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# 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 Henry’s law constant for N2 at 293 K is 76.48 kbar.

Sunil Kumar FP
6 years ago
let the number of mole of N2 dissolved in water be n
Henry’s law constant for N2 at 293 K is 76.48 kbar.
according ti henrys law
partial pressure exerted by N2=Kh*x
x is the mole fraction
mole fraction of N2=x/(x+55.5)
.987=76.48*10^3 *x/(x+55.5)
x=.7 milimole
Kundan
11 Points
4 years ago
The number of moles of N2 dissolver n water is n.The Henerys law constant for N2 at 293K is 76.48 kbar = 76.48*10^3.Partial pressure of N2 = 0.987 bar.According to Henrys lawP=Kh*xHere "x" is the mole fractionMole fraction of N2 = moles of solute(N2) / total number of moles of solute and solvent(N2+H2O).=moles of solvent(H2O)=1000(1litre)/18=55.56Mole fraction of N2=n/n+55.56As , Henrys law,0.987=76.48*10^3*(n/n+55.56)n=.7There are 0.7 millimiles of N2 gas
Rishi Sharma
11 months ago
Hello Students,
The solution is as follows
Let the number of mole of N2 dissolved in water be n
Henry’s law constant for N2 at 293 K is 76.48 kbar
according to it henry’s law partial pressure exerted by N2=Kh*x
where x is the mole fraction
mole fraction of N2=x/(x+55.5)
.987=76.48*10^3 *x/(x+55.5)
x=.7 milimole
I hope the solution will solve all your doubts.
thanks
all the best for your exams