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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.

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.

Grade:11

3 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
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
askIITians Faculty 646 Points
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

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