Chapter 13, Problem 13.46P

(a)

Interpretation Introduction

Interpretation:

The mass of ${\text{O}}_2$ that will dissolve in $2.50\text{L}$ of ${\text{H}}_2\text{O}$ in contact with pure ${\text{O}}_2$ at $1\text{atm}$ is to be determined.

Concept introduction:

Henry’s law gives the quantitative relationship between the pressure of the gas and its solubility. It states that the amount of gas dissolved in a liquid is proportional to the partial pressure of the gas. Higher the partial pressure of the gas, more will be its solubility and vice-versa.

The formula to calculate the solubility of gases according to Henry’s law is as follows:

$S_{\text{gas}}=k_{\text{H}}\times P_{\text{gas}}$ (1)

Here, $S_{\text{gas}}$ is the solubility of the gas.

$k_{\text{H}}$ is Henry’s constant.

$P_{\text{gas}}$ is the partial pressure of the gas.

(b)

Interpretation Introduction

Interpretation:

The mass of ${\text{O}}_2$ that will dissolve in $2.50\text{L}$ of ${\text{H}}_2\text{O}$ in contact with air where the partial pressure of ${\text{O}}_2$ is $0.029\text{atm}$ is to be determined.

Concept introduction:

Henry’s law gives the quantitative relationship between the pressure of the gas and its solubility. It states that the amount of gas dissolved in a liquid is proportional to the partial pressure of the gas. Higher the partial pressure of the gas, more will be its solubility and vice-versa.

The formula to calculate the solubility of gases according to Henry’s law is as follows:

$S_{\text{gas}}=k_{\text{H}}\times P_{\text{gas}}$ (1)

Here, $S_{\text{gas}}$ is the solubility of the gas.

$k_{\text{H}}$ is Henry’s constant.

$P_{\text{gas}}$ is the partial pressure of the gas.