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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. |
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Answer» Solution :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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