If the solubility of a gas in water at $0^{\circ} C$ and 1 atm is $1.5 g / L$ , then calculate the solubility of the same gas in water at $0^{\circ} C$ and a pressure of 2 atm.

3 g / L

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0.75 g / L

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1.5 g / L

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Cannot be determined by the given data.

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Solution

The correct option is A $3 g / L$
The solubility of gas is proportional to the pressure of the gas above the liquid.
$P_{1} = 1 a t m Solubility, S_{1} = 1.5 g / L P_{2} = 2 a t m S_{2} = ?$

From Henry's law,
$P_{g a s} = K_{H} X_{g a s}$
where, $Xgas=mole fraction of gas in liquid$
$P_{g a s} is the partial pressure of gas over liquid surface$
$K_{H}$ is constant at same temperature
So,
$Solubility \propto pressure$

$\frac{S_{1}}{S_{2}} = \frac{P_{1}}{P_{2}} \Rightarrow \frac{1.5 g / L}{S_{2}} = \frac{1}{2}$
$\Rightarrow S_{2} (solubility at 2 atm) = \frac{2}{1} \times 1.5$
$= 3 g / L$

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