Henry's law constant for the solubility of \(\mathrm{N}_{2}\) gas in water at \(298 \mathrm{~K}\) is \(1.0 \times 10^{5} \mathrm{~atm}\). The mole fraction of \(\mathrm{N}_{2}\) in air is \(0.8\). The number of moles of \(\mathrm{N}_{2}\) from air dissolved in 10 moles of water at \(298 \mathrm{~K}\) and 5 atm pressure is

Given, Henry's law constant \(\left(K_{\mathrm{H}}\right)\) for the solubility of \(\mathrm{N}_{2}\) gas in water at \(298 \mathrm{~K}=1 \times 10^{5}\) atm.
Mole fraction of \(\mathrm{N}_{2}\left(\chi_{\mathrm{N}_{2}}\right)=0.8\)
Hence, partial pressure of nitrogen,
\(\begin{aligned}
p_{\mathrm{N}_{2}} &=p_{\text {total }} \cdot \chi_{\mathrm{N}_{2}} \\
&=0.8 \times 5 \mathrm{~atm} \\
&=4 \mathrm{~atm}
\end{aligned}\)
According to Henry's law,
\(\begin{aligned}
p_{\mathrm{N}_{2}} &=K_{\mathrm{H}} \chi_{\mathrm{N}_{2}} \\
4 &=10^{5} \cdot \chi_{\mathrm{N}_{2}} \\
\chi_{\mathrm{N}_{2}} &=4 \times 10^{-5}
\end{aligned}\)
We know that,
\(\begin{aligned}
\chi_{\mathrm{N}_{2}} &=\frac{n_{\mathrm{N}_{2}}}{n_{\mathrm{N}_{2}}+n_{\mathrm{H}_{2} \mathrm{O}}} \\
\Rightarrow \quad 4 \times 10^{-5} &=\frac{n_{\mathrm{N}_{2}}}{10} \quad\left(\because n_{\mathrm{N}_{2}}<< < n_{\mathrm{H}_{2} \mathrm{O}}\right) \\
\Rightarrow \quad n_{\mathrm{N}_{2}} &=4 \times 10^{-4} \text { moles }
\end{aligned}\)
Thus, the number of moles of \(\mathrm{N}_{2}\) from air dissolved in 10 moles of water at \(295 \mathrm{~K}\) and 5 atm pressure is \(4 \times 10^{-4}\).