How many millimoles of nitrogen dissolve when nitrogen is passed in $1 \ L$ water at $293 \ K$ temperature? $[$ The value of $K_H$ is $7.648 \times 10^4 \ bar$ and partial pressure of $N_2$ gas is $0.987 \ bar.]$
According to Henry's Law,$P = K_H \times x$,where $x$ is the mole fraction of the gas in the solution.
$x = P / K_H = 0.987 \ bar / (7.648 \times 10^4 \ bar) = 1.29 \times 10^{-5}$.
Since $1 \ L$ of water contains $1000 \ g$ of water,the number of moles of water is $n_{H_2O} = 1000 \ g / 18 \ g/mol = 55.55 \ mol$.
For dilute solutions,$x = n_{N_2} / n_{H_2O}$,so $n_{N_2} = x \times n_{H_2O} = 1.29 \times 10^{-5} \times 55.55 \ mol = 7.16 \times 10^{-4} \ mol$.
Converting to millimoles: $7.16 \times 10^{-4} \ mol \times 1000 \ mmol/mol = 0.716 \ mmol \approx 0.72 \ mmol$.
$x = P / K_H = 0.987 \ bar / (7.648 \times 10^4 \ bar) = 1.29 \times 10^{-5}$.
Since $1 \ L$ of water contains $1000 \ g$ of water,the number of moles of water is $n_{H_2O} = 1000 \ g / 18 \ g/mol = 55.55 \ mol$.
For dilute solutions,$x = n_{N_2} / n_{H_2O}$,so $n_{N_2} = x \times n_{H_2O} = 1.29 \times 10^{-5} \times 55.55 \ mol = 7.16 \times 10^{-4} \ mol$.
Converting to millimoles: $7.16 \times 10^{-4} \ mol \times 1000 \ mmol/mol = 0.716 \ mmol \approx 0.72 \ mmol$.
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