Chapter 9, Problem 9.76E

Interpretation Introduction

(a)

Interpretation:

The number of equivalents and milliequivalents in $5.00\text{g}$ of given salts is to be calculated.

Concept introduction:

The amount of electrical charge for salts is expressed in the units of equivalents. On dissociation of salt, one equivalent of a salt is equal to the amount of one mole of positive charge or negative charge.

Answer to Problem 9.76E

The number of equivalents and milliequivalents of ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$ are $0.035\text{eq}$ and $35.0\text{meq}$ respectively.

Explanation of Solution

The given amount of ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$ is $5.00\text{g}$. The molar mass of ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$ is $286.00\text{g}/\text{mol}$. The number of moles of $\text{NaCl}$ is calculated by the formula,

$\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values of given mass and molar mass in the given formula.

$\begin{array}{rll}\text{Number}\text{of}\text{Moles}& =& \frac{5.00\text{g}}{286.00\text{g}/\text{mol}}\\ \text{Number}\text{of}\text{Moles}& =& 0.0175\text{moles}\end{array}$

Thus, the number of moles of ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$ is $0.0175\text{moles}$.

On dissociation of $1\text{mol}{\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$, $2\text{mol}{\text{Na}}^{+}$ or $2\text{mol}$ of positive chargesare produced.

Thus, $1\text{mol}{\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}=2\text{eq}{\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$.

Therefore, $0.0175\text{moles}{\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}=2\times 0.0175\text{eq}{\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}=0.035\text{eq}{\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$. Thus, the number of equivalents of ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$ are $0.035\text{eq}$.

For the calculation of milliequivalents the conversion factor is,

$1\text{eq}=1000\text{meq}$

For $0.035\text{eq}{\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$, the number of milliequivalents are,

$\begin{array}{l}\text{Number}\text{of}\text{milliequivalents}=0.035\text{eq}\times \frac{1000\text{meq}}{1\text{eq}}\\ \text{Number}\text{of}\text{milliequivalents}=35.0\text{meq}\end{array}$

Thus, the number of milliequivalents of ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$ are $35.0\text{meq}$.

Conclusion

The number of equivalents and milliequivalents of ${\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_{\text{2}}\text{O}$ are $0.035\text{eq}$ and $35.0\text{meq}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The number of equivalents and milliequivalents in $5.00\text{g}$ of given salts is to be calculated.

Concept introduction:

The amount of electrical charge for salts is expressed in the units of equivalents. On dissociation of salt, one equivalent of a salt is equal to the amount of one mole of positive charge or negative charge.

Answer to Problem 9.76E

The number of equivalents and milliequivalents of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$ are $0.040\text{eq}$ and $40.0\text{meq}$ respectively.

Explanation of Solution

The given amount of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$ is $5.00\text{g}$. The molar mass of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$ is $249.60\text{g}/\text{mol}$. The number of moles of ${\text{NaNO}}_3$ is calculated by the formula,

$\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values of given mass and molar mass in the given formula.

$\begin{array}{rll}\text{Number}\text{of}\text{Moles}& =& \frac{5.00\text{g}}{249.60\text{g}/\text{mol}}\\ \text{Number}\text{of}\text{Moles}& =& 0.020\text{moles}\end{array}$

Thus, the number of moles of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$ is $0.020\text{moles}$.

On dissociation of $1\text{mol}{\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$, $1\text{mol}{\text{Cu}}^{2+}$ or $2\text{mol}$ of positive charges are produced.

Thus, $1\text{mol}{\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}=2\text{eq}{\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$.

Therefore, $0.020\text{moles}{\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}=2\times 0.020\text{eq}{\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}=0.040\text{eq}{\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$. Thus, the number of equivalents of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$ are $0.040\text{eq}$.

For the calculation of milliequivalents, the conversion factor is,

$1\text{eq}=1000\text{meq}$

For $0.040\text{eq}{\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$, the number of milliequivalents are,

$\begin{array}{l}\text{Number}\text{of}\text{milliequivalents}=0.040\text{eq}\times \frac{1000\text{meq}}{1\text{eq}}\\ \text{Number}\text{of}\text{milliequivalents}=40.0\text{meq}\end{array}$

Thus, the number of milliequivalents of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$ are $40.0\text{meq}$.

Conclusion

The number of equivalents and milliequivalents of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_{\text{2}}\text{O}$ are $0.040\text{eq}$ and $40.0\text{meq}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The number of equivalents and milliequivalents in $5.00\text{g}$ of given salts is to be calculated.

Concept introduction:

The amount of electrical charge for salts is expressed in the units of equivalents. On dissociation of salt, one equivalent of a salt is equal to the amount of one mole of positive charge or negative charge.

Answer to Problem 9.76E

The number of equivalents and milliequivalents of ${\text{Li}}_2{\text{CO}}_3$ are $0.1354\text{eq}$ and $135.4\text{meq}$ respectively.

Explanation of Solution

The given amount of ${\text{Li}}_2{\text{CO}}_3$ is $5.00\text{g}$. The molar mass of ${\text{Li}}_2{\text{CO}}_3$ is $73.89\text{g}/\text{mol}$. The number of moles of ${\text{Na}}_3{\text{PO}}_4$ is calculated by the formula,

$\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values of given mass and molar mass in the given formula.

$\begin{array}{rll}\text{Number}\text{of}\text{Moles}& =& \frac{5.00\text{g}}{73.89\text{g}/\text{mol}}\\ \text{Number}\text{of}\text{Moles}& =& 0.0677\text{moles}\end{array}$

Thus, the number of moles of ${\text{Li}}_2{\text{CO}}_3$ is $0.0677\text{moles}$.

On dissociation of $1\text{mol}{\text{Li}}_2{\text{CO}}_3$, $2\text{mol}{\text{Li}}^{+}$ or $2\text{mol}$ of positive charges are produced.

Thus, $1\text{mol}{\text{Li}}_2{\text{CO}}_3=2\text{eq}{\text{Li}}_2{\text{CO}}_3$.

Therefore, $0.0677\text{moles}{\text{Li}}_2{\text{CO}}_3=2\times 0.0677\text{eq}{\text{Li}}_2{\text{CO}}_3=0.1354\text{eq}{\text{Li}}_2{\text{CO}}_3$. Thus, the number of equivalents of ${\text{Li}}_2{\text{CO}}_3$ are $0.1354\text{eq}$.

For the calculation of milliequivalents the conversion factor is,

$1\text{eq}=1000\text{meq}$

For $0.1354\text{eq}{\text{Li}}_2{\text{CO}}_3$, the number of milliequivalents is,

$\begin{array}{l}\text{Number}\text{of}\text{milliequivalents}=0.1354\text{eq}\times \frac{1000\text{meq}}{1\text{eq}}\\ \text{Number}\text{of}\text{milliequivalents}=135.4\text{meq}\end{array}$

Thus, the number of milliequivalents of ${\text{Li}}_2{\text{CO}}_3$ are $135.4\text{meq}$.

Conclusion

The number of equivalents and milliequivalents of ${\text{Li}}_2{\text{CO}}_3$ are $0.1354\text{eq}$ and $135.4\text{meq}$ respectively.

Interpretation Introduction

(d)

Interpretation:

The number of equivalents and milliequivalents in $5.00\text{g}$ of given salts is to be calculated.

Concept introduction:

The amount of electrical charge for salts is expressed in the units of equivalents. On dissociation of salt, one equivalent of a salt is equal to the amount of one mole of positive charge or negative charge.

Answer to Problem 9.76E

The number of equivalents and milliequivalents of ${\text{NaH}}_2{\text{PO}}_4$ are $0.0417\text{eq}$ and $41.7\text{meq}$ respectively.

Explanation of Solution

The given amount of ${\text{NaH}}_2{\text{PO}}_4$ is $5.00\text{g}$. The molar mass of ${\text{NaH}}_2{\text{PO}}_4$ is $120.00\text{g}/\text{mol}$. The number of moles of ${\text{MgSO}}_{\text{4}}\cdot 7{\text{H}}_{\text{2}}\text{O}$ is calculated by the formula,

$\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values of given mass and molar mass in the given formula.

$\begin{array}{rll}\text{Number}\text{of}\text{Moles}& =& \frac{5.00\text{g}}{120.00\text{g}/\text{mol}}\\ \text{Number}\text{of}\text{Moles}& =& 0.0417\text{moles}\end{array}$

Thus, the number of moles of ${\text{NaH}}_2{\text{PO}}_4$ is $0.0417\text{moles}$.

On dissociation of $1\text{mol}{\text{NaH}}_2{\text{PO}}_4$, $1\text{mol}{\text{Na}}^{+}$ or $1\text{mol}$ of positive charges are produced.

Thus, $1\text{mol}{\text{NaH}}_2{\text{PO}}_4=1\text{eq}{\text{NaH}}_2{\text{PO}}_4$.

Therefore, $0.042\text{moles}{\text{NaH}}_2{\text{PO}}_4=0.0417\text{eq}{\text{NaH}}_2{\text{PO}}_4$. Thus, the number of equivalents of ${\text{NaH}}_2{\text{PO}}_4$ are $0.0417\text{eq}$.

For the calculation of milliequivalents the conversion factor is,

$1\text{eq}=1000\text{meq}$

For $0.0417\text{eq}{\text{NaH}}_2{\text{PO}}_4$, the number of milliequivalents is,

$\begin{array}{l}\text{Number}\text{of}\text{milliequivalents}=0.0417\text{eq}\times \frac{1000\text{meq}}{1\text{eq}}\\ \text{Number}\text{of}\text{milliequivalents}=41.7\text{meq}\end{array}$

Thus, the number of milliequivalents of ${\text{NaH}}_2{\text{PO}}_4$ is $41.7\text{meq}$.

Conclusion

The number of equivalents and milliequivalents of ${\text{NaH}}_2{\text{PO}}_4$ are $0.0417\text{eq}$ and $41.7\text{meq}$ respectively.