Q30 / 150 · MHT CET 2024 (15 May Shift 1) · Chemistry

Calculate Henry's law constant if solubility of gas in water at $25^{\circ} \mathrm{C}$ is $5.14 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}$ and partial pressure of the gas is 0.75 bar above solution.

  • Correct answer
    A
    $6.85 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$
  • B
    $5.14 \times 10^{-4} \mathrm{moldm}^{-3} \mathrm{bar}^{-1}$
  • C
    $1.028 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$
  • D
    $1.371 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$

Solution

According to Henry's law, $\mathrm{S}=\mathrm{K}_{\mathrm{H}} \mathrm{P}$ $\begin{aligned} \mathrm{K}_{\mathrm{H}} & =\frac{\mathrm{S}}{\mathrm{P}}=\frac{5.14 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}}{0.75 \mathrm{bar}} \\ & =6.85 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1} \end{aligned}$

Hence, the correct answer is option: 'A'.

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