Chapter 6, Problem 6.107P

Interpretation Introduction

(a)

Interpretation:

To calculate the freezing point of the salt.

Concept Introduction:

Freezing point is depressed by salts due to colligative properties of solvents. When a salt is mixed to a solvent such as water, the freezing point is depressed according to the colligative properties of the solution.

The mathematical expression si given as follows:

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

Where,

$T_{\text{F}-m\text{i}\text{X}}$: freezing point of mixture.

$T_{\text{F}-pur\text{e}}$: Freezing point of pure component.

$K_{\text{F}}$ = Freezing point constant for water.

$m$ =molality.

$\text{i}$ =ions in solution.

Answer to Problem 6.107P

$-1.4651\text{ °C}$.

Explanation of Solution

Given Information:

$\begin{array}{c}{K}_{\text{F}}=-1.86°\text{C}/m\text{O}\text{l}\text{a}\text{l}\\ m=25.0gMg\text{C}{\text{l}}_2\\ m=1000g\\ =1kg\\ MM=95.211g/m\text{O}\text{l}\end{array}$

The depression in freezing point is related to molality as follows:

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

Substitute data:

$\begin{array}{l}{T}_{\text{F}-m\text{i}\text{X}}=0-1.86\frac{(mass/MM)}{kgs\text{O}\text{l}v\text{e}nt}(3)\\ T_{\text{F}-m\text{i}\text{X}}=0-1.86\frac{25/95.211 }{1}(3)\\ T_{\text{F}-m\text{i}\text{X}}=-1.4651°\text{C}\end{array}$

The freezing point is typically depressed directly proportional to the solute amount to solvent.

Interpretation Introduction

(b)

Interpretation:

The boiling point of the salt should be calculated.

Concept Introduction:

Boiling point is increased by salts due to colligative properties of solvents. When a salt is mixed to a solvent such as water, the boiling point is elevated according to the colligative properties of the solution.

The mathematical expression is given as follows:

$T_{b-m\text{i}\text{X}}=T_{b-pur\text{e}}+K_{b}m(\text{i})$

Where,

$T_{b-m\text{i}\text{X}}$: Boiling point of mixture.

$T_{b-pur\text{e}}$: Boiling point of pure component.

$K_b$ = Boiling point constant for water.

$m$ =molality.

$\text{i}$ =ions in solution.

Answer to Problem 6.107P

$100.40331°\text{C}$.

Explanation of Solution

Given Information:

$\begin{array}{l}{K}_b=\text{0}\text{.512 °C/molal}\\ m=2\text{5}\text{.0 g}{\text{MgCl}}_{\text{2}}\\ m=1\text{000 g}=\text{1 kg}\\ MM=\text{95}\text{.211 g/mol}\end{array}$

The elevated boiling point and molality of solution is related to each other as follows:

$T_{b-m\text{i}\text{X}}=T_{b-pur\text{e}}-K_{b}m(\text{i})$

Substitute data:

$\begin{array}{l}{T}_{b-m\text{i}\text{X}}=T_{b-pur\text{e}}-K_{b}m(\text{i})\\ T_{b-m\text{i}\text{X}}=100+0.512\frac{\text{(mass/MM)}}{\text{kg}\text{solvent}}(3)\\ T_{b-m\text{i}\text{X}}=100+0.512\frac{25/95.211 }{1}(3)\\ T_{b-m\text{i}\text{X}}=100.40331\text{°C}\end{array}$.