Question

When a gas is bubbled through water at $298\, K$, a very dilute solution of gas is obtained. Henry's law constant for the gas is $100\, kbar$. If gas exerts a pressure of $1$ bar, the number of moles of gas dissolved in $1$ litre of water is $x \times 10^{-5}$. Find the value of $x$ here? Report your answer upto two decimal places.

Answer 1

$p=K_{H} \times x$
$x=\frac{p}{K_{H}}=\frac{1}{100 \times 10^{3}}=1 \times 10^{-5}$
Mole of fraction $=\frac{\text { Mole of gas }}{\text { Total moles }}$
Moles of $\mathrm{H}_2 \mathrm{O}=\frac{1000}{18}=55.55(\therefore 1 \mathrm{~L}=1000 \mathrm{~g}) \times 10^{-5}$
Mole fraction $=\frac{x}{x+55.55}(55.55>>>x)$
So $10^{-5}=\frac{x}{55.55}$
or $x=55.55 \times 10^{-5}$