# Calculate the molality of a solution made by dissolving a compound in camphor if the freezing point of the solution is 162.33 degrees C: the melting point of pure camphor was measured to be 178.43 degrees C. K sub f for camphor is 37.7 degrees C/m sub c?

## Please explain the equation used for this problem and how to get to the answer. The answer isn't needed to be provided.

May 2, 2017

$0.42 \text{ molal}$

#### Explanation:

To make any sense of this question, let's first write down what we are given.

Given
- $\textcolor{g r e e n}{\text{Pure camphor freezing point} = {178.43}^{\circ} C}$
- $\textcolor{g r e e n}{{K}_{f} = {37.7}^{\circ} \frac{C}{m}}$
- $\textcolor{g r e e n}{\text{Solution's freezing point} = {162.33}^{\circ} C}$

$\textcolor{w h i t e}{a a a a a a} \text{Now, what is this problem really saying and asking?}$

Well first off, we should understand the concept being tested - colligative properties.

Colligative properties describe the physical properties of solutions that depend only on the amount of solutes dissolved in solutions and not the type dissolved. Here, melting point/freezing point is being affected. When you add solutes to a pure solution, the freezing point of the solution decreases.

Clearly, the pure camphor and the camphor with the added compound have different freezing points.

$\textcolor{w h i t e}{a a a a a a a a a a a a a a} \textcolor{b l u e}{\text{pure camphor}} = {178.43}^{\circ} C$
$\textcolor{w h i t e}{a a a a a} \textcolor{red}{\text{camphor with compound}} = {162.33}^{\circ} C$

$\textcolor{w h i t e}{a a a a a a a} \Delta T = {178.43}^{\circ} C - {162.33}^{\circ} C = {16.10}^{\circ} C$

$- - - - - - - - - - - - - - - - - - - - -$

To find the molality of the compound we need the following formula

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a} \textcolor{m a \ge n t a}{\Delta T = i {K}_{f} m}$

Where

• $\Delta T = \text{change in the temperature} \left({\textcolor{w h i t e}{a}}^{\circ} C\right)$
• $i = \text{vant Hoff factor}$
• ${K}_{f} = \text{molal freezing point constant} \left(\frac{{\textcolor{w h i t e}{a}}^{\circ} C}{m}\right)$
• m = "molality" ("moles of solute"/"1 kg of solvent" or m)

The vant Hoff factor of a nonelectrolyte solute is 1.

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a}$

Plugin, rearrange and solve (ignored units for simplicity sake)

• $\textcolor{m a \ge n t a}{\Delta T = i {K}_{f} m}$

• $16.10 = \left(1\right) \left(37.7\right) \left(m\right)$

• $\frac{16.10}{\left(1\right) \left(37.7\right)} = m$

• $\text{molality" = 0.42" m, where m equals molal}$

$\text{Answer": "molality" = 0.42} m$