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What concentration of nitrogen should be present in a glass of water at room temperature. Assume a temperature of 25oC, a total pressure of 1atm and the mole fraction of nitrogen in air of 0.78( KH for nitrogen is 8.42X 10-7 M/mm Hg)           
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Answer :

kvnmurty
The given form of Henry's constant for solubility is
       KH = c_aq / p,        Henry's law
               p =  partial pressure of nitrogen in air above  water
               c_aq = concentration of nitrogen in water

 p = 0.78 atm = 0.78 * 760 mm Hg

c_aq = 0.78 *760 * 8.42 * 10^-7  M = 4.991 * 10^-4 M
           = 1.3796 * 10^-2 gm/litre
AqdasZishan

Given,

Total Pressure,Pt = 1 atm = 760 mm Hg

Mole fraction of N2, XN2= 0.78

KH for N2= 8.42*10^-7 M/mm Hg

By Dalton's law of partial pressure,

Partial pressure of N2, pN2= Pt * XN2

Therefore, pN2= 760 * 0.78 = 592.8 mm Hg

According to Henry's law, solubility is directly proportional to partial pressure,i.e, S = KH * pN2

But here , solubility is for concentration of N2

Hence, For conc. of N2

C = KH * pN2

= 8.42*10^-7 * 592.8

= 4991.37 * 10^-7

= 4.99 * 10^-7 M

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