Chapter 4, Problem 48QP

(a)

Interpretation Introduction

Interpretation:

The number of moles of ${\text{Na}}_2{\text{CO}}_3$ in a sample of $15\text{ g}$ ${\text{Na}}_2{\text{CO}}_3$ is to be determined.

Explanation of Solution

The molar mass of $15\text{ g}$ ${\text{Na}}_2{\text{CO}}_3$ is to be converted into an equivalent number of moles.

The number of moles of ammonia present is determined as follows.

$n=\frac{m}{MM}$ ...(1)

Here, $n$ is the number of moles, $m$ is the given mass and $MM$ is the molar mass.

The molar mass of ${\text{Na}}_2{\text{CO}}_3$ is $105.99\text{g/mol}$ . The weight of ${\text{Na}}_2{\text{CO}}_3$ is $15\text{ g}$ . Substitute these values in equation (1) as follows.

$\begin{array}{c}\text{Number of moles}=\frac{15\cancel{\text{g}}}{\text{105}\text{.99}\cancel{\text{g}}\text{/mol}}\\ =0.142{\text{ mol Na}}_2{\text{CO}}_3\end{array}$

Therefore, there is $0.142\text{ mol}$ of ${\text{Na}}_2{\text{CO}}_3$ in a sample of $15\text{ g}$ ${\text{Na}}_2{\text{CO}}_3$ .

(b)

Interpretation Introduction

Interpretation:

The number of formula units of ${\text{Na}}_2{\text{CO}}_3$ in a sample of $15\text{ g}$ ${\text{Na}}_2{\text{CO}}_3$ is to be determined.

Explanation of Solution

The number of formula units present is determined as follows.

$\text{Number of formula units}=\frac{m}{MM}\times {\text{N}}_{\text{A}}$ ...(2)

Here, ${\text{N}}_{\text{A}}$ is Avogadro’s number with a value of $6.022\times {10}^{23}{\text{mol}}^{-1}$ .

The molar mass of ${\text{Na}}_2{\text{CO}}_3$ is $105.99\text{g/mol}$ . The weight of ${\text{Na}}_2{\text{CO}}_3$ is $15\text{ g}$ . Substitute these values in equation (2) as follows.

$\begin{array}{c}{\text{Formula units Na}}_2{\text{CO}}_3=\frac{15\cancel{\text{g}}}{105.99\cancel{\text{g}}\text{/}\cancel{\text{mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =8.52\times {10}^{22}{\text{ formula units Na}}_2{\text{CO}}_3\end{array}$

Therefore, there are $8.52\times {10}^{22}{\text{ formula units of Na}}_2{\text{CO}}_3$ in a sample of $15\text{ g}$ ${\text{Na}}_2{\text{CO}}_3$ .

(c)

Interpretation Introduction

Interpretation:

The number of ${\text{Na}}^{+}$ ions in a sample of $15\text{ g}$ ${\text{Na}}_2{\text{CO}}_3$ is to be determined.

Explanation of Solution

To calculate the number of ${\text{Na}}^{+}$ ions in the given sample, the ratio between the elements in the chemical formula is to be considered. The ratio of sodium ions in ${\text{Na}}_2{\text{CO}}_3$ is $2{\text{ Na}}^{+}\text{ ions/1 formula units}$ . The conversion is:

$\text{Number of formula units}\stackrel{\text{formula ratio}}{\to }{\text{Number of Na}}^{+}\text{ ions}$

There are $8.52\times {10}^{22}{\text{ formula units of Na}}_2{\text{CO}}_3$ in a sample of $15\text{ g}$ ${\text{Na}}_2{\text{CO}}_3$ . The number of ${\text{Na}}^{+}$ ions is determined as follows.

$\begin{array}{c}{\text{Na}}^{+}\text{ ions}=8.5223\times {10}^{22}\cancel{{\text{formula units Na}}_2{\text{CO}}_3}\times \frac{{\text{2 Na}}^{+}\text{ ions}}{\cancel{{\text{formula units Na}}_2{\text{CO}}_3}}\\ =1.70\times {10}^{23}{\text{Na}}^{+}\text{ ions}\end{array}$

Therefore, there are $1.70\times {10}^{23}{\text{Na}}^{+}\text{ ions}$ in a sample of $15\text{ g}$ ${\text{Na}}_2{\text{CO}}_3$ .

(d)

Interpretation Introduction

Interpretation:

The number of moles of ${\text{CO}}_3{}^{2-}$ ions in a sample of $15\text{ g}$ ${\text{Na}}_2{\text{CO}}_3$ is to be determined.

Explanation of Solution

To calculate the number of moles of ${\text{CO}}_3{}^{2-}$ ions in the given sample, the ratio between the carbonate ions and the formula units of ${\text{Na}}_2{\text{CO}}_3$ is to be considered.

Since one mole of carbonate ion is required for one mole of sodium carbonate hence, the moles of carbonate is the same as that of moles of sodium carbonate that is $0.142\text{ mol}$ .