Chapter F, Problem E.23E

(a)

Interpretation Introduction

Interpretation:

The amount of ${\text{Cu}}^{\text{2+}}$ in moles has to be calculated that is present in $\text{3}\text{.00}\text{g}$ of ${\text{CuBr}}_{\text{2}}$.

Answer to Problem E.23E

Moles of ${\text{Cu}}^{\text{2+}}$ present in $\text{3}\text{.00}\text{g}$ of ${\text{CuBr}}_{\text{2}}$ is $0.0134\text{mol}$.

Explanation of Solution

The mass of copper(II) bromide is given as $\text{3}\text{.00}\text{g}$.  The molar mass of ${\text{CuBr}}_{\text{2}}$ is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}=\text{M(Cu)}+2\times \text{M(Br)}\\ =63.55\text{g}\cdot {\text{mol}}^{-1}+2(80.00\text{g}\cdot {\text{mol}}^{-1})\\ =223.55\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for copper(II) bromide, $223.55\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of copper(II) bromide present in $\text{3}\text{.00}\text{g}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}{\text{CuBr}}_{\text{2}}}{\text{Molar}\text{mass}\text{of}{\text{CuBr}}_{\text{2}}}\\ =\frac{3.00\text{g}}{223.55\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0134\text{mol}\end{array}$

One mole of copper(II) bromide contains one mole of ${\text{Cu}}^{\text{2+}}$ ions.  Therefore, the moles of ${\text{Cu}}^{\text{2+}}$ ions in $0.0134\text{mol}$ of copper(II) bromide is calculated as shown below;

    $\begin{array}{c}\text{Moles}\text{of}{\text{Cu}}^{\text{2+}}=\frac{\text{0}\text{.0134}\text{mol×1}\text{mol}{\text{Cu}}^{\text{2+}}}{\text{1}\text{mol}{\text{CuBr}}_{\text{2}}}\\ =0.0134\text{mol}\end{array}$

Thus the moles of copper(II) ion present in $\text{3}\text{.00}\text{g}$ of copper(II) bromide is $0.0134\text{mol}$.

(b)

Interpretation Introduction

Interpretation:

The amount of ${\text{SO}}_{\text{3}}$ molecules in moles has to be calculated that is present in $\text{7}\text{.00}\times {\text{10}}^{\text{2}}\text{mg}$ of ${\text{SO}}_{\text{3}}$.

Answer to Problem E.23E

Moles of ${\text{SO}}_{\text{3}}$ present in $\text{7}\text{.00}\times {\text{10}}^{\text{2}}\text{mg}$ of ${\text{SO}}_{\text{3}}$ is $0.0087\text{mol}$.

Explanation of Solution

The mass of sulfur trioxide is given as $\text{7}\text{.00}\times {\text{10}}^{\text{2}}\text{mg}$.  The molar mass of ${\text{SO}}_{\text{3}}$ is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}=\text{M(S)}+3\times \text{M(O)}\\ =32.06\text{g}\cdot {\text{mol}}^{-1}+3(16.00\text{g}\cdot {\text{mol}}^{-1})\\ =80.06\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for sulfur trioxide, $80.06\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of ${\text{SO}}_{\text{3}}$ present in $\text{7}\text{.00}\times {\text{10}}^{\text{2}}\text{mg}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}{\text{SO}}_{\text{3}}}{\text{Molar}\text{mass}\text{of}{\text{SO}}_{\text{3}}}\\ =\frac{\text{7}\text{.00}\times {\text{10}}^{\text{2}}\text{mg}}{80.06\text{g}\cdot {\text{mol}}^{-1}}\\ =\frac{0.700\text{g}}{80.06\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0087\text{mol}\end{array}$

Thus the moles of sulfur trioxide present in $\text{7}\text{.00}\times {\text{10}}^{\text{2}}\text{mg}$ of sulfur trioxide is $0.0087\text{mol}$.

(c)

Interpretation Introduction

Interpretation:

The amount of ${\text{F}}^{-}$ ions in moles has to be calculated that is present in $\text{25}\text{.2}\text{kg}$ of ${\text{UF}}_{\text{6}}$.

Answer to Problem E.23E

Moles of ${\text{F}}^{-}$ present in $25.2\text{kg}$ of ${\text{UF}}_{\text{6}}$ is $430\text{mol}$.

Explanation of Solution

The mass of ${\text{UF}}_{\text{6}}$ is given as $\text{25}\text{.2}\text{kg}$.  The molar mass of ${\text{UF}}_{\text{6}}$ is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}=\text{M(U)}+6\times \text{M(F)}\\ =238.03\text{g}\cdot {\text{mol}}^{-1}+6(18.99\text{g}\cdot {\text{mol}}^{-1})\\ =351.97\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for ${\text{UF}}_{\text{6}}$, $351.97\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of ${\text{UF}}_{\text{6}}$ present in $25.2\text{kg}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}{\text{UF}}_{\text{6}}}{\text{Molar}\text{mass}\text{of}{\text{UF}}_{\text{6}}}\\ =\frac{\text{25}\text{.2}\times {\text{10}}^{\text{3}}\text{g}}{351.97\text{g}\cdot {\text{mol}}^{-1}}\\ =71.6\text{mol}\end{array}$

Thus the moles of ${\text{UF}}_{\text{6}}$ present in $\text{25}\text{.2}\text{kg}$ of ${\text{UF}}_{\text{6}}$ is $71.6\text{mol}$.

One mole of ${\text{UF}}_{\text{6}}$ contains six moles of ${\text{F}}^{-}$ ion.  Therefore, the number of moles of ${\text{F}}^{-}$ ion present in $\text{25}\text{.2}\text{kg}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{F}}^{-}=6\times 71.6\text{mol}\\ \text{=429}\text{.6}\text{mol}\\ \text{=430}\text{mol}\end{array}$

Therefore, the number of moles of ${\text{F}}^{-}$ ion present in $\text{25}\text{.2}\text{kg}$ of ${\text{UF}}_{\text{6}}$ is $\text{430}\text{mol}$.

(d)

Interpretation Introduction

Interpretation:

The amount of ${\text{H}}_{\text{2}}\text{O}$ in moles has to be calculated that is present in $\text{2}\text{.00}\text{g}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$.

Answer to Problem E.23E

Moles of ${\text{H}}_{\text{2}}\text{O}$ present in $\text{2}\text{.00}\text{g}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ is $0.0087\text{mol}$.

Explanation of Solution

The mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ is given as $\text{2}\text{.00}\text{g}$.  The molar mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}=2\times \text{M(Na)}+1\times \text{M(C)}+3\times \text{M(O)}+10\times {\text{M(H}}_{\text{2}}\text{O)}\\ =2\times 22.99\text{g}\cdot {\text{mol}}^{-1}+1(12.01\text{g}\cdot {\text{mol}}^{-1})+3(16.00\text{g}\cdot {\text{mol}}^{-1})+10(18.00\text{g}\cdot {\text{mol}}^{-1})\\ =(45.98+12.01+48.00+180.00)\text{g}\cdot {\text{mol}}^{-1}\\ =285.99\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$, $285.99\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ present in $\text{2}\text{.00}\text{g}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}}{\text{Molar}\text{mass}\text{of}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}}\\ =\frac{2.00\text{g}}{285.99\text{g}\cdot {\text{mol}}^{-1}}\\ =0.007\text{mol}\end{array}$

Thus the moles of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ present in $\text{2}\text{.00}\text{g}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ is $0.007\text{mol}$.

One mole of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ contains ten moles of ${\text{H}}_{\text{2}}\text{O}$ molecules.  Therefore, the number of moles of ${\text{H}}_{\text{2}}\text{O}$ molecule present in $\text{2}\text{.00}\text{g}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}=10\times 0.007\text{mol}\\ \text{=0}\text{.07}\text{mol}\end{array}$

Therefore, the number of moles of ${\text{H}}_{\text{2}}\text{O}$ molecules present in $\text{2}\text{.00}\text{g}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ is $\text{0}\text{.07}\text{mol}$.