MHT CET · Chemistry · Solutions
Calculate the solubility of gas in water at \(1.2 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) if Henry's law constant is \(0.45 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1}\) at \(25^{\circ} \mathrm{C}\).
- A \(0.45 \mathrm{~mol} \mathrm{dm}^{-3}\)
- B \(0.54 \mathrm{~mol} \mathrm{dm}^{-3}\)
- C \(0.25 \mathrm{~mol} \mathrm{dm}^{-3}\)
- D \(0.31 \mathrm{~mol} \mathrm{dm}^{-3}\)
Answer & Solution
Correct Answer
(B) \(0.54 \mathrm{~mol} \mathrm{dm}^{-3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Solubility }(\mathrm{s})=\mathrm{k} \times \mathrm{p} \\ & =0.45 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1} \times 1.2 \mathrm{~atm} \\ & =0.54 \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}\)
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