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
$1.38\times 10^{-3}mol/L=K_H\times 1atm$
$K_H=\frac{1.38\times 10^{-3}mol/L}{1atm}$
At $0.21$ atm, $m_A$ will be calculated as
$m_A=\frac{1.38\times 10^{-3}mol/L}{1atm}\times 0.21atm$
$m_A=2.9\times 10^{-4}mol/L$