Chapter 10.3, Problem 43PP

a.

Interpretation Introduction

Interpretation:

The number of sodium ions present in 2.25 g of Na₂SO₃ needs to be determined.

Concept introduction:

In 1 mol of any substance, there are $6.022\times {10}^{23}$ units of that substance. These units can be atoms and molecules. The number of units in 1 mol of any substance is known as Avogadro’s number. It is denoted by symbol N_A.

Answer to Problem 43PP

The number of ${\text{Na}}^{+}$ ions present in the sample is $1.9\times {10}^{22}\text{ ions}$

Explanation of Solution

Given information:

The mass sodium sulfite ${\text{(Na}}_2{\text{SO}}_3)$

$=2.25g$

Calculation:

The molar mass of ${\text{Na}}_2{\text{SO}}_3$ is 142 g/mol. The mass is 2.25 g thus, number of moles will be:

$\begin{array}{l}\text{n=}\frac{m}{M}\\ \text{=}\frac{2.25\text{ g}}{142\text{ g/mol}}\\ \text{=0}\text{.0158 moles}\end{array}$

Now from formula, 1 mol of ${\text{Na}}_2{\text{SO}}_3$ has 2 mol of Na. Also, 1 mol of sodium ion contains $6.022\times {10}^{23}$ sodium ions. Thus, number of sodium ion will be:

$\begin{array}{l}N{a}^{+}\text{ ions in sample }=\text{2}\times \text{0}\text{.0158}\times 6.022\times {10}^{23}\\ =1.9\times {10}^{22}\text{ ions}\end{array}$

b.

Interpretation Introduction

Interpretation:

The number of ${\text{SO}}_3{}^{2-}$ ions present in the sample needs to be determined.

Concept introduction:

In 1 mol of any substance, there are $6.022\times {10}^{23}$ units of that substance. These units can be atoms and molecules. The number of units in 1 mol of any substance is known as Avogadro’s number. It is denoted by symbol N_A.

Answer to Problem 43PP

The number of ${\text{SO}}_3{}^{2-}$ ions present in the sample IS $9.5\times {10}^{21}\text{ ions}$

Explanation of Solution

Given information:

The mass sodium sulfite ${\text{(Na}}_2{\text{SO}}_3)$

$=2.25g$

Calculation:

The molar mass of sodium sulfite is 142 g/mol. Thus, number of moles will be:

$\begin{array}{l}\text{n=}\frac{m}{M}\\ \text{=}\frac{2.25\text{ g}}{142\text{ g/mol}}\\ \text{=0}\text{.0158 moles}\end{array}$

In 1 mol of sodium sulfite, 1 mol of ${\text{SO}}_{\text{3}}^{\text{-2}}$ ion is present. Also, the number of ions will be $6.022\times {10}^{23}$ in 1 mol thus, number of ${\text{SO}}_{\text{3}}^{\text{-2}}$ ions will be:

$\begin{array}{l}{\text{SO}}_{\text{4}}^{\text{-2}}\text{ ions in sample =0}\text{.0158×6}\text{.022}\times {\text{10}}^{23}\\ \text{ =9}{\text{.5×10}}^{\text{21}}\text{ ions}\end{array}$

c.

Interpretation Introduction

Interpretation:

The mass of one formula unit of ${\text{Na}}_2{\text{SO}}_3$ in gram needs to be determined.

Concept introduction:

In 1 mol of any substance, there are $6.022\times {10}^{23}$ units of that substance. These units can be atoms and molecules. The number of units in 1 mol of any substance is known as Avogadro’s number. It is denoted by symbol N_A.

Answer to Problem 43PP

The mass of one formula unit of ${\text{Na}}_2{\text{SO}}_3$is $2.36\times {10}^{-22}\text{grams}$

Explanation of Solution

Given information:

The mass sodium sulfite ${\text{(Na}}_2{\text{SO}}_3)$

$=2.25g$

Calculation:

The molar mass of sodium sulfite is 142 g/mol. Thus, number of moles will be:

$\begin{array}{l}\text{n=}\frac{m}{M}\\ \text{=}\frac{2.25\text{ g}}{142\text{ g/mol}}\\ \text{=0}\text{.0158 moles}\end{array}$

Now, $n=0.0158\text{ moles}$ contains $N_A\times n$ formula units

The number of formula units will be:

$\begin{array}{l}\text{N =0}\text{.0158×6}\text{.022}\times {\text{10}}^{23}\\ \text{ =9}{\text{.5×10}}^{\text{21}}\text{ ions}\end{array}$

Since,

$\text{2}{\text{.25g Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=9}{\text{.5×10}}^{\text{21}}$ formula units ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

Thus, the mass of 1 formula unit will be:

${\text{m}}_{{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{per formula units=}\frac{\text{2}\text{.25 g}}{\text{9}{\text{.95×10}}^{\text{21}}\text{ formula units}}\text{=2}{\text{.36×10}}^{\text{-22}}\text{g /formula unit}$