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Question

The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0×105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atmospheric pressure is x×10y. Find the value of xy.

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Solution

According to Henry's law,
xN2×KH=pN2
(pN2=Partial pressure of N2)
Given,
Total pressure =5 atm
Mole fraction of N2=0.8
Partial pressure of N2=0.8×5=4xN2×1×105=4xN2=4×105Number of moles of H2O,nH2O=10Number of moles of N2,nN2=?nN2nN2+nH2O=xN2=4×105nN210+nH2O=xN2=4×105nN2=4×104Hence,xy=44=1

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