Question

Henry's law constant for oxygen dissolved in water is $4.34\times {10}^4$ atm at ${25}^C$. If the partial pressure of oxygen in air is 0.4atm. Calculate the concentration (in moles per litre) of the dissolved oxygen in water in equilibrium with air at ${25}^{o}C$.

• A. $5.11\times {10}^{-4}$
• B. $5.11\times {10}^{-3}$
• C. $9.2\times {10}^{-6}$
• D. $0.92\times {10}^{-6}$

Answer 1

Answer: (A)

$5.11\times {10}^{-4}$

$P=K_{u}X$

$P\to$ particle pressure of gas is vapour phase

$x\to$ mole fraction of gas in solution

$k_H\to$ Henry's law constant.

$k_H=4.34\times {10}^{4}atm$

$P_{o_2}=0.4atm$

$\Rightarrow X_{o_2}=\frac{P}{k_H}=0.0922\times {10}^{-4}=9.22\times {10}^{-6}$

i.e $9.22\times {10}^{-6}$ moles of oxygen in $1$ mols of water (neglecting moles of oxygen which indeed eligible)

$\Rightarrow 9.22\times {10}^{-6}$ moles oxygen $18g$ of water

$(\frac{18}{18}=1mole)\left(18g/ml\right)$

and density of water $=1gc{e}^{-1}=1000g{l}^{-1}$

(You should know card use this) $=1gm{l}^{-1}$

$\Rightarrow 18g=18ml=18\times {10}^{-3}l$

$\Rightarrow$ Coventration of oxygen in water

$=\frac{9.22\times {10}^{-6}}{18\times {10}^{-3}}$

$=0.512\times {10}^{-3}$

$5.12\times {10}^{-4}\sim 5.11\times {10}^{-4}$

$\left(A\right)5.11\times {10}^{-4}$