Chapter 6, Problem 6.67P

6-67 Calculate the freezing points of solutions made by dissolving 1.00 mole of each of the following ionic solutes in 1000. g of H₂O.

(a) NaCI

(b) MgCI₂

(c) (NH₄)₂CO₃

(d) AI(HCO₃)₃

Interpretation Introduction

(a)

Interpretation:

The freezing point of the solution should be calculated.

Concept Introduction:

The freezing point depression can be related to the total molality of the solution and the normal freezing point of the solvent.

In this specific case the solvent is water, therefore, the freezing point of water as well as the depression constant must be known.

The molality must be calculated.

$m\text{O}\text{l}\text{a}\text{L}\text{i}ty=\frac{m\text{O}\text{l}solute}{kgs\text{O}\text{l}v\text{e}nt}=\frac{1m\text{O}\text{l}}{1kg}=1m\text{O}\text{l}\text{a}\text{l}$

For the given data, note that for that salt we have 2 ions in solution.

$T_{\text{F}-m\text{i}\text{X}}=T_{\text{F}-pur\text{e}}-K_{\text{F}}m\text{i}$

Where,

$\begin{array}{l}{T}_{\text{F}-m\text{i}\text{X}}=n\text{e}w\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}\\ T_{\text{F}-pur\text{e}}=pur\text{e}\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}\\ K_{\text{F}}=\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}d\text{e}pr\text{e}ss\text{i}\text{O}n\text{C}\text{O}ns\tan t\\ m=m\text{O}\text{l}\text{a}\text{L}\text{i}ty\\ \text{i}=numb\text{e}rofsp\text{e}\text{C}\text{i}\text{e}s\end{array}$.

Answer to Problem 6.67P

$-3.7\text{2 °C}$.

Explanation of Solution

Given Information:

$\begin{array}{l}n=1.00m\text{O}\text{l}\\ m=1000g=1kg\\ T_{\text{F}}=0°\text{C}\\ K_{\text{F}}=1.86°\text{C}/m\text{O}\text{l}\text{a}\text{l}\end{array}$

Colligative properties will allow us to relate the freezing point of the pure solvent with the mixture.

For the given data, note that for that salt we have 2 ions in solution.

$\begin{array}{l}{T}_{\text{F}-m\text{i}\text{X}}=T_{\text{F}-pur\text{e}}-K_{\text{F}}m\text{i}\\ T_{\text{F}-m\text{i}\text{X}}=0-(1.86°\text{C}/m\text{O}\text{l}\text{a}\text{l})(1m\text{O}\text{l}\text{a}\text{l})(2)\\ T_{\text{F}-m\text{i}\text{X}}=-3.72°\text{C}\end{array}$.

Interpretation Introduction

(b)

Interpretation:

The freezing point of the solution should be calculated.

Concept Introduction:

The freezing point depression can be related to the total molality of the solution and the normal freezing point of the solvent.

In this specific case the solvent is water, therefore, the freezing point of water as well as the depression constant must be known.

The molality must be calculated.

$m\text{O}\text{l}\text{a}\text{L}\text{i}ty=\frac{m\text{O}\text{l}solute}{kgs\text{O}\text{l}v\text{e}nt}=\frac{1m\text{O}\text{l}}{1kg}=1m\text{O}\text{l}\text{a}\text{l}$

For the given data, note that for that salt we have 2 ions in solution.

$T_{\text{F}-m\text{i}\text{X}}=T_{\text{F}-pur\text{e}}-K_{\text{F}}m\text{i}$

Where,

$\begin{array}{l}{T}_{\text{F}-m\text{i}\text{X}}=n\text{e}w\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}\\ T_{\text{F}-pur\text{e}}=pur\text{e}\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}\\ K_{\text{F}}=\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}d\text{e}pr\text{e}ss\text{i}\text{O}n\text{C}\text{O}ns\tan t\\ m=m\text{O}\text{l}\text{a}\text{L}\text{i}ty\\ \text{i}=numb\text{e}rofsp\text{e}\text{C}\text{i}\text{e}s\end{array}$.

Answer to Problem 6.67P

$-5.\text{58 °C}$.

Explanation of Solution

Given Information:

$\begin{array}{l}n=1.00m\text{O}\text{l}\\ m=1000g=1kg\\ T_{\text{F}}=0°\text{C}\\ K_{\text{F}}=1.86°\text{C}/m\text{O}\text{l}\text{a}\text{l}\end{array}$

For the given data, note that for that salt we have 3 ions in solution.

$\begin{array}{l}{T}_{\text{F}-m\text{i}\text{X}}=T_{\text{F}-pur\text{e}}-K_{\text{F}}m\text{i}\\ T_{\text{F}-m\text{i}\text{X}}=0-(1.86°\text{C}/m\text{O}\text{l}\text{a}\text{l})(1m\text{O}\text{l}\text{a}\text{l})(3)\\ T_{\text{F}-m\text{i}\text{X}}=-5.58°\text{C}\end{array}$.

Interpretation Introduction

(c)

Interpretation:

The freezing point of the solution should be calculated.

Concept Introduction:

The freezing point depression can be related to the total molality of the solution and the normal freezing point of the solvent.

In this specific case the solvent is water, therefore, the freezing point of water as well as the depression constant must be known.

The molality must be calculated.

$\text{molality}=\frac{\text{mol}\text{solute}}{\text{kg}\text{solvent}}=\frac{1\text{mol}}{1\text{kg}}=1\text{molal}$

For the given data, note that for that salt we have 2 ions in solution.

$T_{\text{F}-m\text{i}\text{X}}=T_{\text{F}-pur\text{e}}-K_{\text{F}}m\text{i}$

Where,

$\begin{array}{l}{T}_{\text{F}-m\text{i}\text{X}}=n\text{e}w\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}\\ T_{\text{F}-pur\text{e}}=pur\text{e}\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}\\ K_{\text{F}}=\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}d\text{e}pr\text{e}ss\text{i}\text{O}n\text{C}\text{O}ns\tan t\\ m=m\text{O}\text{l}\text{a}\text{L}\text{i}ty\\ \text{i}=numb\text{e}rofsp\text{e}\text{C}\text{i}\text{e}s\end{array}$.

Answer to Problem 6.67P

$-5.58°\text{C}$.

Explanation of Solution

Given:

$\begin{array}{l}n=1.00m\text{O}\text{l}\\ m=1000g=1kg\\ T_{\text{F}}=0°\text{C}\\ K_{\text{F}}=1.86°\text{C}/m\text{O}\text{l}\text{a}\text{l}\end{array}$

For the given data, note that for that salt we have 3 ions in solution.

$\begin{array}{l}{T}_{\text{F}-m\text{i}\text{X}}=T_{\text{F}-pur\text{e}}-K_{\text{F}}m\text{i}\\ T_{\text{F}-m\text{i}\text{X}}=0-(1.86°\text{C}/m\text{O}\text{l}\text{a}\text{l})(1m\text{O}\text{l}\text{a}\text{l})(3)\\ T_{\text{F}-m\text{i}\text{X}}=-5.58°\text{C}\end{array}$.

Interpretation Introduction

(d)

Interpretation:

The freezing point of the solution should be calculated.

Concept Introduction:

The freezing point depression can be related to the total molality of the solution and the normal freezing point of the solvent.

In this specific case the solvent is water, therefore, the freezing point of water as well as the depression constant must be known.

The molality must be calculated.

$m\text{O}\text{l}\text{a}\text{L}\text{i}ty=\frac{m\text{O}\text{l}solute}{kgs\text{O}\text{l}v\text{e}nt}=\frac{1m\text{O}\text{l}}{1kg}=1m\text{O}\text{l}\text{a}\text{l}$

For the given data, note that for that salt we have 2 ions in solution.

$T_{\text{F}-m\text{i}\text{X}}=T_{\text{F}-pur\text{e}}-K_{\text{F}}m\text{i}$

Where,

$\begin{array}{l}{T}_{\text{F}-m\text{i}\text{X}}=n\text{e}w\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}\\ T_{\text{F}-pur\text{e}}=pur\text{e}\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}\\ K_{\text{F}}=\text{F}r\text{e}\text{e}zingp\text{O}\mathrm{int}d\text{e}pr\text{e}ss\text{i}\text{O}n\text{C}\text{O}ns\tan t\\ m=m\text{O}\text{l}\text{a}\text{L}\text{i}ty\\ \text{i}=numb\text{e}rofsp\text{e}\text{C}\text{i}\text{e}s\end{array}$.

Answer to Problem 6.67P

$-7.44°\text{C}$.

Explanation of Solution

Given Information:

$\begin{array}{l}n=1.00m\text{O}\text{l}\\ m=1000g=1kg\\ T_{\text{F}}=0°\text{C}\\ K_{\text{F}}=1.86°\text{C}/m\text{O}\text{l}\text{a}\text{l}\end{array}$

For the given data, note that for that salt we have 3 ions in solution.

$\begin{array}{l}{T}_{\text{F}-m\text{i}\text{X}}=T_{\text{F}-pur\text{e}}-K_{\text{F}}m\text{i}\\ T_{\text{F}-m\text{i}\text{X}}=0-(1.86°\text{C}/m\text{O}\text{l}\text{a}\text{l})(1m\text{O}\text{l}\text{a}\text{l})(4)\\ T_{\text{F}-m\text{i}\text{X}}=-7.44°\text{C}\end{array}$.