Question

Henry's law constant for $CH_3Br$ is $0.16\ \text{mol}\ \text{L}^{-1}\text{bar}^{-1}$ at $298\ \text{K}$. What pressure is required to have solubility of $0.08\ \text{mol}\ \text{L}^{-1}$?

• A. $0.24\ \text{bar}$
• B. $1.6\ \text{bar}$
• C. $0.5\ \text{bar}$
• D. $4.0\ \text{bar}$

Hint:

Always check the units of Henry's Law constant. If it is given in $\text{mol}\ \text{L}^{-1}\text{bar}^{-1}$, use $S = K_H \times P$. If it is given in bar (or atm), the formula is $P = K_H \times x$ (where $x$ is mole fraction).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the Henry's law constant ($K_H$) and the desired solubility ($S$) of a gas ($CH_3Br$) in a liquid, and we must calculate the required external pressure ($P$).

Step 2: Key Formula or Approach:
According to Henry's Law, the solubility ($S$) of a gas in a liquid is directly proportional to the partial pressure ($P$) of that gas above the liquid. The mathematical relation is:
$$S = K_H \times P$$ Rearranging to solve for pressure ($P$) gives:
$$P = \frac{S}{K_H}$$

Step 3: Detailed Explanation:
Substitute the given values into the rearranged equation:
Solubility $S = 0.08\ \text{mol}\ \text{L}^{-1}$
Henry's law constant $K_H = 0.16\ \text{mol}\ \text{L}^{-1}\text{bar}^{-1}$
$$P = \frac{0.08}{0.16}$$ To simplify, shift the decimals two places to the right for both numbers:
$$P = \frac{8}{16} = \frac{1}{2}$$ $$P = 0.5\ \text{bar}$$

Step 4: Final Answer:
The required pressure is $0.5\ \text{bar}$, matching option (C).