Q3 / 60 · JEE Advanced 2009 (Paper 1) · Chemistry
The Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 105 \mathrm{~atm}$. 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 of $298 \mathrm{~K}$ and $5 \mathrm{~atm}$ pressure is
- Correct answerA$4 \times 10^{-4}$
- B$4.0 \times 10^{-5}$
- C$5.0 \times 10^{-4}$
- D$4.0 \times 10^{-6}$
Solution
$P_{\mathrm{N}_2}=K_{\mathrm{H}} \times$ mole-fraction $\left(\mathrm{N}_2\right)$ mole-fraction
$$
\mathrm{N}_2 \frac{1}{10^5} \times 0.8 \times 5=4 \times 10^{-5} \mathrm{~mol}^{-1}
$$
In 10 mole solubility in $4 \times 10^{-4}$.
Hence, the correct answer is option: 'A'.