Chapter 11, Problem 78A

(a)

Interpretation Introduction

Interpretation: The given table needs to be completed.

Concept Introduction: From the given mass of a substance, the number of moles can be determined using molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Explanation of Solution

The table shows the maximum amount of water absorbed in different quantities of ${\text{CaCl}}_{\text{2}}$ .

${\text{CaCl}}_{\text{2}}$ (g)	${\text{CaCl}}_{\text{2}}$ (mol)	${\text{H}}_{\text{2}}\text{O}$ (g)	${\text{H}}_{\text{2}}\text{O}$ (mol)
17.3		5.62
48.8		15.8
124		40.3
337		109

The above table can be completed by calculating the number of moles of ${\text{CaCl}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ from given masses. The molar mass of ${\text{CaCl}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ is 110.98 g/mol and 18 g/mol respectively. Number of moles can be calculated as follows:

${\text{CaCl}}_{\text{2}}$ (g)	${\text{CaCl}}_{\text{2}}$ (mol)	${\text{H}}_{\text{2}}\text{O}$ (g)	${\text{H}}_{\text{2}}\text{O}$ (mol)
17.3	$\frac{17.3\text{ g}}{110.98\text{ g/mol}}=0.156\text{ mol}$	5.62	$\frac{5.62\text{ g}}{\text{18 g/mol}}=0.312\text{ mol}$
48.8	$\frac{\text{48}\text{.8 g}}{110.98\text{ g/mol}}=0.439\text{ mol}$	15.8	$\frac{15.8\text{ g}}{\text{18 g/mol}}=0.878\text{ mol}$
124	$\frac{\text{124 g}}{110.98\text{ g/mol}}=1.117\text{ mol}$	40.3	$\frac{40.3\text{ g}}{\text{18 g/mol}}=2.24\text{ mol}$
337	$\frac{337\text{ g}}{110.98\text{ g/mol}}=3.04\text{ mol}$	109	$\frac{109\text{ g}}{\text{18 g/mol}}=6.056\text{ mol}$

Thus, the complete table is as follows:

${\text{CaCl}}_{\text{2}}$ (g)	${\text{CaCl}}_{\text{2}}$ (mol)	${\text{H}}_{\text{2}}\text{O}$ (g)	a   ${\text{H}}_{\text{2}}\text{O}$ (mol)
17.3	$0.156$	5.62	$0.312$
48.8	$0.439$	15.8	$0.878$
124	$1.117$	40.3	$2.24$
337	$3.04$	109	$6.056$

(b)

Interpretation Introduction

Interpretation: The number of moles of water absorbed needs to be plotted against the number of moles of ${\text{CaCl}}_{\text{2}}$ .

Concept Introduction: To plot the graph of the number of moles of water absorbed versus the number of moles of ${\text{CaCl}}_{\text{2}}$ , moles of water must be written on the y-axis and moles of ${\text{CaCl}}_{\text{2}}$ on the x-axis. Using excel, the plot can be obtained.

Explanation of Solution

The data is as follows:

${\text{CaCl}}_{\text{2}}$ (mol)	${\text{H}}_{\text{2}}\text{O}$ (mol)
$0.156$	$0.312$
$0.439$	$0.878$
$1.117$	$2.24$
$3.04$	$6.056$

The graph can be plotted as follows:

  

(d)

Interpretation Introduction

Interpretation: Using the graph, the number of molecules of water absorbed in each formula unit of calcium chloride needs to be determined.

Concept Introduction: In 1 mol of a substance, there are $6.022\times {10}^{23}$ units of the substance.

Explanation of Solution

From the graph, it is clear that 0.312 mol of water is absorbed by 0.156 mol of ${\text{CaCl}}_{\text{2}}$ . In 1 mol, there are $6.022\times {10}^{23}$ formula units of ${\text{CaCl}}_{\text{2}}$ similarly, in 1 mol of water, there are $6.022\times {10}^{23}$ molecules of water.

The number of formula units in 0.156 mol of ${\text{CaCl}}_{\text{2}}$ can be calculated as follows:

  $\begin{array}{c}{N}_{{\text{CaCl}}_{\text{2}}}=\left(\frac{6.022\times {10}^{23}}{1\text{ mol}}\right)\left(0.156\text{ mol}\right)\\ =9.44\times {10}^{22}\text{ Formula units}\end{array}$

Similarly, the number of water molecules in 0.312 mol of water will be:

  $\begin{array}{c}{N}_{{\text{H}}_{\text{2}}\text{O}}=\left(\frac{6.022\times {10}^{23}}{1\text{ mol}}\right)\left(0.312\text{ mol}\right)\\ =1.87\times {10}^{23}\text{ molecules}\end{array}$

Thus, $9.44\times {10}^{22}\text{ Formula units}$ of ${\text{CaCl}}_{\text{2}}$ absorb $1.87\times {10}^{23}\text{ molecules}$ of water, the water molecules absorbed by 1 formula unit of ${\text{CaCl}}_{\text{2}}$ will be:

  $\begin{array}{c}X=\frac{1.87\times {10}^{23}}{9.44\times {10}^{22}}\\ =1.98\approx 2\end{array}$

Therefore, 2 water molecules are absorbed by each formula unit of ${\text{CaCl}}_{\text{2}}$