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

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`"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.

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