Chapter 13, Problem 13.133P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the solute is to be calculated.

Concept introduction:

The boiling point of the substance is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure and the liquid changes into a vapor. Liquids can change into vapors at temperatures below the boiling point through evaporation. It is the process that occurs on the liquid surface due to which it changes into vapors. It is a colligative property because it depends on the number of moles of solute particles that are present in the substance.

The formula to calculate the change in boiling point is as follows:

  $\Delta T_{\text{b}}=i{k}_{\text{b}}m$        (1)

Here,

$\Delta T_{\text{b}}$ is the change in boiling point.

$i$ is van’t Hoff factor.

$k_{\text{b}}$ is the boiling point elevation constant.

$m$ is the molality of the solution.

Answer to Problem 13.133P

$\text{68}\text{g/mol}$ is the molar mass of the solute.

Explanation of Solution

The formula to calculate the change in boiling point is as follows:

  $\Delta T_{\text{b}}=T_{\text{b}}{}_{\left(\text{solution}\right)}-T_{\text{b}}{}_{\left(\text{solvent}\right)}$        (2)

Substitute $100.45°\text{C}$ for $T_{\text{b}}{}_{\left(\text{solution}\right)}$ and $100°\text{C}$ for $T_{\text{b}}{}_{\left(\text{solvent}\right)}$ in equation (2).

  $\begin{array}{c}\Delta T_{\text{b}}=100.45°\text{C}-\text{100}°\text{C}\\ =\text{0}\text{.45}°\text{C}\end{array}$

The solute is a nonvolatile non-electrolyte so its van’t Hoff factor is 1.

Rearrange equation (1) to calculate the molarity of the solution as follows:

  $m=\frac{\Delta T_{\text{b}}}{i{k}_{\text{b}}}$        (3)

Substitute 1 for $i$, $\text{0}\text{.45}°\text{C}$ for $\Delta T_{\text{b}}$, $\text{0}\text{.512}°\text{C/}m$ for $k_{\text{b}}$ in equation (3).

  $\begin{array}{c}m=\frac{\text{0}\text{.45}°\text{C}}{\left(1\right)\left(\text{0}\text{.512}°\text{C/}m\right)}\\ =0.878906m\end{array}$

The density of the solution is calculated as follows:

  $\text{Density of solution}\left(\rho \right)=\frac{\text{Mass }\left(M\right)\text{of solution}}{\text{Volume }\left(V\right)\text{of solution}}$        (4)

Rearrange equation (4) to calculate the mass of the solution as follows:

  $\text{Mass of solution}=\left(\text{Density of solution}\right)\left(\text{Volume of solution}\right)$        (5)

Substitute $25\text{mL}$ for the mass of solution and $0.997\text{g/mL}$ for the density of the solution in equation (5) to calculate the mass of water.

  $\begin{array}{c}\text{Mass of water}=\left(25\text{mL}\right)\left(\frac{0.997\text{g}}{1\text{mL}}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\\ =0.0249250\text{kg}\end{array}$

The formula to calculate the molality of the solution is as follows:

  $\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$        (6)

Rearrange equation (6) to calculate the moles of solute as follows:

  $\text{Amount}\text{of}\text{solute}=\left(\text{Molality}\right)\left(\text{mass}\text{of}\text{}\text{solvent}\right)$        (7)

Substitute $0.878906m$ for the molality and $0.0249250\text{kg}$ for the mass of solvent in equation (7) to calculate the amount of solute.

  $\begin{array}{c}\text{Amount of solute}=\left(0.878906m\right)\left(0.0249250\text{kg}\right)\\ =0.0219067\text{mol}\end{array}$

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

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$        (8)

Rearrange equation (8) to calculate the molar mass as follows:

  $\text{Molar mass}=\frac{\text{Given mass}}{\text{Number of moles}}$        (9)

Substitute $1.50\text{g}$ for the given mass and $0.0219067\text{mol}$ for the number of moles in equation (9).

  $\begin{array}{c}\text{Molar mass}=\frac{1.50\text{g}}{0.0219067\text{mol}}\\ =68.4722\text{}\text{g/mol}\\ \approx \text{68}\text{g/mol}\text{.}\end{array}$

Conclusion

Boiling point elevation is a colligative property as it depends on the number of moles of solute.

(b)

Interpretation Introduction

Interpretation:

The molar mass of the solute is to be calculated.

Concept introduction:

The boiling point of the substance is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure and the liquid changes into a vapor. Liquids can change into vapors at temperatures below the boiling point through evaporation. It is the process that occurs on the liquid surface due to which it changes into vapors. It is a colligative property because it depends on the number of moles of solute particles that are present in the substance.

The formula to calculate the change in boiling point is as follows:

  $\Delta T_{\text{b}}=i{k}_{\text{b}}m$        (1)

Here,

$\Delta T_{\text{b}}$ is the change in boiling point.

$i$ is van’t Hoff factor.

$k_{\text{b}}$ is the boiling point elevation constant.

$m$ is the molality of the solution.

Answer to Problem 13.133P

$\text{2}\text{.1}\times {\text{10}}^2\text{g/mol}$ is the  molar mass of the solute.

Explanation of Solution

The solute is ionic with general formula ${\text{AB}}_2$ or ${\text{A}}_{\text{2}}\text{B}$ so it will dissociate into three ions and therefore its van’t Hoff factor is 3.

Substitute 3 for $i$, $\text{0}\text{.45}°\text{C}$ for $\Delta T_{\text{b}}$, $\text{0}\text{.512}°\text{C/}m$ for $k_{\text{b}}$ in equation (3).

  $\begin{array}{c}m=\frac{\text{0}\text{.45}°\text{C}}{\left(3\right)\left(\text{0}\text{.512}°\text{C/}m\right)}\\ =0.29296875m\end{array}$

Substitute $0.29296875m$ for the molality and $0.0249250\text{kg}$ for the mass of solvent in equation (7) to calculate the amount of solute.

  $\begin{array}{c}\text{Amount of solute}=\left(0.29296875m\right)\left(0.0249250\text{kg}\right)\\ =0.00730225\text{mol}\end{array}$

Substitute $1.50\text{g}$ for the given mass and $0.00730225\text{mol}$ for the number of moles in equation (9).

  $\begin{array}{c}\text{Molar mass}=\frac{1.50\text{g}}{0.00730225\text{mol}}\\ =205.416\text{}\text{g/mol}\\ \approx \text{2}\text{.1}\times {\text{10}}^2\text{g/mol}\text{.}\end{array}$

Conclusion

Boiling point elevation depends on the moles of solute and therefore it is a colligative property.

(c)

Interpretation Introduction

Interpretation:

The difference between the actual formula mass and that calculated from the boiling point elevation is to be explained.

Concept introduction:

The boiling point of the substance is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure and the liquid changes into a vapor. Liquids can change into vapors at temperatures below the boiling point through evaporation. It is the process that occurs on the liquid surface due to which it changes into vapors. It is a colligative property because it depends on the number of moles of solute particles that are present in the substance.

Answer to Problem 13.133P

The actual molar mass of ${\text{CaN}}_2{\text{O}}_6$ is $164.10\text{g/mol}$ that lies in between that calculated when the compound is nonelectrolyte and that calculated when it is an electrolyte. This indicates the compound is ionic but dissociation is incomplete.

Explanation of Solution

The molar mass of ${\text{CaN}}_2{\text{O}}_6$ is $164.10\text{g/mol}$. This molar mass is less than that calculated when the compound is considered to be ionic $\left(\text{2}\text{.1}\times {\text{10}}^2\text{g/mol}\right)$ but it is more than that calculated when the compound is considered as a nonelectrolyte $\left(\text{68}\text{g/mol}\right)$. This indicates that the compound dissociates into its ions in the solution and therefore it is an electrolyte. But the ions do not completely dissociate in the solution.

Conclusion

Boiling point elevation is a colligative property as it depends on the number of moles of solute.

(d)

Interpretation Introduction

Interpretation:

The van’t Hoff factor for the solution is to be calculated.

Concept introduction:

The formula to relate the elevation in boiling point and van’t Hoff factor is as follows:

  $\Delta T_{\text{b}}=i{k}_{\text{b}}m$        (1)

Here,

$\Delta T_{\text{b}}$ is the change in boiling point.

$i$ is van’t Hoff factor.

$k_{\text{b}}$ is the boiling point elevation constant.

$m$ is the molality of the solution.

Answer to Problem 13.133P

2.4 is the van’t Hoff factor.

Explanation of Solution

Substitute $1.50\text{g}$ for the given mass and $164.10\text{g/mol}$ for the molar mass in equation (8) to calculate the moles of ${\text{CaN}}_2{\text{O}}_6$.

  $\begin{array}{c}{\text{Moles of CaN}}_2{\text{O}}_6=\frac{1.50\text{g}}{164.10\text{g/mol}}\\ =0.0091408\text{}\text{mol}\end{array}$

Substitute $0.0091408\text{}\text{mol}$ for the moles of solute and $0.0249250\text{kg}$ for the mass of solvent in equation (6) to calculate the molality of ${\text{CaN}}_2{\text{O}}_6$.

  $\begin{array}{c}{\text{Molality of CaN}}_2{\text{O}}_6=\frac{0.0091408\text{}\text{mol}}{0.0249250\text{kg}}\\ =0.3667314m\end{array}$

Rearrange equation (1) to calculate the van’t Hoff factor is as follows:

  $i=\frac{\Delta T_{\text{b}}}{k_{\text{b}}m}$        (10)

Substitute $\text{0}\text{.45}°\text{C}$ for $\Delta T_{\text{b}}$, $\text{0}\text{.512}°\text{C/}m$ for $k_{\text{b}}$, $0.3667314m$ for $m$ in equation (10).

  $\begin{array}{c}i=\frac{\text{0}\text{.45}°\text{C}}{\left(\text{0}\text{.512}°\text{C/}m\right)\left(0.3667314m\right)}\\ =2.39659\\ \approx \mathrm{2.4.}\end{array}$

Conclusion

The van’t Hoff factor for the solution is 2.4.