Question

The solubility of oxygen in water is 2.43×10−32.43×10−3mol/dm3 at 250C and 1 atm pressure. Calculate the concentration of oxygen at 250C and 0.2 atm pressure.​

Answer 1

Explanation:

If the partial pressure of oxygen in air is 0.2 atm , under atmospheric conditions, calculate the concentration (in moles per litre ) of dissolved

Answer 2

Answer:

The concentration of oxygen at $25^{\cdot }C$ and $0.2$ atm pressure is $4.86\times10^{-4}$ mole/dm

Explanation:

Given that $M=2.43\times10^{-3}$  and $P=1$

According to Henry's law:-

$M=K_H.P$

Where $M=$ the mass of gas dissolved in a unit volume of solvent

$K_H=$ Henry's constant

$P=$ the pressure of the gas in equilibrium

Now substitute all the values in henry's law:

$2.43\times10^{-3}=K_H.1\\K_H=2.43\times 10^-3$

Now the concentration of oxygen  when $P=0.2$ atm is

$M_A=K_H.P_A\\=2.43\times10^-3\times0.2\\=4.86\times 10^-4 mole/dm$