What is Henry's law constant of a gas if solubility of gas in water at $25^{\circ} \mathrm{C}$ is $0.028…

What is Henry's law constant of a gas if solubility of gas in water at $25^{\circ} \mathrm{C}$ is $0.028 \mathrm{~mol} \mathrm{dm}^{-3}$ ? [Partial pressure of gas $=0.346 \mathrm{bar}]$
  1. $0.081 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$
  2. $0.075 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$
  3. $0.093 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$
  4. $0.049 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$

Solution

$\mathrm{K}_{\mathrm{H}}=\frac{\mathrm{S}}{\mathrm{P}}=\frac{0.028}{0.346}=0.081 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$

Asked in: MHT CET 2023 (14 May Shift 2)

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