Chapter 15, Problem 50QAP

Interpretation Introduction

(a)

Interpretation:

The number of moles of each ion in the given solution is to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{in}\text{Liters}\right)}$.

Answer to Problem 50QAP

The number of moles of ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ ions in solution is $0.0046\text{moles}$ and $0.0138\text{moles}$ respectively.

Explanation of Solution

It is given that $10.2\text{mL}$ of $0.451\text{M}$

${\text{AlCl}}_{\text{3}}$ solution is prepared.

The conversion of units of volume into $\text{L}$ is done as,

$\begin{array}{c}10.2\text{mL}=\frac{10.2}{1000}\text{L}\\ =0.0102\text{L}\end{array}$

The number of moles of a solute is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$

Substitute the values of volume of solution and molarity of ${\text{AlCl}}_{\text{3}}$ solution in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=0.0102\text{L}\times 0.451\text{M}\\ =0.0046\text{moles}\end{array}$

The compound ${\text{AlCl}}_{\text{3}}$ has one mole of ${\text{Al}}^{3+}$ ions and three moles of ${\text{Cl}}^{-}$ ions.

Thus, the number of moles of ${\text{Al}}^{3+}$ is calculated as shown below.

$\begin{array}{c}\text{Moles}\text{of}{\text{Al}}^{3+}=1\times 0.0046\text{moles}\text{of}{\text{Al}}^{3+}\\ =0.0046\text{moles}\end{array}$

The number of moles of ${\text{Cl}}^{-}$ is calculated as shown below.

$\begin{array}{c}\text{Moles}\text{of}{\text{Cl}}^{-}=3\times 0.0046\text{moles}\text{of}{\text{Cl}}^{-}\\ =0.0138\text{moles}\end{array}$

Therefore, the number of moles of ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ ions in solution is $0.0046\text{moles}$ and $0.0138\text{moles}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The number of moles of each ion in the given solution is to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{in}\text{Liters}\right)}$.

Answer to Problem 50QAP

The number of moles of ${\text{Na}}^{+}$ and ${\text{PO}}_4^{3-}$ ions in solution is $1.704\text{moles}$ and $0.568\text{moles}$ respectively.

Explanation of Solution

It is given that $5.51\text{L}$ of $0.103\text{M}$

${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is prepared.

The number of moles of a solute is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$

Substitute the values of volume of solution and molarity of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=5.51\text{L}\times 0.103\text{M}\\ =0.568\text{moles}\end{array}$

The solution of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ consists of ${\text{Na}}^{+}$ and ${\text{PO}}_4^{3-}$ ions. The compound ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ has three moles of ${\text{Na}}^{+}$ ions and one mole of ${\text{PO}}_4^{3-}$ ions.

Thus, the number of moles of ${\text{Na}}^{+}$ is calculated as shown below.

$\begin{array}{c}\text{Moles}\text{of}{\text{Na}}^{+}=3\times 0.568\text{moles}\\ =1.704\text{moles}\end{array}$

The number of moles of ${\text{PO}}_4^{3-}$ is calculated as shown below.

$\begin{array}{c}\text{Moles}\text{of}{\text{PO}}_4^{3-}=1\times 0.568\text{moles}\\ =0.568\text{moles}\end{array}$

Therefore, the number of moles of ${\text{Na}}^{+}$ and ${\text{PO}}_4^{3-}$ ions in solution is $1.704\text{moles}$ and $0.568\text{moles}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The number of moles of each ion in the given solution is to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{in}\text{Liters}\right)}$.

Answer to Problem 50QAP

The number of moles of ${\text{Cu}}^{2+}$ and ${\text{Cl}}^{-}$ ions in solution is $\text{2}\text{.19}\times {\text{10}}^{-3}\text{moles}$ and $\text{4}\text{.38}\times {\text{10}}^{-3}\text{moles}$ respectively.

Explanation of Solution

It is given that $1.75\text{mL}$ of $1.25\text{M}$

${\text{CuCl}}_2$ solution is prepared.

The conversion of units of volume into $\text{L}$ is done as,

$\begin{array}{c}1.75\text{mL}=\frac{1.75}{1000}\text{L}\\ =0.00175\text{L}\end{array}$

The number of moles of a solute is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$

Substitute the values of volume of solution and molarity of ${\text{CuCl}}_2$ solution in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=0.00175\text{L}\times 1.25\text{M}\\ =0.00219\text{moles}\end{array}$

The solution of ${\text{CuCl}}_2$ consists of ${\text{Cu}}^{2+}$ and ${\text{Cl}}^{-}$ ions. The compound ${\text{CuCl}}_2$ has one mole of ${\text{Cu}}^{2+}$ ions and two moles of ${\text{Cl}}^{-}$ ions.

Thus, the number of moles of ${\text{Cu}}^{2+}$ is calculated as shown below.

$\begin{array}{c}\text{Moles}\text{of}{\text{Cu}}^{2+}=1\times 0.00219\text{moles}\\ =0.00219\text{moles}\\ =\text{2}\text{.19}\times {\text{10}}^{-3}\text{moles}\end{array}$

The number of moles of ${\text{Cl}}^{-}$ is calculated as shown below.

$\begin{array}{c}\text{Moles}\text{of}{\text{Cl}}^{-}=2\times 0.00219\text{moles}\\ =0.00438\text{moles}\\ =\text{4}\text{.38}\times {\text{10}}^{-3}\text{moles}\end{array}$

Therefore, the number of moles of ${\text{Cu}}^{2+}$ and ${\text{Cl}}^{-}$ ions in solution is $\text{2}\text{.19}\times {\text{10}}^{-3}\text{moles}$ and $\text{4}\text{.38}\times {\text{10}}^{-3}\text{moles}$ respectively.

Interpretation Introduction

(d)

Interpretation:

The number of moles of each ion in the given solution is to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{in}\text{Liters}\right)}$.

Answer to Problem 50QAP

The number of moles of ${\text{Ca}}^{2+}$ and ${\text{OH}}^{-}$ ions in solution is $3.96\times {10}^{-5}\text{moles}$ and $7.92\times {\text{10}}^{-5}\text{moles}$ respectively.

Explanation of Solution

It is given that $25.2\text{mL}$ of $0.00157\text{M}$

$\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ solution is prepared.

The conversion of units of volume into $\text{L}$ is done as,

$\begin{array}{c}25.2\text{mL}=\frac{25.2}{1000}\text{L}\\ =0.0252\text{L}\end{array}$

The number of moles of a solute is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{volume}\text{of}\text{solution}\times \text{Molarity}$

Substitute the values of volume of solution and molarity of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ solution in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=0.0252\text{L}\times 0.00157\text{M}\\ =3.96\times {10}^{-5}\text{moles}\end{array}$

The solution of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ consists of ${\text{Ca}}^{2+}$ and ${\text{OH}}^{-}$ ions. The compound $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ has one mole of ${\text{Ca}}^{2+}$ ions and two moles of ${\text{OH}}^{-}$ ions.

Thus, the number of moles of ${\text{Ca}}^{2+}$ is calculated as shown below.

$\begin{array}{c}\text{Moles}\text{of}{\text{Ca}}^{2+}=1\times 3.96\times {10}^{-5}\text{moles}\\ =3.96\times {10}^{-5}\text{moles}\end{array}$

The number of moles of ${\text{OH}}^{-}$ is calculated as shown below.

$\begin{array}{c}\text{Moles}\text{of}{\text{OH}}^{-}=2\times 3.96\times {10}^{-5}\text{moles}\\ =7.92\times {\text{10}}^{-5}\text{moles}\end{array}$

Therefore, the number of moles of ${\text{Ca}}^{2+}$ and ${\text{OH}}^{-}$ ions in solution is $3.96\times {10}^{-5}\text{moles}$ and $7.92\times {\text{10}}^{-5}\text{moles}$ respectively.