Calculate the solubility of gas in water at 152 mm of Hg and at 300 K, if Henry's law constant for same gas at same temperature is 0.16 mol L − 1 atm − 1 .

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TestHub · Accepted Answer

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

$S = K_{H} \times P$

Given:

$K_{H} = 0.16 mol L^{- 1} atm^{- 1}$

$P = 152 mm of Hg = \frac{152}{760} atm = 0.2 atm$

Substituting the values:

$S = 0.16 \times 0.2 = 0.032 mol L^{- 1}$

$S = 3.2 \times 1 0^{- 2} mol L^{- 1}$

Answer: $3.2 \times 1 0^{- 2} mol/L$

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