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 $55.55\times 10^{- x}$ . Find the value of $\text{x}$ here?

Answer 1

$p=K_{H}\times x$
$x=\frac{p}{K_{H}}=\frac{1}{100 \times 1 0^{3}}=1\times 10^{- 5}$
Mole of fraction $=\frac{\text{Mole of gas}}{\text{Total moles}}$
Moles of $H _{2} O =\frac{1000}{18}=55.55(\therefore 1 L =1000 g ) \times 10^{-5}$
Mole fraction $=\frac{x}{x + 55.55}$ $\left(55.55 > > > x\right)$
So $10^{- 5}=\frac{x}{55.55}$
or $x=55.55\times 10^{- 5}$