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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 \(6.85 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1}\) ]

  1. A \(5.48 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
  2. B \(3.94 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
  3. C \(6.858 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
  4. D \(2.74 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(5.48 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & S=P k \\ & =0.8 \times 6.85 \times 10^{-4} \\ & =5.48 \times 10^{-4} \mathrm{~mol} / \mathrm{dm}^{-3}\end{aligned}\)