Chapter 16, Problem 73A

(a)

Interpretation Introduction

Interpretation: The freezing point of $0.20m$ ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ in an aqueous solution is to be calculated.

Concept Introduction:

The formula used for calculating the depression in the freezing point is,

  $\Delta T_f=i\times K_f\times m\text{}\text{}$

Here $\Delta T_b$ is the depression in freezing point, $K_f$ is the molal freezing point depression constant, $i$ is the Van’t Hoff factor, and $m$ is molality.

Answer to Problem 73A

The freezing point of $0.20m$ ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{in}\text{aqueous}\text{solution}$ is $-1.1°\text{C}$

Explanation of Solution

The Van’t Hoff factor of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $3$ because when it dissociates in water it yields the following ions, that is., ${\text{2K}}^{\text{+}}\text{and}{\text{SO}}_{\text{4}}^{\text{2-}}$

The molality of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is given as $0.20m$ .

The molal freezing point depression constant of the aqueous solution is $1.86°{\text{Cmol}}^{-1}$ .

Substitute $1.86°{\text{Cmol}}^{-1}$ for $K_a$ , $3$ for $i$ , and $0.20$ for $m$ in the equation of freezing point depression.

  $\begin{array}{c}\Delta T_f=3\times 1.86°\text{C}\text{mol/Kg}\times 0.20m\\ =-1.1°\text{C}\end{array}$

(b)

Interpretation Introduction

Interpretation: The freezing point of $0.20m$ ${\text{CsNO}}_3$ in aqueous solution is to be calculated.

Concept Introduction:

The formula used for calculating the depression in the freezing point is,

  $\Delta T_f=i\times K_f\times m\text{}\text{}$

Here $\Delta T_b$ is the depression in freezing point, $K_f$ is the molal freezing point depression constant, $i$ is the Van’t Hoff factor, and $m$ is molality.

Answer to Problem 73A

The freezing point of $0\cdot 20m$ ${\text{CsNO}}_3$ in aqueous solution is $-0.74°\text{C}$ .

Explanation of Solution

The Van’t Hoff factor of ${\text{CsNO}}_{\text{3}}$ is 2 because when Cesium Nitrate dissociates into water it yields the following ions, that is, ${\text{Cs}}^{\text{+}}\text{and}{\text{NO}}_{\text{3}}^{\text{3-}}$ .

The molality of ${\text{CsNO}}_{\text{3}}$ is given as $0.20m$ .

The molal freezing point depression constant of the aqueous solution is $1.86°{\text{Cmol}}^{-1}$ .

Substitute $1.86°{\text{Cmol}}^{-1}$ for $K_a$ , $2$ for $i$ , and $0.20$ for $m$ in the equation of freezing point depression.

  $\begin{array}{c}\Delta T_f=2\times 1.86°\text{C}\text{mol/Kg}\times 0.20m\\ =-0.74°\text{C}\end{array}$

(c)

Interpretation Introduction

Interpretation: The freezing point of $0.20m$ $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ in aqueous solution is to be calculated.

Concept Introduction:

The formula used for calculating the depression in the freezing point is,

  $\Delta T_f=i\times K_f\times m\text{}\text{}$

Here $\Delta T_b$ is the depression in freezing point, $K_f$ is the molal freezing point depression constant, $i$ is the Van’t Hoff factor, and $m$ is molality.

Answer to Problem 73A

The freezing point of $0\cdot 20m$ $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ in aqueous solution is $-1.5°\text{C}$ .

Explanation of Solution

The Van’t Hoff factor of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is $4$ because when Aluminium Nitrate dissociates into water it yields the following ions i.e., ${\text{Al}}^{\text{3+}}\text{and}{\text{3NO}}_{\text{3}}^{\text{-}}$ .

The molality of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is given as $0.20m$ .

The molal freezing point depression constant of the aqueous solution is $1.86°{\text{Cmol}}^{-1}$ .

Substitute $1.86°{\text{Cmol}}^{-1}$ for $K_a$ , $4$ for $i$ , and $0.20$ for $m$ in the equation of freezing point depression.

  $\begin{array}{c}\Delta T_f=4\times 1.86°\text{C}\text{mol/Kg}\times 0.20m\\ =-1.5°\text{C}\end{array}$