Chapter 16, Problem 119E

A student dissolves $\text{27}\text{.2 g}$ of aniline, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$, in $\text{1}\text{.20}\times {\text{10}}^2\text{g}$ of water. At what temperatures will the solution freeze and boil?

Aniline is used in the polyurethane manufacturing process.

Materials made from polyurethane include foams (as pictured), spandex, hardwood floor coatings, dolly wheels, and many other end products.

Interpretation Introduction

Interpretation:

The freezing and boiling temperature of solution on dissolving aniline in water is to be stated.

Concept introduction:

Freezing temperature of any liquid is the temperature at which vapor pressure of solid becomes equal to the vapor pressure of liquid. Addition of a solute into any other solution decreases the freezing point of solution. Boiling point is the temperature at which vapor pressure of liquid is equal to atmospheric pressure.

Answer to Problem 119E

Freezing temperature when aniline is dissolved in $\text{1}\text{.20}\times {\text{10}}^2\text{g}$ of water is $-4.52°\text{C}$.

Boiling temperature of solution when aniline is dissolved in $\text{1}\text{.20}\times {\text{10}}^2\text{g}$ of water is $101.3°\text{C}$.

Explanation of Solution

The equation used to calculate freezing temperature of the solution is stated below.

$\text{Δ}{\text{T}}_{\text{f}}=\text{m}\times {\text{K}}_{\text{f}}$…(1)

$\text{Δ}{\text{T}}_{\text{f}}=({\text{T}}_1-{\text{T}}_{\text{2}})$…(2)

Where,

• $\text{Δ}{\text{T}}_{\text{f}}$ is the depression in freezing point.

• ${\text{T}}_1$ is the initial freezing point in $°\text{C}$.

• ${\text{T}}_{\text{2}}$ is the final freezing point in $°\text{C}$.

• $\text{m}$ is the molality of the solution.

• ${\text{K}}_{\text{f}}$ is the molal freezing point depression constant in $\text{°C}\cdot \text{kg }/\text{mol}$.

The formula used to calculate the molality is stated below.

$\text{m}=\frac{\text{Mass of aniline}\text{(w in g)}\times \text{1000}}{\text{Molar mass of aniline}\text{(}\text{g}/\text{mol})\times \text{mass of water (in g)}}$…(3)

The given value of mass of aniline is $\text{27}\text{.2 g}$.

The molar mass of aniline is $93.13\text{g}/\text{mol}$.

The given mass of water is $\text{1}\text{.20}\times {\text{10}}^2\text{g}$.

Substitute the value of mass of aniline, mass of water and molar mass of aniline in equation (3) as shown below.

$\begin{array}{rll}\text{m}& =& \frac{\text{Mass of aniline}\text{(w in g)}\times \text{1000}}{\text{Molar mass of aniline}\left(\text{g}/\text{mol}\right)\times \text{Mass of water (in g)}}\\ & =& \frac{27.2\text{g}\times \text{1000}}{93.13\left(\text{g}/\text{mol}\right)\times 1.2\times 100\text{g}}\\ & =& \text{2}\text{.43 m}\end{array}$

Calculation of freezing temperature is shown below.

The given value of ${\text{K}}_{\text{f}}$ is $1.86°\text{C}/\text{m}$.

Substitute the value of $\text{m}$ and ${\text{K}}_{\text{f}}$ in equation (1) as shown below.

$\begin{array}{rll}\text{Δ}{\text{T}}_{\text{f}}& =& \text{m}\times {\text{K}}_{\text{f}}\\ & =& 2.43\text{m}\times 1.86°\text{C}/\text{m}\\ & =& 4.52°\text{C}\end{array}$

Given value of ${\text{T}}_1$ is $0°\text{C}$.

Substitute the value of $\text{Δ}{\text{T}}_{\text{f}}$ and ${\text{T}}_1$ in equation (2) as shown below.

$\begin{array}{l}\text{Δ}{\text{T}}_{\text{f}}=({\text{T}}_1-{\text{T}}_{\text{2}})\\ 4.52°\text{C}=(0-{\text{T}}_{\text{2}})°\text{C}\end{array}$

Therefore, the ${\text{T}}_2$ is calculated as shown below.

$\begin{array}{rll}{\text{T}}_{\text{2}}& =& \left(0-4.52\right)°\text{C}\\ & =& -4.52°\text{C}\end{array}$

Therefore, freezing temperature of the solution is $-4.52°\text{C}$.

The equation used to calculate boiling temperature of the solution is stated below.

$\text{Δ}{\text{T}}_{\text{b}}=\text{m}\times {\text{K}}_{\text{b}}$…(4)

$\text{Δ}{\text{T}}_{\text{b}}=({\text{T}}_2-{\text{T}}_1)$…(5)

Where,

• $\text{Δ}{\text{T}}_{\text{b}}$ is the elevation in boiling point.

• ${\text{T}}_1$ is the initial boiling point in $°\text{C}$.

• ${\text{T}}_{\text{2}}$ is the final boiling point in $°\text{C}$.

• $\text{m}$ is the molality of the solution.

• ${\text{K}}_{\text{b}}$ is the molal boiling point elevation constant in $\text{°C}\cdot \text{kg }/\text{mol}$.

Calculation of boiling temperature is shown below.

The given value of ${\text{K}}_{\text{b}}$ is $0.52°\text{C}/\text{m}$.

Calculated value of $\text{m}$ is $\text{2}\text{.43 m}$.

Substitute the value of $\text{m}$ and ${\text{K}}_{\text{b}}$ in equation (4) as shown below.

$\begin{array}{rll}{\text{ΔT}}_{\text{b}}& =& \text{m}\times {\text{K}}_{\text{b}}\\ & =& \text{2}\text{.43}\text{m}\times \text{0}\text{.52}°\text{C}/\text{m}\\ & =& 1.3°\text{C}\end{array}$

Given value of ${\text{T}}_1$ is $100°\text{C}$.

Substitute the value of ${\text{ΔT}}_{\text{b}}$ and ${\text{T}}_1$ in equation (5) as shown below.

$\begin{array}{rll}{\text{ΔT}}_{\text{b}}& =& {\text{(T}}_{\text{2}}-{\text{T}}_{\text{1}}\text{)}\\ \text{1}\text{.3}°\text{C}& =& {\text{(T}}_{\text{2}}-100)°\text{C}\end{array}$

Therefore, the ${\text{T}}_2$ is calculated as shown below.

$\begin{array}{l}{\text{T}}_{\text{2}}=\left(100+\text{1}\text{.3}\right)°\text{C}\\ {\text{T}}_{\text{2}}=101.3°\text{C}\end{array}$

Therefore, boiling temperature of the solution is $101.3°\text{C}$.

Conclusion

Freezing and boiling temperature of the solution is $-4.52°\text{C}$ and $101.3°\text{C}$ respectively.