JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
At 298 K, the mole percentage of \(N_{2}(g)\) in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of \(N_{2}(g)\) in water at 298 K ? (\(K_{H}\) for \(N_{2}\) is \(6.5 \times 10^{7}\) mm Hg)
- A \(1.23 \times 10^{-7}\)
- B \(1.17 \times 10^{-4}\)
- C \(9.35 \times 10^{5}\)
- D \(9.35 \times 10^{-5}\)
Answer & Solution
Correct Answer
(D) \(9.35 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
\(P _{ N _2}= K _{ H } \cdot X _{ N _2}\) \(P_{N_{2}} = 0.8 \times 10 = 8\) atm \(8 \times 760 = 6.5 \times 10^{7} \times X_{N_{2}}\) \(X _{ N _2}=\frac{8 \times 760}{6.5 \times 10^7}\) \(X _{ N _2}=9.35 \times 10^{-5}\)
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