Chapter 13, Problem 13.109P

(a)

Interpretation Introduction

Interpretation:

The molality and van’t Hoff factor for the aqueous $\text{KCl}$ solution are to be calculated.

Concept introduction:

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by $\text{m}$ and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

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

$\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$ (1)

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

$\Delta T_{\text{f}}=i{k}_{\text{f}}m$ (2)

Here, $\Delta T_{\text{f}}$ is the change in freezing point.

$i$ is van’t Hoff factor.

$k_{\text{f}}$ is the freezing point depression constant.

$m$ is the molality of the solution.

The van’t Hoff factor is a measure of the effect of solute on the colligative properties. It is represented by $i$. It is defined as the ratio of measured values for the electrolytic solution to the expected values for non-electrolytic solutions. It is a unitless quantity.

Answer to Problem 13.109P

The molality and van’t Hoff factor for the aqueous $\text{KCl}$ solution are $0.000068m$ and 2.

Explanation of Solution

Consider the mass of the solution to be. $100\text{}\text{g}$.

The formula to calculate the mass of the compound is as follows:

$\text{Mass of compound}=\left(\text{Mass of solution}\right)\left(\frac{\text{mass }\%\text{of compound}}{100\text{}\%}\right)$ (3)

Substitute $100\text{}\text{g}$ for the mass of solution and $\text{0}\text{.500 }\%$ for the mass percent of the compound in equation (3) to calculate the mass of $\text{KCl}$.

$\begin{array}{c}\text{Mass of KCl}=\left(\text{100 g}\right)\left(\frac{\text{0}\text{.500 }\%}{100\text{}\%}\right)\\ =0.500\text{g}\end{array}$

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

$\text{Moles of compound}=\frac{\text{Given mass}}{\text{Molar mass}}$ (4)

Substitute $\text{0}\text{.500}\text{g}$ for the given mass and $74.55\text{g/mol}$ for the molar mass in equation (4) to calculate the moles of $\text{KCl}$.

$\begin{array}{c}\text{Moles of KCl}=\left(0.500\text{g}\right)\left(\frac{1\text{mol}}{74.55\text{g}}\right)\\ =0.0067069081\text{mol}\end{array}$

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

$\text{Mass of solution}=\text{Mass of solute}+\text{Mass of solvent}$ (5)

Rearrange equation (5) to calculate the mass of the solvent as follows:

$\text{Mass of solvent}=\text{Mass of solution}\text{}-\text{Mass of solute}$ (6)

Substitute $\text{100}\text{g}$ for the mass of solution and $\text{0}\text{.500}\text{g}$ for the mass of solute in equation (6) to calculate the mass of water.

$\begin{array}{c}\text{Mass of water}=\text{100}\text{g}-\text{0}\text{.500 g}\\ =\text{99}\text{.500 g}\end{array}$

Substitute $0.0067069081\text{mol}$ for the amount of solute and $\text{99}\text{.500 g}$ for the mass of solvent in equation (1) to calculate the molality of $\text{KCl}$.

$\begin{array}{c}\text{Molality of KCl}=\left(\frac{0.0067069081\text{mol}}{99.500\text{g}}\right)\left(\frac{{\text{10}}^{\text{3}}\text{g}}{\text{1}\text{kg}}\right)\\ =0.0000674061m\\ =0.000068m\end{array}$

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

$\Delta T_{\text{f}}=T_{\text{f}}{}_{\left(\text{solvent}\right)}-T_{\text{f}}{}_{\left(\text{solution}\right)}$ (7)

Substitute $0^{°}\text{C}$ for $T_{\text{f}}{}_{\left(\text{solvent}\right)}$ and $-{0.234}^{°}\text{C}$ for $T_{\text{f}}{}_{\left(\text{solution}\right)}$ in equation (7).

$\begin{array}{c}\Delta T_{\text{f}}=0^{°}\text{C}-\left(-{0.234}^{°}\text{C}\right)\\ ={0.234}^{°}\text{C}\end{array}$

Rearrange equation (2) to calculate the van’t Hoff factor of the solution as follows:

$i=\frac{\Delta T_{\text{f}}}{m{k}_{\text{f}}}$ (8)

Substitute ${0.234}^{°}\text{C}$ for $\Delta T_{\text{f}}$, $0.0000674061m$ for $m$ and ${1.86}^{°}\text{C/}m$ for $k_{\text{f}}$ in equation (8).

$\begin{array}{c}i=\frac{{0.593}^{°}\text{C}}{\left(0.0000674061m\right)\left({1.86}^{°}\text{C/}m\right)}\\ =1.866398\\ =1.87\end{array}$

$\text{KCl}$ dissociates to give one ${\text{K}}^{+}$ and one ${\text{Cl}}^{-}$ so its van’t Hoff factor should be close to 2.

Conclusion

The molality and van’t Hoff factor for the aqueous $\text{KCl}$ solution are $0.000068m$ and 2.

(b)

Interpretation Introduction

Interpretation:

The molality and van’t Hoff factor for the aqueous ${\text{H}}_2{\text{SO}}_4$ solution are to be calculated.

Concept introduction:

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by $\text{m}$ and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

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

$\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$ (1)

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

$\Delta T_{\text{f}}=i{k}_{\text{f}}m$ (2)

Here, $\Delta T_{\text{f}}$ is the change in freezing point.

$i$ is van’t Hoff factor.

$k_{\text{f}}$ is the freezing point depression constant.

$m$ is the molality of the solution.

The van’t Hoff factor is a measure of the effect of solute on the colligative properties. It is represented by $i$. It is defined as the ratio of measured values for the electrolytic solution to the expected values for non-electrolytic solutions. It is a unitless quantity.

Answer to Problem 13.109P

The molality and van’t Hoff factor for the aqueous ${\text{H}}_2{\text{SO}}_4$ solution are $0.10298746m$ and 3.

Explanation of Solution

Consider the mass of the solution to be. $100\text{}\text{g}$.

Substitute $100\text{}\text{g}$ for the mass of solution and $\text{1 }\%$ for the mass percent of the compound in equation (3) to calculate the mass of ${\text{H}}_2{\text{SO}}_4$.

$\begin{array}{c}{\text{Mass of H}}_2{\text{SO}}_4=\left(\text{100 g}\right)\left(\frac{\text{1 }\%}{100\text{}\%}\right)\\ =\text{1}\text{g}\end{array}$

Substitute $\text{1}\text{g}$ for the given mass and $98.08\text{g/mol}$ for the molar mass in equation (4) to calculate the moles of ${\text{H}}_2{\text{SO}}_4$.

$\begin{array}{c}{\text{Moles of H}}_2{\text{SO}}_4=\left(1\text{g}\right)\left(\frac{1\text{mol}}{98.08\text{g}}\right)\\ =0.0101957586\text{mol}\end{array}$

Substitute $\text{100}\text{g}$ for the mass of solution and $\text{1}\text{g}$ for the mass of solute in equation (6) to calculate the mass of water.

$\begin{array}{c}\text{Mass of water}=\text{100}\text{g}-\text{1 g}\\ =\text{99 g}\end{array}$

Substitute $0.0101957586\text{mol}$ for the amount of solute and $\text{99 g}$ for the mass of solvent in equation (1) to calculate the molality of ${\text{H}}_2{\text{SO}}_4$.

$\begin{array}{c}{\text{Molality of H}}_2{\text{SO}}_4=\left(\frac{0.0101957586\text{mol}}{99\text{g}}\right)\left(\frac{{\text{10}}^{\text{3}}\text{g}}{\text{1}\text{kg}}\right)\\ =0.10298746m\\ =0.1023m\end{array}$

Substitute $0^{°}\text{C}$ for $T_{\text{f}}{}_{\left(\text{solvent}\right)}$ and $-{0.423}^{°}\text{C}$ for $T_{\text{f}}{}_{\left(\text{solution}\right)}$ in equation (7).

$\begin{array}{c}\Delta T_{\text{f}}=0^{°}\text{C}-\left(-{0.423}^{°}\text{C}\right)\\ ={0.423}^{°}\text{C}\end{array}$

Substitute ${0.423}^{°}\text{C}$ for $\Delta T_{\text{f}}$, $0.10298746m$ for $m$ and ${1.86}^{°}\text{C/}m$ for $k_{\text{f}}$ in equation (8).

$\begin{array}{c}i=\frac{{0.423}^{°}\text{C}}{\left(0.10298746m\right)\left({1.86}^{°}\text{C/}m\right)}\\ =2.20842\\ =2.21\end{array}$

${\text{H}}_2{\text{SO}}_4$ is a strong acid so it completely dissociates to give two ${\text{H}}^{+}$ and one ${\text{SO}}_4^{2-}$ and therefore its van’t Hoff factor is 3.

Conclusion

The molality and van’t Hoff factor for the aqueous ${\text{CH}}_3\text{COOH}$ solution are $0.10298746m$ and 3.