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Question
The Henry′s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is
Options
A.4.0 × 10-4
B.4.0 × 10-5
C.5.0 × 10-4
D.4.0 × 10-6
Solution
P = KH χN2
0.8 × 5 = 1 × 105 × χN2
χN2 = 4 × 10-5 (in 10 moles of water)
⇒ 4 × 10-5 =
nN2 × 5 × 10-5 + 4 × 10-4 = n N2
⇒ nN2 × = 4 × 10-4
0.8 × 5 = 1 × 105 × χN2
χN2 = 4 × 10-5 (in 10 moles of water)
⇒ 4 × 10-5 =
nN2 × 5 × 10-5 + 4 × 10-4 = n N2
⇒ nN2 × = 4 × 10-4
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