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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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Verified Answer
The correct answer is:
0.04
Pressure, $p=2$ bar
Mole fraction,
$$
\begin{aligned}
& \chi=0.02 \\
& p=K_{\mathrm{H}} \times \chi (Henry law)\\
& 2=K_{\mathrm{H}} \times 0.02...(i)
\end{aligned}
$$
Pressure doubled.
$$
\begin{aligned}
& p=K_{\mathrm{H}} \times \chi \\
& 4=K_{\mathrm{H}} \times \chi...(ii)
\end{aligned}
$$
From Eqs. (i) and (ii)
$$
\frac{1}{2}=\frac{0.02}{\chi} \Rightarrow \chi=0.04
$$
Mole fraction,
$$
\begin{aligned}
& \chi=0.02 \\
& p=K_{\mathrm{H}} \times \chi (Henry law)\\
& 2=K_{\mathrm{H}} \times 0.02...(i)
\end{aligned}
$$
Pressure doubled.
$$
\begin{aligned}
& p=K_{\mathrm{H}} \times \chi \\
& 4=K_{\mathrm{H}} \times \chi...(ii)
\end{aligned}
$$
From Eqs. (i) and (ii)
$$
\frac{1}{2}=\frac{0.02}{\chi} \Rightarrow \chi=0.04
$$
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