Question

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

• 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,
$p_{o_{2}} =P_{\text {total }} X_{O_{2}}^{\text {gas }}$
$=0.20 \times 760=152 \,mm\, Hg$
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 $n _{ O 2}$
$=55.56 \times 5 \times 10^{-6}=2.77 \times 10^{-4}$
$\left[ O _{2}\right] =2.77 \times 10^{-4} M$