If n2 gas is bubbled through water at 298k how many millilitres of n2 would dissolve in 1 litre of water
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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
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
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