Question

The Henry law constant for dissolution of a gas in aqueous medium is 3x$10^2$ atm. At what partial pressure of the gas (in atm), the molality of gas in aqueous solution will be $\frac{5}{9}$m.

Answer 1

$\frac{300}{101}$

Answer 2

The Henry's law constant for the dissolution of a gas in an aqueous medium is given as $K_H' = 3 \times 10^2$ atm. The unit of the constant (atm) indicates that Henry's law should be used in the form $P_{gas} = K_H' \chi_{gas}$, where $P_{gas}$ is the partial pressure of the gas and $\chi_{gas}$ is the mole fraction of the gas in the solution.

We are given the molality of the gas in the aqueous solution as $\frac{5}{9}$ m. Molality ($m$) is defined as the number of moles of solute per kilogram of solvent.

Molality = $\frac{n_{gas}}{W_{water} \text{ (in kg)}}$

Given molality is $\frac{5}{9}$ m, this means that there are $\frac{5}{9}$ moles of the gas dissolved in 1 kg of water.

The mass of the solvent (water) is 1 kg = 1000 g. The molar mass of water ($H_2O$) is approximately 18 g/mol. The number of moles of water in 1 kg is $n_{water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} = \frac{500}{9}$ moles.

The number of moles of the gas (solute) is $n_{gas} = \frac{5}{9}$ moles.

The mole fraction of the gas in the solution is given by:

$\chi_{gas} = \frac{n_{gas}}{n_{gas} + n_{water}}$

$\chi_{gas} = \frac{5/9}{5/9 + 500/9}$

$\chi_{gas} = \frac{5/9}{(5 + 500)/9}$

$\chi_{gas} = \frac{5/9}{505/9}$

$\chi_{gas} = \frac{5}{505} = \frac{1}{101}$

Now, we can use Henry's Law:

$P_{gas} = K_H' \chi_{gas}$

$P_{gas} = (3 \times 10^2 \text{ atm}) \times \frac{1}{101}$

$P_{gas} = \frac{300}{101}$ atm

The partial pressure of the gas is $\frac{300}{101}$ atm.