N2 gas is bubbled through water at 293 K and exerts a partial pressure of 0.987 bar. If the Henry's law constant for N2 at 293 K is 76,480 bar, the mole fraction of N2 in the solution is:
A1.29 x 10^-7
B7.75 x 10^4
C1.29 x 10^-3
D1.29 x 10^-5
Answer & Solution
Correct answer: D. 1.29 x 10^-5
1. By Henry's law, $x = p / K_H$.
2. $x = 0.987 \text{ bar} / 76{,}480 \text{ bar}$.
3. $x = 1.29 \times 10^{-5}$.
_Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.7_
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