Question

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

• A. $1.8\, g$
• B. $2.5\, g$
• C. $4\, g$
• D. $5\, g$

Answer 1

Answer: (A)

Given, $K_{H}= 1.67 \times 10^{8}\, Pa$
$P_{CO_{2}} = 2.5\,$ atm $= 2.5 \times 101325\, Pa$
Applying Henry’s law, $P_{CO_{2}} = K_{H} \times x_{CO_{2}}$
$\therefore x_{CO_2} = \frac{P_{CO_2}}{K_{H}}$
$= \frac{2.5 \times 101325\, Pa}{1.67 \times 10^{8}\, Pa} = 1.517 \times 10^{-3}$
i, e. $x_{CO_2} = \frac{n_{CO_2}}{n_{H_2O} + n_{CO_2}}$
$= \frac{n_{CO_2}}{n_{H_2O}} = 1.517 \times 10^{-3}$
[$n_{CO_{2}}$ is negligible]
For $500\, mL$ of soda water,
volume of water $= 500\, mL$; mass of water $ = 500\, g$,
moles of water $\frac{500}{18} = 27.78$
i.e., $n_CO{_{2}} = 27.78$
$\therefore \frac{n_{CO_{2}}}{27.78} = 1.517 \times 10^{-3}$
or $n_{CO_{2}} = 42 .14 \times 10^{-3}$ mole
Mass of $CO_{2} = 42.14 \times 10^{-3} \times 44\, g = 1.854\, g$