MHT CET · Chemistry · Solutions
Calculate the solubility of gas in water at \(260 \mathrm{~mm} \mathrm{Hg}\) and \(25^{\circ} \mathrm{C}\) if Henry's law constant of gas is \(0.159 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1}\) at \(25^{\circ} \mathrm{C}\)
- A \(3.8 \times 10^{-2} \mathrm{~mol} \mathrm{dm}-3\)
- B \(5.4 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3}\)
- C \(2.7 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3}\)
- D \(1.2 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3}\)
Answer & Solution
Correct Answer
(B) \(5.4 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & S=p K \\ & =\frac{260}{760} \times 0.159 \\ & =5.4 \times 10^{-2} \frac{\mathrm{mol}}{\mathrm{dm}^3}\end{aligned}\)
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