Question

If 120.3 mL of water is shaken with oxygen gas at 2.1 atm, it will dissolve 0.0043 g O2. Estimate the Henry's law constant for the oxygen gas in water in units of g mL-1atm-1.

Answer 1

Answer: The Henry's law constant for oxygen gas in water is $1.702\times 10^{-5}g/mL.atm$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

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

where,

$K_H$ = Henry's constant = ?

$C_{O_2}$ = solubility of oxygen gas = $0.0043g/120.3mL$

$p_{O_2$ = partial pressure of oxygen gas = 2.1 atm

Putting values in above equation, we get:

$0.0043g/120.3mL=K_H\times 2.1atm\\\\K_H=\frac{0.0043g}{120.3mL\times 2.1atm}=1.702\times 10^{-5}g/mL.atm$

Hence, the Henry's law constant for oxygen gas in water is $1.702\times 10^{-5}g/mL.atm$