$\mathrm{K}_{\mathrm{H}}$ value for some gases at the same temperature ' $\mathrm{T}$ ' are given:…

$\mathrm{K}_{\mathrm{H}}$ value for some gases at the same temperature ' $\mathrm{T}$ ' are given: $\begin{array}{c|c}
\text {gas } & \mathrm{K}_{\mathrm{H}} / \mathbf{k} \text { bar } \\
\hline \mathrm{Ar} & 40.3 \\
\mathrm{CO}_2 & 1.67 \\
\mathrm{HCHO} & 1.83 \times 10^{-5} \\
\mathrm{CH}_4 & 0.413
\end{array}$ where $\mathrm{K}_{\mathrm{H}}$ is Henry's Law constant in water. The order of their solubility in water is :
  1. $\mathrm{Ar} < \mathrm{CO}_2 < \mathrm{CH}_4 < \mathrm{HCHO}$
  2. $\mathrm{Ar} < \mathrm{CH}_4 < \mathrm{CO}_2 < \mathrm{HCHO}$
  3. $\mathrm{HCHO} < \mathrm{CO}_2 < \mathrm{CH}_4 < \mathrm{Ar}$
  4. $\mathrm{HCHO} < \mathrm{CH}_4 < \mathrm{CO}_2 < \mathrm{Ar}$

Solution

$\mathrm{K}_{\mathrm{H}}=$ Henry Law constant of gas in water, released to solubility as: Solubility of a gas $\propto \frac{1}{\mathrm{~K}_{\mathrm{H}} \text { Value }}$ This means higher is the value of $\mathrm{K}_{\mathrm{H}}$, lower will be the solubility of gas in water. $\therefore$ Solubility order of given gas in water $=\mathrm{HCHO}>\mathrm{CH}_4>\mathrm{CO}_2>\mathrm{Ar}$

Asked in: NEET 2022 (Phase 2)

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