MHT CET · Chemistry · Solutions
What is Henry's law constant if solubility of a gas in water at \(298 \mathrm{~K}\) and 1 bar pressure is \(7 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
- A \(2.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{bar}^{-1}\)
- B \(7.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{bar}^{-1}\)
- C \(3.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{bar}^{-1}\)
- D \(3.1 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{bar}^{-1}\)
Answer & Solution
Correct Answer
(B) \(7.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{bar}^{-1}\)
Step-by-step Solution
Detailed explanation
Solubility \(=\mathrm{K}_{\mathrm{H}} \times\) pgas \(7 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}=\mathrm{K}_{\mathrm{H}} \times 1\) bar \(\mathrm{K}_{\mathrm{H}}=7 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \cdot \mathrm{bar}^{-1}\)
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