the following is the CORRECT value of the mole fraction of \(\mathrm{CO_2}\) dissolved in water?

• A. \(2.9 \times 10^{-4}\)
• B. \(0.9 \times 10^{-4}\)
• C. \(2.5 \times 10^{-4}\)
• D. \(0.5 \times 10^{-4}\)

Hint:

When \(H\) is given in pressure units, use \(x_A=p_A/H\); keep units consistent (kPa–kPa or Pa–Pa).
For gas mixtures behaving ideally, \(p_A=y_A P\).

Answer 1

The Correct Option is C

Step 1: Find the partial pressure of \(\mathrm{CO_2}\) in the gas: \(p_{CO_2}=y_{CO_2}P=0.03\times 100\ \text{kPa}=3\ \text{kPa}\).
Step 2: Apply Henry’s law in the form \(x_{CO_2}=\dfrac{p_{CO_2}}{H}\) with \(H=12\ \text{MPa}=12000\ \text{kPa}\): \[ x_{CO_2}=\frac{3}{12000}=2.5\times 10^{-4}. \] Hence option (C).