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's law constant for $N_2$ at $293 \ K$ is $76.48 \ kbar$.

(0.716) The solubility of a gas is related to its mole fraction in an aqueous solution. According to Henry's law,$p = K_H \times x$.
First,calculate the mole fraction of $N_2$ $(x(N_2))$:
$x(N_2) = \frac{p(N_2)}{K_H} = \frac{0.987 \ bar}{76.48 \times 10^3 \ bar} = 1.29 \times 10^{-5}$.
Since $1 \ L$ of water contains $55.5 \ mol$ of water,and assuming $n$ is the number of moles of $N_2$ dissolved:
$x(N_2) = \frac{n}{n + 55.5} \approx \frac{n}{55.5} = 1.29 \times 10^{-5}$.
Solving for $n$:
$n = 1.29 \times 10^{-5} \times 55.5 \ mol = 7.16 \times 10^{-4} \ mol$.
Converting to millimoles:
$7.16 \times 10^{-4} \ mol \times 1000 \ mmol/mol = 0.716 \ mmol$.