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MHT CET · Chemistry · Solutions

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. A \(0.081 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
  2. B \(0.075 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
  3. C \(0.093 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
  4. D \(0.049 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.081 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)

Step-by-step Solution

Detailed explanation

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