Chapter 16, Problem 82A

Interpretation Introduction

Interpretation: The freezing point and the boiling point of a solution that contains $15\text{g}$ of urea and $250\text{g}$ of water are to be calculated.

Concept introduction: The addition of solute particles to a particular solvent will decrease the freezing point and raise the boiling point of that solvent.

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

  $\Delta T_f=K_{f}m$

Here, $\Delta T_f$ is the freezing point depression, $K_f$ is the molal freezing point depression constant, and $m$ is molality.

The formula used for calculating the elevation in boiling point is,

  $\Delta T_b=K_{b}m$

Here, $\Delta T_b$ is the elevation in boiling point, $K_b$ is the molal boiling point elevation constant, and $m$ is molality.

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}$   ...... (1)

The expression for molality is as follows:

  $m=\frac{\text{moles of solute}}{\text{Mass of solvent(kg)}}$   ...... (2)

Answer to Problem 82A

The boiling point of the solution is $100.51°\text{C}$ and the freezing point of the solution is - $1.86°\text{C}$ .

Explanation of Solution

Given:

The molar mass of urea is $60\text{g/mol}$ .

The solution contains $15\text{g}$ of urea and $250\text{g}$ of water.

Convert $\text{g}$ to $\text{kg}$ .

  $\begin{array}{c}250.0\text{g}=250\times \frac{1\text{kg}}{1000\text{g}}\\ =0.25\text{kg}\end{array}$

Substitute, $15\text{g}$ for the mass of urea $\left({\text{CH}}_4{\text{N}}_2\text{O}\right)$ and $60\text{g/mol}$ for molar mass of $\left({\text{CH}}_4{\text{N}}_2\text{O}\right)$ in equation (1).

  $\begin{array}{c}\text{moles}\left({\text{CH}}_4{\text{N}}_2\text{O}\right)=\frac{\text{ 15 g}}{\text{60 g/mol}}\\ =0.25\text{ mol}\end{array}$

Substitute $0.25moles$ for $\left({\text{CH}}_4{\text{N}}_2\text{O}\right)$ and $\text{0}\text{.25}\text{kg}$ for the mass of solvent in equation (2).

  $\begin{array}{c}m=\frac{0.25}{0.25}\\ =0.1\text{m}\end{array}$

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

So,

Substitute $1.86°{\text{Cmol}}^{-1}$ for $K_f$ and $\text{1}\text{.0}\text{m}$ for $m$ in the equation of freezing point depression.

  $\begin{array}{c}\Delta T_f=1.86°{\text{Cmol}}^{-1}\times \text{ 1}\text{.0}\text{mol/kg}\\ =1.86°\text{C}\end{array}$

So,

The difference in the depression point of solvent and solution is $1.86°\text{C}$ .

The freezing point of water is $0°\text{C}$

So, the depression point of the solution is,

  $\text{0}°\text{C}-1.86°\text{C}=-1.86°\text{C}$

So, the freezing point of urea is $-1.86°\text{C}$ .

The molal boiling point elevation constant of urea is $0.51°{\text{Cmol}}^{-1}$ .

So,

Substitute $0.51°{\text{Cmol}}^{-1}$ for $K_b$ and $\text{1}\text{.0}\text{m}$ for $m$ in the equation of freezing point depression.

  $\begin{array}{c}\Delta T_b=0.51°{\text{Cmol}}^{-1}\times \text{ 1}\text{.0}\text{mol/kg}\\ =0.51°\text{C}\end{array}$

So,

The difference in elevation point of solvent and solution is $0.51°\text{C}$ .

The boiling point of water is $100°\text{C}$

So, the depression point of the solution is,

  $\text{100}°\text{C}+1.86°\text{C}=100.51°\text{C}$

So, the boiling point of urea is $100.51°\text{C}$ .

Conclusion

So, the freezing point of the given solution is $-1.86°\text{C}$ and the boiling point is $100.51°\text{C}$ .