Chapter 4, Problem 4.44E

Interpretation Introduction

(a)

Interpretation:

The number of positive ions and negative ions contained in $1.00\text{mol}$ of given compound is to be calculated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The binary ionic compounds are composed of only one cation and one anion.

Answer to Problem 4.44E

The number of positive ions and negative ions contained in $1.00\text{mol}$ of $\text{NaBr}$ is $6.02\times {10}^{23}$ and $6.02\times {10}^{23}$ respectively.

Explanation of Solution

The given compound is $\text{NaBr}$.

One mole of $\text{NaBr}$ compound contains one mole of ${\text{Na}}^{+}$ cation and one mole of ${\text{Br}}^{-}$ anion. The number of atoms present in one mole is equal to Avogadro’s number $\left(6.02\times {10}^{23}\right)$. Therefore,

The number of ${\text{Na}}^{+}$ cations is $\begin{array}{rll}& =& 1\times 6.02\times {10}^{23}\\ & =& 12.04\times {10}^{23}\end{array}$

The number of ${\text{Br}}^{-}$ anions is $\begin{array}{rll}& =& 1\times 6.02\times {10}^{23}\\ & =& 6.02\times {10}^{23}\end{array}$

The number of positive ions and negative ions contained in $1.00\text{mol}$ of $\text{NaBr}$ is $6.02\times {10}^{23}$ and $6.02\times {10}^{23}$ respectively.

Conclusion

The number of positive ions and negative ions contained in $1.00\text{mol}$ of $\text{NaBr}$ is $6.02\times {10}^{23}$ and $6.02\times {10}^{23}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The number of positive ions and negative ions contained in $1.00\text{mol}$ of given compound is to be calculated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The binary ionic compounds are composed of only one cation and one anion.

Answer to Problem 4.44E

The number of positive ions and negative ions contained in $1.00\text{mol}$ of ${\text{CaF}}_2$ is $6.02\times {10}^{23}$ and $12.04\times {10}^{23}$ respectively.

Explanation of Solution

The given compound is ${\text{CaF}}_2$.

One mole of ${\text{CaF}}_2$ compound contains one mole of ${\text{Ca}}^{2+}$ cation and two moles of ${\text{F}}^{-}$ anion. The number of atoms present in one mole is equal to Avogadro’s number $\left(6.02\times {10}^{23}\right)$. Therefore,

The number of ${\text{Ca}}^{2+}$ cations is $\begin{array}{rll}& =& 1\times 6.02\times {10}^{23}\\ & =& 6.02\times {10}^{23}\end{array}$

The number of ${\text{F}}^{-}$ anions is $\begin{array}{rll}& =& 2\times 6.02\times {10}^{23}\\ & =& 12.04\times {10}^{23}\end{array}$

The number of positive ions and negative ions contained in $1.00\text{mol}$ of ${\text{CaF}}_2$ is $6.02\times {10}^{23}$ and $12.04\times {10}^{23}$ respectively.

Conclusion

The number of positive ions and negative ions contained in $1.00\text{mol}$ of ${\text{CaF}}_2$ is $6.02\times {10}^{23}$ and $12.04\times {10}^{23}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The number of positive ions and negative ions contained in $1.00\text{mol}$ of given compound is to be calculated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The binary ionic compounds are composed of only one cation and one anion.

Answer to Problem 4.44E

The number of positive ions and negative ions contained in $1.00\text{mol}$ of ${\text{Cu}}_2\text{S}$ is $12.04\times {10}^{23}$ and $6.02\times {10}^{23}$ respectively.

Explanation of Solution

The given compound is ${\text{Cu}}_2\text{S}$.

One mole of ${\text{Cu}}_2\text{S}$ compound contains two moles of ${\text{Cu}}^{+}$ cations and one mole of ${\text{S}}^{2-}$ anion. The number of atoms present in one mole is equal to Avogadro’s number $\left(6.02\times {10}^{23}\right)$. Therefore,

The number of ${\text{Cu}}^{+}$ cations is $\begin{array}{rll}& =& 2\times 6.02\times {10}^{23}\\ & =& 12.04\times {10}^{23}\end{array}$

The number of ${\text{S}}^{2-}$ anions is $\begin{array}{rll}& =& 1\times 6.02\times {10}^{23}\\ & =& 6.02\times {10}^{23}\end{array}$

The number of positive ions and negative ions contained in $1.00\text{mol}$ of ${\text{Cu}}_2\text{S}$ is $12.04\times {10}^{23}$ and $6.02\times {10}^{23}$ respectively.

Conclusion

The number of positive ions and negative ions contained in $1.00\text{mol}$ of ${\text{Cu}}_2\text{S}$ is $12.04\times {10}^{23}$ and $6.02\times {10}^{23}$ respectively.

Interpretation Introduction

(d)

Interpretation:

The number of positive ions and negative ions contained in $1.00\text{mol}$ of given compound is to be calculated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. The binary ionic compounds are composed of only one cation and one anion.

Answer to Problem 4.44E

The number of positive ions and negative ions contained in $1.00\text{mol}$ of ${\text{Li}}_{\text{3}}\text{N}$ is $18.06\times {10}^{23}$ and $6.02\times {10}^{23}$ respectively.

Explanation of Solution

The given compound is ${\text{Li}}_{\text{3}}\text{N}$.

One mole of ${\text{Li}}_{\text{3}}\text{N}$ compound contains three moles of ${\text{Li}}^{+}$ cations and one mole of ${\text{N}}^{3-}$ anion. The number of atoms present in one mole is equal to Avogadro’s number $\left(6.02\times {10}^{23}\right)$. Therefore,

The number of ${\text{Li}}^{+}$ cations is $\begin{array}{rll}& =& 3\times 6.02\times {10}^{23}\\ & =& 18.06\times {10}^{23}\end{array}$

The number of ${\text{N}}^{3-}$ anions is $\begin{array}{rll}& =& 1\times 6.02\times {10}^{23}\\ & =& 6.02\times {10}^{23}\end{array}$

The number of positive ions and negative ions contained in $1.00\text{mol}$ of ${\text{Li}}_{\text{3}}\text{N}$ is $18.06\times {10}^{23}$ and $6.02\times {10}^{23}$ respectively.

Conclusion

The number of positive ions and negative ions contained in $1.00\text{mol}$ of ${\text{Li}}_{\text{3}}\text{N}$ is $18.06\times {10}^{23}$ and $6.02\times {10}^{23}$ respectively.