Question

If N2 he bubbled through water at 293K,how many millimoles of N2 gas would dissolve in 1 litre of water?Assume that N2 exerts partial pressure of 0.987 bar.Given that henry's law constant for N2 at 293K is 76.48K bar.​

Answer 1

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Answer 2

Answer:

According to Henry's law

PN2 = KH . XN2

P = 0.987 bar

KH = 76.48Kbar = 76480 bar

so XN2 = PN2/KH = 0.987/76480 = 0.0000129 = 1.29 x 10^-5

No of moles of water in 1L = 1000/18 = 55.56

Let us take the moles of nitrogen as n

total moles = 55.56+n

XN2 = n/(n+55.56)

since  XN2 = PN2/KH so  1.29x10^-5 = n/(n+55.56)

so n= 0.7 x 10^-3 = 0.7 milimoles

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