Chapter 13, Problem 21A

Interpretation Introduction

Interpretation:

The volume of oxygen gas has to be calculated.

Concept Introduction:

Avogadro’s law: The volume of a gas is directly proportional to the number of moles at constant temperature and pressure.

The relationship between the volume of a gas and the number of moles present in the gas is expressed as-

  $\begin{array}{l}V\propto n\\ or,\frac{V}{n}=\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\end{array}$

When T and P are constant.

When the number of moles of gas is changed from an initial amount to another amount at constant T and P , Avogadro’s law can be represented as -

  $\frac{V_1}{n_1}=\frac{V_2}{n_2}$

Initial volume = V₁ Final volume = V₂

Initial moles = n₁ Final moles = n₂

Answer to Problem 21A

The volume of oxygen gas is 10.823 L.

Explanation of Solution

Given Information:

Mass of oxygen gas m₁= 46.2 g

V₁ = 100 L

Mass of oxygen gas m₂= 5.00 g

  $\frac{V_1}{n_1}=\frac{V_2}{n_2}$

By putting all the values in the above equation, calculate V₂

  $\begin{array}{c}{V}_2=\frac{V_1{n}_2}{n_1}\end{array}$

    $\begin{array}{c}=\frac{V_1\left(\raisebox{1ex}{$m_2$}\!\left/ \!\raisebox{-1ex}{$M$}\right.\right)}{\left(\raisebox{1ex}{$m_1$}\!\left/ \!\raisebox{-1ex}{$M$}\right.\right)}=\frac{V_1{m}_2}{m_1}\end{array}$

   $\begin{array}{c}=\frac{\text{100}\text{L}\text{×}\text{5}\text{.00}\text{gm}}{\text{46}\text{.2}\text{gm}}\end{array}$

    $\begin{array}{c}\text{=}\text{10}\text{.823}\text{L}\end{array}$