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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 \(25^{\circ} \mathrm{C}\).

  1. A \(0.45 \mathrm{~mol} \mathrm{dm}^{-3}\)
  2. B \(0.54 \mathrm{~mol} \mathrm{dm}^{-3}\)
  3. C \(0.25 \mathrm{~mol} \mathrm{dm}^{-3}\)
  4. D \(0.31 \mathrm{~mol} \mathrm{dm}^{-3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.54 \mathrm{~mol} \mathrm{dm}^{-3}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \text { Solubility }(\mathrm{s})=\mathrm{k} \times \mathrm{p} \\ & =0.45 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1} \times 1.2 \mathrm{~atm} \\ & =0.54 \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}\)