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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 \(7.16 \times 10^{-5}\)
- C \(7.16 \times 10^{-4}\)
- D \(7.16 \times 10^{-3}\)
Answer & Solution
Correct Answer
(C) \(7.16 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
Henry law :- \(\quad \mathrm{P}=\mathrm{K}_{\mathrm{H}} \mathrm{X}\)
\(\mathrm{XN}_2=\frac{\mathrm{PN}_2}{\mathrm{~K}_4}=\frac{0.987}{76.48 \times 10^3}=1.29 \times 10^{-5}\)
\(\begin{aligned} & \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=\frac{1000}{18}=55.5 \\ & \mathrm{XN}_2=\frac{\mathrm{n}_{\mathrm{N}_2}}{\mathrm{n}_{\mathrm{N}_2}+\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}} \\ & 1.29 \times 10^{-5}=\frac{\mathrm{n}}{\mathrm{n}+55.5}=\frac{\mathrm{n}}{55.5} \\ & \mathrm{n}=1.29 \times 10^{-5} \times 55.5 \\ & \mathrm{n}_{\mathrm{N}_2}=7.16 \times 10^{-4}\end{aligned}\)
\(\mathrm{XN}_2=\frac{\mathrm{PN}_2}{\mathrm{~K}_4}=\frac{0.987}{76.48 \times 10^3}=1.29 \times 10^{-5}\)
\(\begin{aligned} & \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=\frac{1000}{18}=55.5 \\ & \mathrm{XN}_2=\frac{\mathrm{n}_{\mathrm{N}_2}}{\mathrm{n}_{\mathrm{N}_2}+\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}} \\ & 1.29 \times 10^{-5}=\frac{\mathrm{n}}{\mathrm{n}+55.5}=\frac{\mathrm{n}}{55.5} \\ & \mathrm{n}=1.29 \times 10^{-5} \times 55.5 \\ & \mathrm{n}_{\mathrm{N}_2}=7.16 \times 10^{-4}\end{aligned}\)
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