Chapter U4, Problem C14.5RE

Interpretation Introduction

Interpretation: The following compounds should be listed in increasing order of moles of metal:

  $5.0\text{ g NaCl, 5}\text{.0 g AgCl, 5}\text{.0 g LiCl}$

Concept Introduction: The mathematical relationship between number of moles, mass of substance and molar mass is:

  $\text{number of moles = }\frac{\text{mass of the substance}}{\text{Molar mass of the substance}}$ - (1)

Answer to Problem C14.5RE

  $\text{AgCl}<\text{NaCl}<\text{LiCl}$ .

Explanation of Solution

The given compounds are:

  $5.0\text{ g NaCl, 5}\text{.0 g AgCl, 5}\text{.0 g LiCl}$

The number of moles of each compound is determined using equation (1) as:

For $5.0\text{ g NaCl}$:

The molar mass of $\text{NaCl}$ is $\text{58}\text{.44 g/mol}$ .

Substituting the values:

  $\begin{array}{l}\text{number of moles = }\frac{\text{5}\text{.0 g}}{\text{58}\text{.44 g/mol}}\\ \text{number of moles = 0}\text{.086 mol}\end{array}$

Since, 1 mole of $\text{NaCl}$ contains 1 mole of ${\text{Na}}^{+}$ metal so, $\text{0}\text{.086 mol}$ of $\text{NaCl}$ will contain $\text{0}\text{.086 mol}$ of ${\text{Na}}^{+}$ metal.

For $5.0\text{ g AgCl}$:

The molar mass of $\text{AgCl}$ is $\text{143}\text{.32 g/mol}$ .

Substituting the values:

  $\begin{array}{l}\text{number of moles = }\frac{\text{5}\text{.0 g}}{\text{143}\text{.32 g/mol}}\\ \text{number of moles = 0}\text{.035 mol}\end{array}$

Since, 1 mole of $\text{AgCl}$ contains 1 mole of ${\text{Ag}}^{+}$ metal so, $\text{0}\text{.035 mol}$ of $\text{AgCl}$ will contain $\text{0}\text{.035 mol}$ of ${\text{Ag}}^{+}$ metal.

For $5.0\text{ g LiCl}$:

The molar mass of $\text{LiCl}$ is $\text{42}\text{.394 g/mol}$ .

Substituting the values:

  $\begin{array}{l}\text{number of moles = }\frac{\text{5}\text{.0 g}}{\text{42}\text{.394 g/mol}}\\ \text{number of moles = 0}\text{.118 mol}\end{array}$

Since, 1 mole of $\text{LiCl}$ contains 1 mole of ${\text{Li}}^{+}$ metal so, $\text{0}\text{.118 mol}$ of $\text{LiCl}$ will contain $\text{0}\text{.118 mol}$ of ${\text{Li}}^{+}$ metal.

Thus, the given compounds can be listed in increasing moles of metal in each compound as:

  $\text{AgCl}<\text{NaCl}<\text{LiCl}$