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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Answered by
138
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
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
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Answered by
39
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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