Question

Calculate the partial pressure of oxygen if its solubility in water is 3.2mgd * m ^ - 3 ( K_{H} = 2 * 10 ^ - 3 mol d * m ^ - 3 * at * m ^ - 1 ) M NaCl

• A. 0.03 atm
• B. 0.04 atm
• C. 0.05 atm
• D. 0.06 atm

Answer 1

Answer: (C)

0.05 atm

Answer 2

To calculate the partial pressure of oxygen, we use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The formula for Henry's Law is:

$C = k_H * p$

Where:

• $C$ is the solubility of the gas in the liquid
• $k_H$ is Henry's Law constant
• $p$ is the partial pressure of the gas

Given:

• Solubility $C = 3.2 \, \text{mg/dm}^3 = 0.0032 \, \text{g/dm}^3$
• Henry's Law constant $k_H = 2 \times 10^{-3} \, \text{mol/dm}^3\text{atm}$

First, convert the solubility to moles per liter (mol/dm³):

$$\text{Moles of } O_2 = \frac{0.0032 \, \text{g}}{32 \, \text{g/mol}} = 0.0001 \, \text{mol/dm}^3$$

Next, rearrange Henry's Law to solve for $p$:

$$p = \frac{C}{k_H} = \frac{0.0001 \, \text{mol/dm}^3}{2 \times 10^{-3} \, \text{mol/dm}^3\text{atm}} = 0.05 \, \text{atm}$$

Thus, the partial pressure of oxygen is 0.05 atm.