Question

The amount of dissolved oxygen in $1$ litre water in equilibrium with air at $1atm$ pressure at $25^{\circ }C$ will be (assume that air contains $20$ mole $\%$ oxygen, Henry's constant $(kH)$ for oxygen is $3.04\times 10^{7}mmHg$ and density of $H_{2}O$ at $25^{\circ }C$ is $1g/cc$ )

• A. the mole fraction of $O_2$ in solution is $5\times 10^{-6}$
• B. the mole fraction of $O_2^2$ in solution is $5\times 10^{-7}$
• C. the molarity of $O_2$ in solution is $2.77\times 10^{-5}M$
• D. the molarity of $O_2$ in solution is $2.77\times 10^{-4}M$

Answer 1

Answer: (A), (C)

According to Dalton's law,
$p_{o_2}=P_{\text{total }}{X}_{O_2}^{\text{gas }}$
$=0.20\times 760=152mmHg$
According to Henry's law,
$p_{o_2}=K_H\times X_{O_2}^{\text{liq. }}$
$152=3.04\times 10^7{X}_{O_2}^{\text{liq. }}$
$X_{O_2}=5\times 10^{-6}$
$X_{O_2}=\frac{n{o}_2}{n_{O_2}+n_{H_2}O}$
$-\frac{n{o}_2}{n_{O_2}+1000/18}$
$=\frac{n{o}_2}{n_{O_2}+55.56}\simeq \frac{n{o}_2}{55.56}$
Number of moles of $O_2$ dissolved $n_{O2}$
$=55.56\times 5\times 10^{-6}=2.77\times 10^{-4}$
$\left[O_2\right]=2.77\times 10^{-4}M$