Q22 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Chemistry
The Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1 \times 10^{+5} \mathrm{~atm}$. The mole fraction of 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
- A$4 \times 10^{-5}$
- Correct answerB$4 \times 10^{-4}$
- C$5 \times 10^{-4}$
- D$4 \times 10^{-6}$
Solution
At total pressure $5 \mathrm{~atm}$, the partial pressure of $\mathrm{N}_2=5 \times 0.8=4 \mathrm{~atm}$
According to Henry's Law,
$$
\mathrm{P}_{\mathrm{N}_2}=\mathrm{K}_{\mathrm{H}} \mathrm{x}_{\mathrm{N}_2}
$$
$\left(\mathrm{x}_{\mathrm{N}_2}\right)$ is the mole fraction of nitrogen gas dissolved in water.
or, $\quad \mathrm{x}_{\mathrm{N}_2}=\frac{4 \mathrm{~atm}}{1 \times 10^5 \mathrm{~atm}}$ or, $\mathrm{x}_{\mathrm{N}_2}=4 \times 10^{-5}$
$$
\begin{aligned}
& \frac{\mathrm{n}_{\mathrm{N}_2}}{\mathrm{n}_{\mathrm{N}_2}+\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}} \simeq \frac{\mathrm{n}_{\mathrm{N}_2}}{\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}}=4 \times 10^{-5}\left[\text { since } \mathrm{n}_{\mathrm{H}_2 \mathrm{O}} \gg \mathrm{n}_{\mathrm{N}_2}\right] \\
& \therefore \quad \mathrm{n}_{\mathrm{N}_2}=4 \times 10^{-5} \times 10=4 \times 10^{-4} \mathrm{~mol}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.