Question

The solubility of pure oxygen in water at $ 20^{\circ}C $ and $ 1.0 $ atmosphere pressure is $ 1.38 \times 10^{-3} $ mol/litre. What will be the concentration of oxygen at $ 20^{\circ}C $ and partial pressure of $ 0.21 $ atmosphere?

• A. $ 2.9 \times 10^{-4} $ mol/litre
• B. $ 5.8 \times 10^{-4} $ mol/litre
• C. $ 7.6 \times 10^{-4} $ mol/litre
• D. $ 11.6 \times 10^{-4} $ mol/litre

Answer 1

Answer: (A)

The solubility of gas in liquid at particular
temperature is given by Henry's law
i.e. $m_{A}=K_{H}P_{A}$
where, $m_{A}$ = mass of gas dissolved in a unit volume of solvent
$p_{A}$= pressure of the gas in equilibrium
$K_{H}$= Henry's law constant
On putting the given values in above expression
$1.38\times10^{-3}\, mol/L=K_{H}\times 1\,atm$
$K_{H}=\frac{1.38\times 10^{-3}\,mol/L}{1\,atm}$
At $0.21$ atm, $m_{A}$ will be calculated as
$m_{A}=\frac{1.38\times10^{-3}\,mol/L}{1\,atm} \times 0.21\,atm$
$m_{A}=2.9\times 10^{-4}\,mol/L$