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 \(=0.346 \mathrm{bar}]\)
- A \(0.081 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
- B \(0.075 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
- C \(0.093 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
- D \(0.049 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
Answer & Solution
Correct Answer
(A) \(0.081 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{K}_{\mathrm{H}}=\frac{\mathrm{S}}{\mathrm{P}}=\frac{0.028}{0.346}=0.081 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}\)
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