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

Step 1: Understanding Henry’s Law.
Henry's Law states that the solubility of a gas in a liquid at a constant temperature is directly proportional to the partial pressure of the gas above the liquid.
This is represented by the equation: \[ p = K_H \, x \] where:

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

Step 2: Rearranging the formula to solve for \( x \).
To find the mole fraction \( x \), we can rearrange Henry's Law equation: \[ x = \frac{p}{K_H} \]
Step 3: Substitute the given values into the formula.
Substitute the values provided into the equation to calculate \( x \): \[ x = \frac{700}{1.25 \times 10^6} \]
Step 4: Simplify the expression.
Now, simplify the expression: \[ x = \frac{7 \times 10^2}{1.25 \times 10^6} \] \[ x = \frac{7}{1.25} \times 10^{-4} \] After performing the division: \[ x = 5.6 \times 10^{-4} \]
Step 5: Final Answer.
Thus, the mole fraction of CO₂ in the solution is: \[ \boxed{\text{Mole fraction of CO}_2 = 5.6 \times 10^{-4}} \]