Question

Calculate the Henry's law constant at $25^\circ\text{C}$ if solubility of gas in liquid is $2.1 \times 10^{-2} \text{ moldm}^{-3}$ at $0.18 \text{ bar}$.

• A. $0.1166 \text{ moldm}^{-3}\text{bar}^{-1}$
• B. $0.1445 \text{ moldm}^{-3}\text{bar}^{-1}$
• C. $0.1730 \text{ moldm}^{-3}\text{bar}^{-1}$
• D. $0.2014 \text{ moldm}^{-3}\text{bar}^{-1}$

Hint:

Henry's law: $k_H = \frac{\text{Solubility}}{\text{Pressure}}$

Answer 1

The Correct Option is A

Concept: Henry's law: \[ C = k_H \cdot P \Rightarrow k_H = \frac{C}{P} \]

Step 1: Substitute values. \[ k_H = \frac{2.1 \times 10^{-2}}{0.18} \]

Step 2: Calculate. \[ k_H = \frac{0.021}{0.18} = 0.1166 \]

Step 3: Conclusion.
Thus, Henry's constant = $0.1166 \text{ moldm}^{-3}\text{bar}^{-1}$. Final Answer: Option (A)