A gas X is dissolved in water at 2 bar press | vedclass

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(C) According to Henry's Law,$p = K_{H} 	imes \chi_{X}$,where $p$ is the partial pressure of the gas and $\chi_{X}$ is its mole fraction in the solut

Vedclass · Accepted Answer

$0.96$

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(C) According to Henry's Law,$p = K_{H} 	imes \chi_{X}$,where $p$ is the partial pressure of the gas and $\chi_{X}$ is its mole fraction in the solution. For the initial condition: $2 = K_{H} 	imes 0.02$ ...$(i)$ When the pressure is doubled,$p' = 4 \ bar$. $4 = K_{H} 	imes \chi'_{X}$ ...$(ii)$ Dividing equation $(ii)$ by $(i)$: $\frac{4}{2} = \frac{\chi'_{X}}{0.02}$ $\Rightarrow 2 = \frac{\chi'_{X}}{0.02}$ $\Rightarrow \chi'_{X} = 0.04$. The sum of mole fractions in a binary solution is $1$. Therefore,the mole fraction of water is $\chi_{water} = 1 - \chi'_{X} = 1 - 0.04 = 0.96$.

$A$ gas $X$ is dissolved in water at $2 \ bar$ pressure. Its mole fraction in the solution is $0.02$. Find the mole fraction of water in the solution when the pressure of the gas is doubled at the same temperature.

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