Henry's law constant for $CO_{2}$ in water is $1.67 \times 10^{8} \ Pa$ at $298 \ K$. Calculate the quantity of $CO_{2}$ in $500 \ mL$ of soda water when packed under $2.5 \ atm$ $CO_{2}$ pressure at $298 \ K$.

(D) Given:
$K_{H} = 1.67 \times 10^{8} \ Pa$
$P_{CO_{2}} = 2.5 \ atm = 2.5 \times 1.01325 \times 10^{5} \ Pa = 2.533125 \times 10^{5} \ Pa$
According to Henry's law,$p_{CO_{2}} = K_{H} x$,where $x$ is the mole fraction of $CO_{2}$.
$x = \frac{p_{CO_{2}}}{K_{H}} = \frac{2.533125 \times 10^{5}}{1.67 \times 10^{8}} = 0.00152$
Assuming $n_{CO_{2}}$ is negligible compared to $n_{H_{2}O}$,$x \approx \frac{n_{CO_{2}}}{n_{H_{2}O}}$.
For $500 \ mL$ of water,mass $= 500 \ g$,so $n_{H_{2}O} = \frac{500}{18} = 27.78 \ mol$.
$n_{CO_{2}} = x \times n_{H_{2}O} = 0.00152 \times 27.78 = 0.0422 \ mol$.
Mass of $CO_{2} = n_{CO_{2}} \times \text{Molar mass} = 0.0422 \times 44 = 1.857 \ g$ (approx $1.85 \ g$).