Question

The Henry’s law constant for $O_2$ dissolved in water is $4.34 \times 10^4$ atm at certain temperature. If the partial pressure of $O_2$ in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of $O_2$ in the solution?

• A. $1 \times 10^{-5}$
• B. $1 \times 10^{-4}$
• C. $2 \times 10^{-5}$
• D. $1 \times 10^{-6}$
• E. $\times 10^{-6}$

Answer 1

Answer: (A)

Given, $K_{H}$ for $O _{2}$ dissolved in water

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

$PO _{2} =0.434 \,atm$

$x\left(\right.$ mole fraction of $O _{2}$ in solution $)=?$

According to Henry's law,

$P =K_{H} \times x $

$\frac{0.434}{4.34 \times 10^{4}} =x_{ O _{2}} $

$1 \times 10^{-5} =x_{ O _{2}}$

$\therefore $ Mole fraction of $O _{2}$ in water $=1 \times 10^{-5}$.