Question

If $He\left(g\right)$ is bubbled through the water at $298K$, how many millimoles of $He\left(g\right)$ would dissolve in one litre of water?
Assume that $He\left(g\right)$ exerts a partial pressure of $0.5bar$
Take Henry's Law constant for $He\left(g\right)$ at $298K$ as $145kBar$

• A. $0.85mmol$
• B. $1.52mmol$
• C. $0.19mmol$
• D. $2.2mmol$

Answer 1

The correct option is C $0.19mmol$

The solubility of gas is realted to the mole fraction in aqueous solution. The mole fraction of the gas in solution is calculated by applying Henry's Law
Thus :

According to Henry's law :
${\chi }_{He}=\frac{p_{He}}{K_H}=\frac{0.5bar}{145000bar}=3.45\times {10}^{-6}$

As 1 litre of water contains $55.55mol$ of it, therefore if $n$ represents number of moles of $He\left(g\right)$ in solution
${\chi }_{He}=\frac{n}{n+55.55}$
Since , $n<<55.55\therefore \frac{n}{55.55}=3.45\times {10}^{-6}n=1.91\times {10}^{-4}moln=1.91\times {10}^{-4}\times 1000mmoln=0.191mmol$