Calculate the pressure at which solubility of a certain gas in water is 0.165 mol L^ -1 at 25\ ^ ^ C. (Henry's law constant for this gas in water is 0.034 mol L^ -1 bar^ -1 at 25\ ^ ^ C.)

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 4.85 bar

According to Henry's law, the solubility of a gas in a liquid is given by:

S = K_{H} \times P

Given:

S = 0.165 \text{ mol L}^{-1}

K_{H} = 0.034 \text{ mol L}^{-1} \text{ bar}^{-1}

Substituting the values:

0.165 = 0.034 \times P

P = \dfrac{0.165}{0.034}

P = 4.85 \text{ bar}

Answer:

4.85

bar

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