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If N_(2) gas is bubbled through water at 293 K, how many millimoles of N_(2) gas would dissolve in 1 litre of water ? Assume that N_(2) exerts 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 :The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry.s law. Thus : x (Nitrogen) `= ("p (nitrogen)")/(K_(H))` `=(0.987" bar")/(76480" bar")=1.29xx10^(-5)` As 1 LITRE of water contains 55.5 MOL of it, therefore if n represents number of MOLES of `N_(2)` in solution, x(Nitrogen) `= ("n mol")/(n mol +55.5 mol)` `= (n)/(55.5)=1.29xx10^(-5)` (n in denominator is neglected as it is `LT lt 55.5`) Thus, `n = 1.29xx10^(-5)xx55.5 mol = 7.16xx10^(-4)` mol `= (7.16xx10^(-4)mol xx 1000 m mol)/(1 mol)` = 0.716 m mol |
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