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If N(2) gas is bubbled through water at ...

If `N_(2)` gas is bubbled through water at 293 K, how many of millimoles of `N_(2)` gas would dissolve in 1 litre of water 7 Assume that `N_(1)` exerls a partial pressure of 0.987 bar. Given that Henry's law constant for `N_(2)` at 293 K is 76.48 K bar.

Text Solution

Verified by Experts

According to Henry.s law
x(Nitrogen ) = `(p("Nitrogen"))/(K_(H))`
= `(0.987)/(74,480("bar"))`
= `1.29 xx 10^(-5)`
As 1 litre of water contains 55.5 mole of it, therefore if n represents number of moles of `N_(2)` in solution.
x (Nitrogen ) = `(n"mole")/("n mole + 55.5 mol")`
= `(n)/(55.5) = 1.29 xx 10^(-5) = 1.29 xx 10^(-5)`
(n in denominator is neglecti!d as it is `lt lt `55.5)
So, n= 1.29` xx 10^(-5) xx 55.5 `mol
= `7.16 xx 10^(-4)` mol
`= (7.16 xx 10^(-4)"mol" xx 1000 "mol")/(1 "mol")`
= 0.716 m/mol
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