Q48 / 180 · KCET 2025 · Chemistry
If $\mathrm{N}_2$ gas is bubbled through water at 293 K , how many moles of $\mathrm{N}_2$ gas would dissolve in 1 litre of water? Assume that $\mathrm{N}_2$ exerts a partial pressure of 0.987 bar.
[Given $\mathrm{K}_{\mathrm{H}}$ for $\mathrm{N}_2$ at 293 K is 76.48 K bar]
- A$0.716 \times 10^{-3}$
- B$7.16 \times 10^{-5}$
- Correct answerC$7.16 \times 10^{-4}$
- D$7.16 \times 10^{-3}$
Solution
Henry law :- $\quad \mathrm{P}=\mathrm{K}_{\mathrm{H}} \mathrm{X}$
$\mathrm{XN}_2=\frac{\mathrm{PN}_2}{\mathrm{~K}_4}=\frac{0.987}{76.48 \times 10^3}=1.29 \times 10^{-5}$
$\begin{aligned} & \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=\frac{1000}{18}=55.5 \\ & \mathrm{XN}_2=\frac{\mathrm{n}_{\mathrm{N}_2}}{\mathrm{n}_{\mathrm{N}_2}+\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}} \\ & 1.29 \times 10^{-5}=\frac{\mathrm{n}}{\mathrm{n}+55.5}=\frac{\mathrm{n}}{55.5} \\ & \mathrm{n}=1.29 \times 10^{-5} \times 55.5 \\ & \mathrm{n}_{\mathrm{N}_2}=7.16 \times 10^{-4}\end{aligned}$
Hence, the correct answer is option: 'C'.