If `N_(2)` gas is bubbled throgh water at `20^(@)C`, how many millilitres (at stp) of `N_(2)` would dissolve in a litre of water ? Given that the partial pressure of `N_(2)` is equal to 742.5 torr and `K_(H)` for nitroge= `5.75xx10^(7)` torr.

Explanation

`"Mole fraction of "N_(2)(x_(N_2))=("Partial pressure of" N_(2))/(K_(H)"for"N_(2))`
`=((0.987 bar))/((76.48 k bar))=((0.987 bar))/((76480 bar))=1.29xx10^(-5)`
`x_(N_2)=n_(N_2)/(n_(N_2)+n_(H_2)o)=n_(N_2)/(n_(H_2)o)`
`(N_(2))" in the denominator has been negleeted as the gas is very little soubel in water").`
`x_(N_2)=((n))/((55.5 mol))=n/(55.5),`
`n=x_(N_2)xx(55.5 mol)=7.16xx10^(4) mol`
` =(7.16xx10^(-4))xx(1000mL)=0.716 milli mol .`
`n_(H_2)=x_(N_(2))xxn_(H_2)o=(1.29xx10^(-5)55.55 mol)=7.16xx10^(-4) mol.`
Calculation of valume of `N_(2)` at S.T.P.
`"1 mole of" N_(2)" at S.T.P. occupy"=22400 mL`
`7.16xx10^(-4) "mole of" N_(2) "at S.T.P. occupy"=((2240mL))/((1 mol))xx(7.16xx10^(-4)mol)=16.04 mL`