Question

O2is bubbled through water at 293K. Assuming that O2 exerts a partial pressure of 0.98 bar, calculate the solubility of O2 in grams per litre. The value of Henry's Law constant (KH) for O2 is 34.84 kbar​

Answer 1

Answer:

The solubility of the oxygen gas is 0.05 g/L.

Explanation:

Partial pressure of the oxygen gas,$p__{O_2}$ = 0.98 bar

Henry's Law constant $K_H$ for oxygen gas = 34.84 kbar = 34840 bar​

$p_{O_2}=K_H\times \chi_{O_2}$

$\chi_{O_2}=\frac{0.98 bar}{34840 bar}=2.81\times 10^{-5}$

$\chi_w=0.9999$

Moles of water = 0.999

Mass of water = $18 g/mol\times 0.999 mol=17.98 g$

Density of water = 1000 g/L

Volume of water =$\frac{17.98 g}{1000 g/L}=0.01798 L$

Moles of oxygen gas = $2.81\times 10^{-5}$

Mass of oxygen gas = $32\times g/mol\times 2.81\times 10^{-5}=89.92\times 10^{-5} g$

Solubility of oxygen gas:$\frac{89.92\times 10^{-5} g}{0.01798 L}=0.05 g/L$

The solubility of the oxygen gas is 0.05 g/L.

Answer 2

Explanation:

so this is the correct answer