State Henry's law. Calculate the mole fraction of $CO_2$ in water at $298\,K$ under $760\,mm\,Hg$. (Given : $K_H$ for $CO_2$ in $H_2O$ at $298\,K = 1.25 \times 10^6\,mm\,Hg$)

Question

Given below are two statements. Assertion (A): Molar solubility of a gas in a liquid decreases with increase in pressure at constant temperature. Reason (R): Henry's law describes that at constant temperature, the partial pressure of a gas in vapour phase is directly proportional to the mole fraction of the gas in solution. Choose the most appropriate answer.

Answer 1

Step 1: State Henry's law.
At constant temperature, the partial pressure of a gas above a solution is directly proportional to the mole fraction of the gas dissolved in it: $p = K_H x$, where $K_H$ is Henry's law constant. Higher pressure means greater solubility.
Step 2: Identify the given values.
Partial pressure of $CO_2$: $p = 760\,mm\,Hg$. Henry's constant at $298\,K$: $K_H = 1.25 \times 10^6\,mm\,Hg$.
Step 3: Calculate the mole fraction and interpret.
Rearranging Henry's law: \[ x_{CO_2} = \frac{p}{K_H} = \frac{760}{1.25 \times 10^6} = 6.08 \times 10^{-4} \] This very small value confirms that $CO_2$ dissolves sparingly in water at atmospheric pressure; solubility would increase proportionally with higher pressure.
\[ \boxed{x_{CO_2} = 6.08 \times 10^{-4}} \]

State Henry's law. Calculate the mole fraction of \(CO_2\) in water at \(298\,K\) under \(760\,mm\,Hg\). (Given : \(K_H\) for \(CO_2\) in \(H_2O\) at \(298\,K = 1.25 \times 10^6\,mm\,Hg\))

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