Henry's law constant for $\mathrm{CO}_2$ in water is $1.67 \times 10^8 \mathrm{~Pa}$. Calculate the…
Henry's law constant for $\mathrm{CO}_2$ in water is $1.67 \times 10^8 \mathrm{~Pa}$. Calculate the approximate quantity of $\mathrm{CO}_2$ in $500 \mathrm{~mL}$ of soda water when packed under $5 \mathrm{~atm} \mathrm{CO}_2$ at $298 \mathrm{~K}$.
$3.7 \mathrm{~g}$
$1.84 \mathrm{~g}$
$2.2 \mathrm{~g}$
$4.4 \mathrm{~g}$
Solution
Henry's law is applicable for gas, it states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.
We know that, formula of Henry law is
$
\begin{aligned}
p & =K_{\mathrm{H}} \times \chi \\
\chi & =\frac{5 \times 1.013 \times 10^5}{1.67 \times 10^8} \\
\frac{n_{\mathrm{OO}_2}}{500 / 18} & =\frac{5 \times 1.013}{1.67 \times 1000} \\
n_{\mathrm{CO}_2} & =\frac{5 \times 1.013 \times 500}{1.67 \times 18} \\
n_{\mathrm{CO}_2} & =\frac{2500}{1.67} \times \frac{1.013}{18}
\end{aligned}
$
Now, mass of $\mathrm{CO}_2$ dissolved $=\frac{2500}{1.67} \times \frac{1.013 \times 44}{18}$
$
=3.7 \mathrm{~g} \text {. }
$