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

Answer: 5

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