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.

_{H}= 76.48kbar = 76480bar

P= 0.989bar

P = K

_{H}

_{ $\times $}x

x= $\frac{0.989}{76480}=1.293\times {10}^{-5}$. This is the mole fraction of N

_{2}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 N

_{2}+ moles of water = x + 55.56

mole fraction of N

_{2}

$1.293\times {10}^{-5}=\frac{x}{x+55.56}1.293\times 10-5x+1.293\times 10-5\times 55.56=xx(1-1.293\times 10-5)=1.293\times 10-5\times 55.56x=7.17\times 10-4mol=0.717mmoles$

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