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Henry's Law

Concentration...

Question

Concentration of $C O_{2}$ at $25^{\circ} C$ in a can is equal to 0.068 molal. Thus, pressure of $C O_{2}$ gas in the can is:
( $K_{H} = 0.034 m o l / k g . b a r$ )

1 b a r

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0.5 b a r

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1.5 b a r

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2 b a r

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Solution

The correct option is D $2 b a r$
Henry Law : The solubility of a gas in a liquid at a given temperature is directly proportional to its partial pressure of the gas present above the solution.
Solubility of gas can be expressed in mole fraction based on the unit of $K_{H}$ .
$S = K_{H} P_{C O_{2}}$
Where,
S is solubility of gas
$K_{H}$ is Henry's constant
$P_{C O_{2}}$ is partial pressure of the gas
$∴ P_{C O_{2}} = \frac{Solubility}{K_{H}} = \frac{0.068 m o l k g^{- 1}}{0.034 m o l k g^{- 1} b a r^{- 1}} P_{C O_{2}} = 2 b a r$

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