Question

When a gas is bubbled through water at $298K$, a very dilute solution of gas is obtained. Henry's law constant for the gas is $100kbar$. 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

Answer: 55.55

$p=K_H\times x$
$x=\frac{p}{K_H}=\frac{1}{100\times 10^3}=1\times 10^{-5}$
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 $x=55.55\times 10^{-5}$