Henry's law constants for aqueous solution of $CO, O_{2}, {CO}_{2}$ and $C_{2} H_{2}$ gases are respectively at $25 ° C$ as
$58 \times 10^{3}, 43 \times 10^{3}, 1.61 \times 10^{3} and 1.34 \times 10^{3}$ . The solubility of these gases decreases in the order

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$O_{2} > {CO}_{2} > CO > C_{2} H_{2}$

$O_{2} > {CO}_{2} > C_{2} H_{2} > CO$

$C_{2} H_{2} > {CO}_{2} > O_{2} > CO$

$CO > O_{2} > {CO}_{2} > C_{2} H_{2}$

answer is C.

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Detailed Solution

Greater the Henry's constant $(K_{H}),$ lower is the solubility at the same pressure.

$∴$ Order of $K_{H} : CO > O_{2} > {CO}_{2} > C_{2} H_{2}$

$∴$ Order of solubility $: C_{2} H_{2} > {CO}_{2} > O_{2} > CO$

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