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Standard XII

Chemistry

Molality

O2 is bubbled...

Question

O2 is bubbled through water at 293K. How many moles of O2 will dissolve in 1L of water, if partial pressure of 02 is 2 kilobar and Kh is 34.86 kilobar?

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Solution

pO2=KHχO2

Thus
χO2=pO2/KH=0.098/(34.84103) =2.8110^-5

No. of moles of H2O in 1L =55.55
χO2 = nO2/(nO2 + nH2O) = nO2/(nO2 +55.55)
APPROXIMATELY
χO2 =nO2/55.55

nO2 =2.8110^-555.55 =1.5610^-3 mol

Thus solubility of O2 is = 1.5610^-3 32 g/L = 0.05 g/L

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O2 is bubbled through water at 293K. How many moles of O2 will dissolve in 1L of water, if partial pressure of 02 is 2 kilobar and Kh is 34.86 kilobar?

pO2=KH*χO2 Thus χO2=pO2/KH=0.098/(34.84*103) =2.81*10^-5 No. of moles of H2O in 1L =55.55 χO2 = nO2​/(nO2 + nH2O​) = nO2/​(nO2 +55.55​) APPROXIMATELY χO2 =​nO2​/55.55 nO2 =2.81*10^-5*55.55 =1.56*10^-3 mol Thus solubility of O2 is = ​1.56*10^-3 *32 g/L = 0.05 g/L

pO2=KHχO2

Thus
χO2=pO2/KH=0.098/(34.84103) =2.8110^-5

No. of moles of H2O in 1L =55.55
χO2 = nO2/(nO2 + nH2O) = nO2/(nO2 +55.55)
APPROXIMATELY
χO2 =nO2/55.55

nO2 =2.8110^-555.55 =1.5610^-3 mol

Thus solubility of O2 is = 1.5610^-3 32 g/L = 0.05 g/L