Chapter 3, Problem 3.115P

(a)

Interpretation Introduction

Interpretation:

The larger quantity between $0.4\text{ mol}$ of ${\text{O}}_3\text{ molecules}$ and $0.4\text{ mol}$ of $\text{O atoms}$ is to be identified.

Concept introduction:

$1\text{ mol}$ of any substance contains Avogadro’s number of entities that is $6.022\times {10}^{23}$.

The formula to calculate the number of moles is:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molar mass (g/mol)}}$        (1)

The formula to calculate mass is as follows:

  $\text{Mass of a given sample}\left(\text{g}\right)=\left(\text{molar mass g/mol}\right)\left(\text{number of moles}\right)$        (2)

Answer to Problem 3.115P

Both the quantities have an equal number of entities.

Explanation of Solution

Since an equal number of moles of various substances contains an equal number of entities. Both ${\text{O}}_{\text{3}}$ molecules and $\text{O}$ atoms have an equal number of moles being $\text{0}\text{.4 mol}$ hence the number of entities is equal for both the samples of ${\text{O}}_{\text{3}}$ molecules and $\text{O}$ atoms.

Conclusion

Ozone molecules present in $0.4\text{ mol}$ is equal to the number of oxygen atoms present in $0.4\text{ mol}$.

(b)

Interpretation Introduction

Interpretation:

The larger of the two given quantities in terms of grams is to be identified.

Concept introduction:

$1\text{ mol}$ of any substance contains Avogadro’s number of entities that is $6.022\times {10}^{23}$.

The formula to calculate the number of moles is:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molar mass (g/mol)}}$        (1)

The formula to calculate mass is as follows:

  $\text{Mass of a given sample}\left(\text{g}\right)=\left(\text{molar mass g/mol}\right)\left(\text{number of moles}\right)$        (2)

Answer to Problem 3.115P

$0.4{\text{ mol O}}_3\text{ molecules}$ has a larger mass than $0.4\text{ mol O atoms}$.

Explanation of Solution

The molar mass of O is $16.00\text{ g/mol}$.so the molar mass of ${\text{O}}_{\text{3}}$ would be calculated as follows:

  $\begin{array}{c}{\text{Molar mass of O}}_3=3\left(16.00\text{ g/mol}\right)\\ =48.00\text{ g/mol}\end{array}$

The molar mass of $\text{O}$ would be calculated as follows:

  $\begin{array}{c}{\text{Molar mass of O}}_{}=1\left(16.00\text{ g/mol}\right)\\ =16.00\text{ g/mol}\end{array}$

Since ${\text{O}}_{\text{3}}$ has larger molar mass $\left(48.00\text{ g/mol}\right)$ compared to the molar mass of $\text{O}$ hence the former has greater mass than the latter. This can be verified as follows:

Substitute $48.00\text{ g/mol}$ for molar mass of ${\text{O}}_{\text{3}}$ and $0.4\text{ mol}$ for the number of moles in equation (2).

  $\begin{array}{c}{\text{Mass of O}}_{\text{3}}\text{ in given sample}\left(\text{g}\right)=\left(\text{48}\text{.00 g/mol}\right)\left(\text{0}\text{.4 mol}\right)\\ =19.2\text{ g}\end{array}$

Substitute $16.00\text{ g/mol}$ for molar mass of $\text{O}$ and $0.4\text{ mol}$ for the number of moles in equation (2).

  $\begin{array}{c}{\text{Mass of O}}_{\text{3}}\text{ in given sample}\left(\text{g}\right)=\left(\text{16}\text{.00 g/mol}\right)\left(\text{0}\text{.4 mol}\right)\\ =6.4\text{ g}\text{.}\end{array}$

Conclusion

For the same number of moles, the quantity with larger molar mass will have a larger mass.

(c)

Interpretation Introduction

Interpretation:

Whether the number of moles will be larger in $4.0{\text{ g N}}_{\text{2}}{\text{O}}_{\text{4}}$ or $3.2{\text{ g SO}}_2$ is to be determined.

Concept introduction:

The formula to calculate the number of moles is:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molar mass (g/mol)}}$        (1)

The formula to calculate mass is as follows:

  $\text{Mass of a given sample}\left(\text{g}\right)=\left(\text{molar mass g/mol}\right)\left(\text{number of moles}\right)$        (2)

Answer to Problem 3.115P

In terms of moles ${\text{SO}}_{\text{2}}$ is larger compared to ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

Explanation of Solution

The molar mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is $92.02\text{ g/mol}$. Substitute $92.02\text{ g/mol}$ for molar mass and $4.0\text{ g}$ for given mass in equation (1) to calculate the moles of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

  $\begin{array}{c}{\text{Moles of N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{mol}\right)\text{=}\frac{\text{ 4}\text{.0 g }}{\text{92}\text{.02 g/mol }\text{}}\\ \text{= 0}\text{.043 mol}\end{array}$

Substitute $\text{3}\text{.3 g}$ for the mass of ${\text{SO}}_{\text{2}}$ and $64.06\text{ g/mol}$ for molar mass of ${\text{SO}}_{\text{2}}$ in equation (1) to calculate the moles of ${\text{SO}}_{\text{2}}$ as follows:

  $\begin{array}{c}{\text{Moles of SO}}_{\text{2}}\left(\text{mol}\right)\text{=}\frac{\text{ 3}\text{.3 g }}{\text{64}\text{.06 g/mol }\text{}}\\ \text{= 0}\text{.052 mol}\text{.}\end{array}$

Hence ${\text{SO}}_{\text{2}}$ is a larger quantity in terms of moles.

Conclusion

The number of moles in $3.2{\text{ g SO}}_2$  is greater than the number of moles in $4.0{\text{ g N}}_{\text{2}}{\text{O}}_{\text{4}}$.

(d)

Interpretation Introduction

Interpretation:

Whether the mass of $0.6{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}$ or $0.6{\text{ mol F}}_2$ is larger is to be determined.

Concept introduction:

The formula to calculate the number of moles is:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molar mass (g/mol)}}$        (1)

The formula to calculate mass is as follows:

  $\text{Mass of a given sample}\left(\text{g}\right)=\left(\text{molar mass g/mol}\right)\left(\text{number of moles}\right)$        (2)

Answer to Problem 3.115P

In terms of mass in grams ${\text{F}}_{\text{2}}$ is larger compared to ${\text{C}}_2{\text{H}}_4$.

Explanation of Solution

Substitute $\text{28}\text{.05 g/mol}$ for molar mass of ${\text{C}}_2{\text{H}}_4$ and $0.6\text{ mol}$ for moles of ${\text{C}}_2{\text{H}}_4$ in equation (2).

  $\begin{array}{c}{\text{Mass of C}}_2{\text{H}}_4\text{ in given sample}\left(\text{g}\right)=\left(\text{ 28}\text{.05 g/mol}\right)\left(\text{0}\text{.6 mol}\right)\\ =17\text{ g}\end{array}$

Substitute $\text{38}\text{.00 g/mol}$ for molar mass of ${\text{F}}_{\text{2}}$ and $0.6\text{ mol}$ for moles of ${\text{F}}_{\text{2}}$ in equation (2).

  $\begin{array}{c}{\text{Mass of F}}_{\text{2}}\text{ in given sample}\left(\text{g}\right)=\left(\text{ 38}\text{.00 g/mol}\right)\left(\text{0}\text{.6 mol}\right)\\ =23\text{ g}\text{.}\end{array}$

Hence ${\text{F}}_{\text{2}}$ has larger mass compared to ${\text{C}}_2{\text{H}}_4$.

Conclusion

For the same number of moles, the quantity with larger molar mass will have a larger mass.

(e)

Interpretation Introduction

Interpretation:

Whether the number of ions in $\text{2}{\text{.3 mol NaClO}}_3$ or $\text{2}{\text{.2 mol MgCl}}_2$ is larger is to be determined.

Concept introduction:

The electrolyte is the substance that produces ions when it is dissolved in a polar solvent. It breaks into positively and negatively charged ions that spread uniformly through the solvent. The electrolytic solution, as a whole, is electrically neutral. Sodium chloride, potassium chloride, calcium phosphate are some of the examples of electrolytes.

Strong electrolytes are those electrolytes that completely dissociates into its ions. These have a very high value of electrical conductance. Sodium chloride and potassium chloride are strong electrolytes.

Weak electrolytes are those electrolytes that partially dissociates into its ions. They are poor conductors of electricity. Acetic acid and carbonic acid are weak electrolytes.

Answer to Problem 3.115P

The number of ions is larger in $\text{2}{\text{.2 mol MgCl}}_2$ compared to $\text{2}{\text{.3 mol NaClO}}_3$.

Explanation of Solution

The chemical formulas for sodium chlorate and magnesium chloride are ${\text{NaClO}}_3$ and ${\text{MgCl}}_2$ respectively.

On dissociation ${\text{NaClO}}_3$ gives ${\text{Na}}^{1+}$ and ${\text{ClO}}_3{}^{2-}$ that is $2\text{ mol}$ of ions while ${\text{MgCl}}_2$ on dissociation gives ${\text{Mg}}^{\text{2+}}$ and ${\text{2Cl}}^{-}$ ions that is $\text{3 mol}$ of ions.

Hence total moles of ions present in $\text{2}\text{.3 mol}$ of ${\text{NaClO}}_3$ is calculated as follows:

  $\begin{array}{c}\text{Total moles of ions in 2}{\text{.3 mol NaClO}}_3=\left(2.3\right)\left(\frac{\text{2 mol }}{\text{1 mol }}\right)\\ =4.6\text{ mol}\end{array}$

Total moles of ions present in $\text{2}\text{.2 mol}$ of ${\text{MgCl}}_2$ is calculated as follows:

  $\begin{array}{c}\text{Total moles of ions in 2}{\text{.2 mol MgCl}}_2=\left(2.2\right)\left(\frac{\text{3 mol }}{\text{1 mol }}\right)\\ =6.6\text{ mol}\text{.}\end{array}$

Hence ${\text{MgCl}}_2$ is a larger quantity in terms of the total number of ions compared to ${\text{NaClO}}_3$.

Conclusion

Magnesium chloride and sodium chlorate are strong electrolyte and give corresponding ions in the solution. Magnesium chloride gives more number of ions per mole compared to sodium chlorate.

(f)

Interpretation Introduction

Interpretation:

Whether the number of molecules in $1.0{\text{ g H}}_{\text{2}}\text{O}$ or $1.0{\text{ g H}}_{\text{2}}{\text{O}}_2$ is larger is to be determined.

Concept introduction:

The formula to calculate the number of molecules from the number of moles is as follows:

  $\text{Number}\text{of molecules }=\left[\begin{array}{l}\left(\text{number of moles }\right)\\ \left(6.022\times {10}^{23}\text{molecules}\right)\end{array}\right]$        (3)

Answer to Problem 3.115P

${\text{H}}_{\text{2}}\text{O}$ has a greater number of molecules compared to ${\text{H}}_2{\text{O}}_2$.

Explanation of Solution

Substitute $\text{18}\text{.02 g/mol}$ for molar mass of ${\text{H}}_{\text{2}}\text{O}$ and $1.0\text{ g}$ for the mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (1).

  $\begin{array}{c}{\text{Moles of H}}_{\text{2}}\text{O}\left(\text{mol}\right)=\frac{\text{ 1}\text{.0 g }}{\text{18}\text{.02 g/mol }\text{}}\\ =\text{0}\text{.05549 mol}\end{array}$

Substitute $0.05549\text{ mol}$ for the number of moles in equation (3) to calculate the number of molecules of ${\text{H}}_{\text{2}}\text{O}$.

  $\begin{array}{c}\text{Number}{\text{of H}}_{\text{2}}\text{O molecules}=\left(0.05549\text{ mol}\right)\left(6.022\times {10}^{23}\text{molecules}\right)\\ =0.33416\times {10}^{23}\text{ molecules}\\ \approx 0.3342\times {10}^{23}\text{ molecules}\end{array}$

Substitute $\text{34}\text{.02 g/mol}$ for molar mass of ${\text{H}}_2{\text{O}}_2$ and $1.0\text{ g}$ for the mass of ${\text{H}}_2{\text{O}}_2$ in equation (1).

  $\begin{array}{c}{\text{Moles of H}}_{\text{2}}{\text{O}}_2\left(\text{mol}\right)=\frac{\text{ 1}\text{.0 g }}{\text{34}\text{.02 g/mol }\text{}}\\ =\text{0}\text{.02939 mol}\end{array}$

Substitute $0.02939\text{ mol}$ for the number of moles in equation (3) to calculate the number of molecules of ${\text{H}}_2{\text{O}}_2$.

  $\begin{array}{c}\text{Number}{\text{of H}}_{\text{2}}{\text{O}}_2\text{ molecules }=\left(0.05549\text{ mol}\right)\left(6.022\times {10}^{23}\text{molecules}\right)\\ =0.11701\times {10}^{23}\text{ molecules}\\ \approx 0.1170\times {10}^{23}\text{ molecules}\end{array}$

Hence ${\text{H}}_{\text{2}}\text{O}$ has a greater number of molecules compared to ${\text{H}}_2{\text{O}}_2$.

Conclusion

Greater the number of molecules present in the given mass of the sample larger will be the number of molecules present.

(g)

Interpretation Introduction

Interpretation:

Whether the mass of $6.022\times {10}^{23}{\text{atoms }}^{235}\text{U}$ or $6.022\times {10}^{23}{\text{atoms }}^{238}\text{U}$ is larger is to be determined.

Concept introduction:

$1\text{ mol}$ of any substance contains Avogadro’s number of entities that is $6.022\times {10}^{23}$.

The formula to calculate the number of moles is:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molar mass (g/mol)}}$        (1)

The formula to calculate mass is as follows:

  $\text{Mass of a given sample}\left(\text{g}\right)=\left(\text{molar mass g/mol}\right)\left(\text{number of moles}\right)$        (2)

The formula to calculate the number of molecules is as follows:

  $\text{Number}\text{of molecules }=\left[\begin{array}{l}\left(\text{number of moles }\right)\\ \left(6.022\times {10}^{23}\text{molecules}\right)\end{array}\right]$        (3)

Answer to Problem 3.115P

${}^{238}\text{U}$ has a greater mass number (238) would have a greater total mass in grams compared to ${}^{235}\text{U}$.

Explanation of Solution

For an equal number of atoms of each isotope of uranium that is $6.022\times {10}^{23}$, there would an equal number of moles too and since $6.022\times {10}^{23}$ atoms are equivalent to $1\text{ mol}$ hence each of the given sample of ${}^{235}\text{U}$ and ${}^{238}\text{U}$ are $1\text{ mol}$.

Equation (2) implies that for an equal number of moles the one with greater molar mass or whichever isotope is heavier would have greater mass in grams.

Hence ${}^{238}\text{U}$ that has greater mass number 238 would have a greater total mass in grams compared to ${}^{235}\text{U}$.

Conclusion

For an equal number of particles $\left(6.022\times {10}^{23}\right)$ the one having greater mass number should have greater mass. Hence ${}^{238}\text{U}$ having greater mass number 238 would have a greater total mass in grams compared to ${}^{235}\text{U}$.