Chapter 3, Problem 3.36QP

(a)

Interpretation Introduction

Interpretation: The substance which contains greater number of ions from the given pairs of compound is to be selected.

Concept introduction: The mole is unit specially discovered for the atoms or molecules due to extreme small size and large numbers. One mole is defined as the amount of the substance which contains exactly the same number of particles as in $12\text{g}$ of carbon-12 isotope. One mole of anything consists of $N_{\text{A}}$ number of things. This number is known as Avogadro number $\left(N_{\text{A}}\right)$ and is numerically equal to $6.023\times {10}^{23}$. Every molecular formula contains exactly the same number of moles of elements as given the stoichiometric ratio in the molecular formula.

To determine: The substance which contains greater number of ions from the given pairs of compound.

Answer to Problem 3.36QP

Solution

The substance which contains greater number of ions from the given pairs of compound is $\text{KCl}$.

Explanation of Solution

Explanation

Given

The given pair of compound is $\text{NaBr}$ or $\text{KCl}$.

To calculate the number of ions in the given pair of compounds the mass of both the compound is assumed to be $100\text{g}$. The molar mass of both the compounds is given as,

$\begin{array}{ccc}\text{NaBr}& =& 102.89\text{g/mol}\\ \text{KCl}& =& 74.54\text{g/mol}\end{array}$

Now, for the assumed mass, the moles of both the compound is calculated by the formula,

$n=\frac{m}{M}$

Where,

• $n$ is the number of moles.
• $m$ is the given mass of the molecule.
• $M$ is the molar mass of the molecule.

Substitute the values in the above formula to calculate the moles.

For $\text{NaBr}$ compound,

$\begin{array}{c}n=\frac{m}{M}\\ n=\frac{100\text{g}}{102.89\text{g/mol}}\\ n=0.971\text{mol}\end{array}$

For $\text{KCl}$ compound,

$\begin{array}{c}n=\frac{m}{M}\\ n=\frac{100\text{g}}{74.54\text{g/mol}}\\ n=1.34\text{mol}\end{array}$

The compound $\text{NaBr}$ contains two ions that are ${\text{Na}}^{\text{+}}$ and ${\text{Br}}^{-}$. Therefore, one mole of $\text{NaBr}$ contains one mole of ${\text{Na}}^{\text{+}}$ and one mole of ${\text{Br}}^{-}$ that is two moles of ions.

The compound $\text{KCl}$ contains two ions that are ${\text{K}}^{\text{+}}$ and ${\text{Cl}}^{-}$. Therefore, one mole of $\text{KCl}$ contains one mole of ${\text{K}}^{\text{+}}$ and one mole of ${\text{Cl}}^{-}$ that is two moles of ions.

As already given in concept that one mole of anything contains the $N_{\text{A}}$ number of things. Therefore, for the calculated moles, the number of ions, $N_{\text{ions}}$ is calculated by the formula,

$N_{\text{ions}}=2n\times N_{\text{A}}$

Substitute the values for $\text{NaBr}$ compound,

$\begin{array}{l}{N}_{\text{ions}}=2n\times N_{\text{A}}\\ N_{\text{ions}}=2\times 0.927\times 6.023\times {10}^{23}\\ N_{\text{ions}}=11.708\times {10}^{23}\text{Ions}\end{array}$

Substitute the values for $\text{KCl}$ compound,

$\begin{array}{l}{N}_{\text{ions}}=2n\times N_{\text{A}}\\ N_{\text{ions}}=2\times 1.34\times 6.023\times {10}^{23}\\ N_{\text{ions}}=16.14\times {10}^{23}\text{Ions}\end{array}$

The calculated values of number of ions clearly reveals that $\text{KCl}$ compound contains more number of ions than $\text{NaBr}$.

(b)

Interpretation Introduction

To determine: The substance which contains greater number of ions from the given pairs of compound.

Answer to Problem 3.36QP

Solution

The substance which contains greater number of ions from the given pairs of compound is $\text{NaCl}$.

Explanation of Solution

Explanation

Given

The given pair of compound is $\text{NaCl}$ or ${\text{MgCl}}_2$.

To calculate the number of ions in the given pair of compounds the mass of both the compound is assumed to be $100\text{g}$. The molar mass of both the compounds is given as,

$\begin{array}{cc}\text{NaCl}& 58.43\text{g/mol}\\ {\text{MgCl}}_2& 95.2\text{g/mol}\end{array}$

Now, for the assumed mass, the moles of both the compound is calculated by the formula,

$n=\frac{m}{M}$

Substitute the values in the above formula to calculate the moles.

For $\text{NaCl}$ compound,

$\begin{array}{c}n=\frac{m}{M}\\ n=\frac{100\text{g}}{52.43\text{g/mol}}\\ n=1.71\text{mol}\end{array}$

For ${\text{MgCl}}_2$ compound,

$\begin{array}{c}n=\frac{m}{M}\\ n=\frac{100\text{g}}{95.2\text{g/mol}}\\ n=1.05\text{mol}\end{array}$

The compound $\text{NaCl}$ contains two ions that are ${\text{Na}}^{\text{+}}$ and ${\text{Cl}}^{-}$. Therefore, one mole of $\text{NaCl}$ contains one mole of ${\text{Na}}^{\text{+}}$ and one mole of ${\text{Cl}}^{-}$ that is two moles of ions.

As already given in concept that one mole of anything contains the $N_{\text{A}}$ number of things. Therefore, for the calculated moles of $\text{NaCl}$, the number of ions, $N_{\text{ions}}$ is calculated by the formula,

$N_{\text{ions}}=2n\times N_{\text{A}}$

Substitute the values for $\text{NaCl}$ compound,

$\begin{array}{l}{N}_{\text{ions}}=2n\times N_{\text{A}}\\ N_{\text{ions}}=2\times 1.71\times 6.023\times {10}^{23}\\ N_{\text{ions}}=20.6\times {10}^{23}\text{Ions}\end{array}$

The compound ${\text{MgCl}}_2$ contains three ions that are ${\text{Mg}}^{\text{+}}$ and ${\text{Cl}}^{-}$. Therefore, one mole of ${\text{MgCl}}_2$ contains one mole of ${\text{Mg}}^{\text{+}}$ and two mole of ${\text{Cl}}^{-}$ that is three moles of ions.

For the calculated moles of ${\text{MgCl}}_2$, the number of ions, $N_{\text{ions}}$ is calculated by the formula,

$N_{\text{ions}}=3n\times N_{\text{A}}$

Substitute the values for ${\text{MgCl}}_2$ compound,

$\begin{array}{l}{N}_{\text{ions}}=3n\times N_{\text{A}}\\ N_{\text{ions}}=3\times 1.05\times 6.023\times {10}^{23}\\ N_{\text{ions}}=18.97\times {10}^{23}\text{Ions}\end{array}$

The calculated values of number of ions clearly reveals that $\text{NaCl}$ compound contains more number of ions than ${\text{MgCl}}_2$.

(c)

Interpretation Introduction

To determine: The substance which contains greater number of ions from the given pairs of compound.

Answer to Problem 3.36QP

Solution

The substance which contains greater number of ions from the given pairs of compound is ${\text{Li}}_2{\text{CO}}_3$.

Explanation of Solution

Explanation

Given

The given pair of compound is ${\text{BaCl}}_2$ or ${\text{Li}}_2{\text{CO}}_3$.

To calculate the number of ions in the given pair of compounds the mass of both the compound is assumed to be $100\text{g}$. The molar mass of both the compounds is given as,

$\begin{array}{cc}{\text{BaCl}}_2& 208.2\text{g/mol}\\ {\text{Li}}_2{\text{CO}}_3& 73.887\text{g/mol}\end{array}$

Now, for the assumed mass, the moles of both the compound is calculated by the formula,

$n=\frac{m}{M}$

Substitute the values in the above formula to calculate the moles.

For ${\text{BaCl}}_2$ compound,

$\begin{array}{c}n=\frac{m}{M}\\ n=\frac{100\text{g}}{208.2\text{g/mol}}\\ n=0.48\text{mol}\end{array}$

For ${\text{Li}}_2{\text{CO}}_3$ compound,

$\begin{array}{c}n=\frac{m}{M}\\ n=\frac{100\text{g}}{73.887\text{g/mol}}\\ n=1.35\text{mol}\end{array}$

The compound ${\text{BaCl}}_2$ contains three ions that are ${\text{Ba}}^{\text{+}}$ and ${\text{Cl}}^{-}$. Therefore, one mole of ${\text{BaCl}}_2$ contains one mole of ${\text{Ba}}^{\text{+}}$ and two moles of ${\text{Cl}}^{-}$ that is three moles of ions.

As already given in concept that one mole of anything contains the $N_{\text{A}}$ number of things. Therefore, for the calculated moles of ${\text{BaCl}}_2$, the number of ions, $N_{\text{ions}}$ is calculated by the formula,

$N_{\text{ions}}=3n\times N_{\text{A}}$

Substitute the values for ${\text{BaCl}}_2$ compound,

$\begin{array}{l}{N}_{\text{ions}}=3n\times N_{\text{A}}\\ N_{\text{ions}}=3\times 0.48\times 6.023\times {10}^{23}\\ N_{\text{ions}}=8.67\times {10}^{23}\text{Ions}\end{array}$

For the calculated moles of ${\text{Li}}_2{\text{CO}}_3$, the number of ions, $N_{\text{ions}}$ is calculated as,

$\begin{array}{l}{N}_{\text{ions}}=3n\times N_{\text{A}}\\ N_{\text{ions}}=3\times 1.35\times 6.023\times {10}^{23}\\ N_{\text{ions}}=24.39\times {10}^{23}\text{Ions}\end{array}$

The calculated values of number of ions clearly reveals that ${\text{Li}}_2{\text{CO}}_3$ compound contains more number of ions than ${\text{BaCl}}_2$.

Conclusion

1. a) The substance which contains greater number of ions from the given pairs of compound is $\text{KCl}$.
2. b) The substance which contains greater number of ions from the given pairs of compound is $\text{NaCl}$.
3. c) The substance which contains greater number of ions from the given pairs of compound is ${\text{Li}}_2{\text{CO}}_3$