How many millimoles of $CO_2$ gas will dissolve when $CO_2$ gas is passed in $900 \ mL$ water at $298 \ K$ temperature? $[$ The value of $K_H$ is $6.02 \times 10^{-4} \ bar$ and partial pressure of $CO_2$ gas is $2 \times 10^{-8} \ bar.]$

(A) According to Henry's Law,$P = K_H \times X$,where $X$ is the mole fraction of the gas in the solution.
Given: $P = 2 \times 10^{-8} \ bar$,$K_H = 6.02 \times 10^{-4} \ bar$.
$X = \frac{P}{K_H} = \frac{2 \times 10^{-8}}{6.02 \times 10^{-4}} \approx 3.322 \times 10^{-5}$.
Since the amount of gas is very small,the number of moles of water $n_{H_2O} = \frac{900 \ g}{18 \ g/mol} = 50 \ mol$.
$X = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}}$.
$n_{CO_2} = X \times n_{H_2O} = 3.322 \times 10^{-5} \times 50 = 1.661 \times 10^{-3} \ mol$.
$n_{CO_2} = 1.661 \ \text{millimoles}$.