Question

The Rast method for determining molar masses uses camphor as the solvent. Camphor melts at $\mathbf{178}\mathbf{.}{\mathbf{4}}^{\mathbf{0}}\mathbf{\text{C}}\mathbf{,}$and it’s large ${\mathit{K}}_{\mathbf{\text{f}}}\left(37.7\text{K}\text{kg}{\text{mol}}^{-1}\right)$makes it especially useful for accurate work. A sample of an unknown substance that weighs 0.840 greduces the freezing point of 25.0 gcamphor to $\mathbf{170}\mathbf{.}{\mathbf{8}}^{\mathbf{0}}\mathbf{\text{C}}\mathbf{.}$What is its molar mass?

Short answer

The molar mass is $166.34\text{g}/\text{mol}$.

Step-by-step solution

Given data.

Initial freezing point$\left({\text{T}}_{\text{i}}\right){\text{= 178.4}}^0\text{C}$

Depressed freezing point$\left({\text{T}}_{\text{f}}\right){\text{= 170.8}}^0\text{C}$

Freezing point depression constant $\left({\text{K}}_{\text{f}}\right)\text{= 37.7} \text{Kkg/mol}$

Mass of the unknown substance$=0.840 \text{g}$

Mass of camphor = 25 g

Concept of molar mass.

The mass of one mole of a substance is defined as the molar mass of the substance. It is expressed in g/mol.

formula to calculate depression of freezing point.

The depression in the freezing point due to addition of a solute is given by the following expression.

$\text{∆T =} \text{- m} \times {\text{K}}_{\text{f}}$ …. (i)

Where,

m is molality and${\text{K}}_{\text{f}}$ is Freezing point depression constant.

Calculation of molality

The molality can be calculated from equation (i) as given below.

$$\begin{gathered} \text{m}=-\frac{170.8-178.4}{37.7} \\ m=0.202mol/Kg \end{gathered}$$

Calculation of moles of an unknown sample.

The number of moles of solute dissolved in the solvent can be calculated from the molarity of the solution. Moles of an unknown sample can be determined as follows:

$\text{molarity} \text{=}\frac{\text{number} \text{of} \text{moles} \text{of} \text{solute}}{\text{Mass} \text{of} \text{solvent} \text{in} \text{kilograms}}$

$$\begin{gathered} \text{m =}\frac{\text{n( solute )}}{\text{M( solvent )}} \\ \text{n = m×M(camphor)} \\ {\text{n = 0.202×25×10}}^{\text{- 3}} \\ {\text{n=5.05×10}}^{-3}\text{mol} \end{gathered}$$

With the help of the mass of the unknown substance dissolved and also the number of moles present in the solution, we can calculate the molar mass of the unknown substance..

The value of the molar mass of the unknown sample is determined by:

$\text{Molar} \text{mass} \text{=}\frac{\text{Mass} \text{of} \text{the} \text{substance}}{\text{Number} \text{of} \text{moles}}$

$$\begin{gathered} \text{M =}\frac{\text{m}}{\text{n}} \\ \text{M =}\frac{\text{0.840}}{{\text{5.05×10}}^{\text{- 3}}} \\ M=166.34g/mol \end{gathered}$$