Question

The Henry's law constant for oxygen dissolved in water is 4.34×104 atm at 25∘C.If the partial pressure of oxygen in air is 0.2atm,under atmospheric conditions,calculate the concentration (in moles per litre) of dissolve oxygen in water in equilibrium with air at 25∘C.

Answer 1

Here is your answer (-: 2.55×10^-4 M

Answer 2

Answer: The concentration of $O_2$ in water in equilibrium with air at 25∘C is 271.25 moles/L

Explanation:  

Henry's law states that the mass of a gas dissolved in a given volume of the liquid at constant temperature is directly proportional to the pressure of the gas present in equilibrium with the liquid.

Mathematically,

$m\propto p$

or,

$m=Kp$

Where, m = mass of the gas dissolved in a unit volume of the solvent

K = Proportionality constant

p = Pressure of the gas in equilibrium with the solvent

$m=4.34\times 10^4\times 0.2$

$m=8680g/L$

Solubility of $O_2$ in moles per liter=$\frac{8680g/L}{32g/mol}=271.25moles/L$