Chapter 4, Problem 151QP

(a)

Interpretation Introduction

Interpretation:

Whether the statement for $2.0\text{ L}$ of $0.100\text{ M}$ solution of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is true or not.

Explanation of Solution

The number of moles of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ can be calculated by using the expression of molarity as follows:

$M=\frac{n}{V}$ …… (1)

Here, $M$ is the molarity, $n$ is the number of moles, and $V$ is the volume of the solution.

Substitute $2.0\text{ L}$ for $V$ and $0.100\text{ M}$ for $M$ in equation (1). The number of moles of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is calculated as follows:

$\begin{array}{c}0.100\text{ M}=\frac{n}{2.0\text{ L}}\\ n=0.100\text{ M}\times 2.0\text{ L}\\ =\text{0}\text{.2 mol}\end{array}$

Therefore, $0.2\text{ mol}$ is required to make $2.0\text{ L}$ of $0.100\text{ M}$ solution of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ .

(b)

Interpretation Introduction

Interpretation:

Whether the statement for $2.0\text{ L}$ of $0.100\text{ M}$ solution of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is true or not.

Explanation of Solution

One mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ contains three moles of calcium, two moles of phosphorus, and eight moles of oxygen.

In $2.0\text{ L}$ of $0.100\text{ M}$ solution, $0.2\text{ mol}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is required, so the number of moles of oxygen atoms is calculated as follows:

$\begin{array}{c}\text{moles of O atoms}=0.2{\text{ mol Ca}}_3{\left({\text{PO}}_4\right)}_2\times 8\\ =1.60\text{ mol}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Whether the statement for $2.0\text{ L}$ of $0.100\text{ M}$ solution of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is true or not.

Explanation of Solution

The number of moles of $0.100\text{ M}$ solution of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ in $1.0\text{ L}$ is calculated as follows:

$\begin{array}{c}0.100\text{ M}=\frac{n}{1.0\text{ L}}\\ n=0.100\text{ M}\times 1.0\text{ L}\\ =0.1\text{ mol}\end{array}$

One mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ contains three moles of calcium ions. Thus, the number of moles of calcium ions required in $0.1\text{ mol}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is calculated as follows:

$\begin{array}{c}{\text{Moles of Ca}}^{2+}=0.1\text{ mol}\times \text{3}\\ =\text{0}\text{.3 mol}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Whether the statement for $2.0\text{ L}$ of $0.100\text{ M}$ solution of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is true or not.

Explanation of Solution

The number of moles of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ in $0.1\text{ M}$ in $500\text{ mL}$ is calculated by using equation (1) as follows:

$\begin{array}{c}\text{0}\text{.1 M}=\frac{n}{500\text{ mL}}\times 1000\\ n=\frac{\text{0}\text{.1 M}\times 500\text{ mL}}{1000}\\ =0.05\text{ mol}\end{array}$

One mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ contains two moles of phosphorus atoms and also contains $6.022\times {10}^{23}\text{ atoms}$ . The number of phosphorus atoms in $0.05\text{ mol}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ are calculated as follows:

$\begin{array}{c}\text{Number of P atoms}=0.05\text{ mol}\times 2\text{ mol P}\times \text{6}\text{.022}\times {\text{10}}^{23}\text{ atoms}\\ =\text{6}\text{.022}\times {\text{10}}^{22}\text{ atoms}\end{array}$

(e)

Interpretation Introduction

Interpretation:

Whether the statement for $2.0\text{ L}$ of $0.100\text{ M}$ solution of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is true or not.

Explanation of Solution

One mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ contains three moles of calcium ions. The number of moles required in $2.0\text{ L}$ of $0.100\text{ M}$ solution of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is $0.2\text{ mol}$ . The number of moles of calcium ions in $0.2\text{ mol}$ is calculated as follows:

$\begin{array}{c}{\text{Moles of Ca}}^{2+}=3\text{ mol}\times \text{0}\text{.2 mol}\\ =\text{0}\text{.600 mol}\end{array}$