MHT CET · Chemistry · Solutions
Calculate the solubility of a gas in water at \(0.8 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\)
[Henry's law constant is \(6.85 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1}\) ]
- A \(5.48 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
- B \(3.94 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
- C \(6.858 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
- D \(2.74 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
Answer & Solution
Correct Answer
(A) \(5.48 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & S=P k \\ & =0.8 \times 6.85 \times 10^{-4} \\ & =5.48 \times 10^{-4} \mathrm{~mol} / \mathrm{dm}^{-3}\end{aligned}\)
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