Question

What is Henry's law constant of a gas if solubility of gas in water at 25$\circ$C is 0.028 mol dm$^-3$? [Partial pressure of gas = 0.346 bar]}

• A. 0.081 mol dm$^-3$bar$^-1$
• B. 0.075 mol dm$^-3$bar$^-1$
• C. 0.093 mol dm$^-3$bar$^-1$
• D. 0.049 mol dm$^-3$bar$^-1$

Hint:

Remember Henry's Law: $S = k_H P$. To find $k_H$, simply divide the solubility ($S$) by the partial pressure ($P$). Ensure units are consistent ($S$ in mol dm$^{-3}$, $P$ in bar, so $k_H$ in mol dm$^{-3}$bar$^{-1}$).

Answer 1

The Correct Option is D

Step 1: State Henry's Law equation. Henry's Law states that the solubility of a gas ($S$) in a liquid at a particular temperature is directly proportional to the partial pressure ($P$) of the gas above the liquid. The mathematical expression is: \[ S = k_H P \] Where:

• $S$ is the solubility of the gas (in mol dm$^{-3}$).
• $P$ is the partial pressure of the gas (in bar).
• $k_H$ is Henry's law constant (in mol dm$^{-3}$bar$^{-1}$).

Step 2: Rearrange the equation to solve for Henry's law constant ($k_H$). To find $k_H$, we can rearrange the equation: \[ k_H = \frac{S}{P} \]
Step 3: Substitute the given values into the equation. Given:

• Solubility ($S$) = 0.028 mol dm$^{-3}$
• Partial pressure ($P$) = 0.346 bar

Substitute these values into the rearranged equation: \[ k_H = \frac{0.028 \text{ mol dm}^{-3{0.346 \text{ bar \]
Step 4: Calculate the value of $k_H$. Perform the division: \[ k_H \approx 0.0809248... \text{ mol dm}^{-3}\text{bar}^{-1} \]
Step 5: Compare the calculated value with the given options. The calculated value $k_H \approx 0.08092 \text{ mol dm}^{-3}\text{bar}^{-1}$ is closest to option A, which is $0.081 \text{ mol dm}^{-3}\text{bar}^{-1}$.
Step 6: Conclude the correct answer. Therefore, the Henry's law constant is $0.081 \text{ mol dm}^{-3}\text{bar}^{-1}$.