\( 5.5 \mathrm{mg} \) of nitrogen gas dissolves in \( 180 \mathrm{~g} \) of water at \( 273 \mathrm{~K} \) and one atm pressure due to
nitrogen gas. The mole fraction of nitrogen in \( 180 \mathrm{~g} \) of water at \( 5 \mathrm{~atm} \) nitrogen pressure is
approximately

No. of moles of \( N_{2}=\frac{5.5 \times 10^{-3}}{28} \) moles
\[
\begin{array}{l}
=2.36 \times 10^{-4} \text { moles } \\
p_{N_{2}}=k_{H} \times \frac{n_{N_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}}\left(x_{N_{2}}=\frac{n_{N_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}} ; n_{\mathrm{H}_{2} \mathrm{O}} \gg n_{N_{2}}\right) \\
1=k_{H} \times \frac{2.36 \times 10^{-4}}{10} \\
k_{H}=\frac{10 \times 10^{4}}{2.36}=4.2 \times 10^{4}
\end{array}
\]
Now, at \( 5 \mathrm{~atm} \) pressure of nitrogen gas, the mole fraction of nitrogen gas is,
\[
\begin{array}{l}
x_{N_{2}}=\frac{p_{N_{2}}}{k_{H}} \\
\Rightarrow x_{N_{2}}=\frac{5}{4.2 \times 10^{4}} \\
\Rightarrow x_{N_{2}}=1.12 \times 10^{-4} \approx 1 \times 10^{-4}
\end{array}
\]