Question

State Henry’s law. Calculate the mole fraction of CO\(_2\) in water at 298 K under 700 mm Hg pressure. (Given: Henry’s constant for CO\(_2\) in water at 298 K = \( 1.25 \times 10^6 \) mm Hg)

Hint:

Henry’s Law Shortcut: Higher Henry constant → Lower solubility of gas.

Answer 1

Henry’s Law: At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. \[ p = K_H \, x \] where:

• \( p \) = partial pressure of gas
• \( K_H \) = Henry’s constant
• \( x \) = mole fraction of gas in solution

Step 1: Write the formula. \[ x = \frac{p}{K_H} \]
Step 2: Substitute given values. \[ x = \frac{700}{1.25 \times 10^6} \]
Step 3: Simplify. \[ x = \frac{7 \times 10^2}{1.25 \times 10^6} = \frac{7}{1.25} \times 10^{-4} \] \[ x = 5.6 \times 10^{-4} \]
Final Answer: \[ \therefore \text{Mole fraction of CO}_2 = 5.6 \times 10^{-4} \]