How many millimoles of N2 gas would dissolve in 1 litre of water if N2 gas is bubbled through water at 293 K, and exerts a partial pressure of 0.987 bar. Given that Henry�s law constant for N2 at 293 K is 76.48 kbar.

Given:

P= 0.987 bar

K= 76.48 Kbar= 76480

As per the Henry’s law

P=Kx
x is the mole fraction
K is the henry’s law constant
P is the pressure in atm.

mole fraction of N2 = x / (x = 55.5)

Hence 
0.987= 76480 * x / (x = 55.5)
x = 0.7 millimole

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