# If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 litre of water. Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henrys law constant for N2 at 293 K is 76.48 kbar.

KH = 76.48kbar = 76480bar

P= 0.989bar

P = KH $×$ x
x= . This is the mole fraction of N2 gas dissolved in water

Volume of water =  1L = 1000mL
Taking density as 1g/mL,  mass of 1L =1000g
Moles of water in 1L =  1000/18=55.56
Let us take the moles of nitrogen as x
Total moles = moles of N2 + moles of water = x + 55.56
mole fraction of N2

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