# Determination of molar mass (by the depression of freezing point)

## Molality

Molality is a unit of concentration defined by:

$\mu$ $=$ $\frac{n}{m}$
with:
$n$ = number of moles of solute
$m$ = mass of solvent (in kg)

## Loi de Raoult

## The law of Raoult

The melting (or freezing) point of a solution $tf_{solution} $ is lower than that of the pure solvent $tf_{solvent} $ .
The law of Raoult connects this depression of the temperature of the solution to the molality of the solute:

$\Delta T_f$ $=$ $K_f \cdot \mu$
where:
$\Delta T_f$ $=$ $tf_{solution} - tf_{solvent}$
$ \mu $ = molality
$ K_f $ = cryoscopic constant depending only on the solvent.

## Measure of a melting point

Click on the image !

In the test tube, we will introduce the solution of the subtance whose molar mass we want to determine.

Images: →

## Determination of molar masses

1) Cryoscopic constant $K_f$
In $m \;g$ of a given solvent is introduced $y \;g$ of a solute of known molar mass . The melting points of the solution and of the pure solvent are measured (2 steps !)
We calculate:
Number of moles of solute =
$n$ $=$ $\frac{y}{M}$
$\mu$ =
$\frac{n}{m}$ $=$
$\frac{y}{m\cdot M}$
$K_f$ =
$\frac{\Delta T}{\mu}$
2) Molar mass $M$
In $m_1\; g$ of the same solvent are introduced $y_1 \; g$ of a solute whose molar mass $M_1 \;$ is to be found.
The melting point of this solution is determined.
We calculate:
$\mu$ $=$ $\frac{\Delta T}{K_f} $
Number of moles of solute =
$n_1 $ $=$ $m_1 \cdot \mu$
$M_1 =\frac{y_1 }{n_1 }$

## Example

A solution $(S)$ of $2.40\;g$ of a substance supposed to be biphenyl ($ C_{12}H_{10}$) is introduced in $75.0\;g$ benzene.
The melting point of this solution is $4.39\;^oC$
Moreover, the melting point of pure benzene was determined : $5.45\;^oC$ and that of a solution $(S_1)$ of $7.24\;g$ of $C_2Cl_4H_2$ in $115.3\;g$ benzene: $3.55\;^oC$
It is asked to verify the molecular weight of biphenyl

1) Determination of $K_f$
$n_{C_2Cl_4H_2}$ $=$ $\frac{7.24}{168}$
$\mu_{S_1 }$ $=$ $\frac{7.24}{168\cdot 0.1153}$
$K_f$ $=$ $\frac{5.45-3.55}{\mu}$ $=$ $5.08 \frac{^oC}{mole}$
2) Determination of the molar mass
$\mu_S$ $ =$ $\frac{5.45-4.39}{5.08}$ $=$ $0.208 \frac{mol}{kg}$
Number of moles of biphenyl =
$n_1 $ $=$ $0.075 \cdot 0,208$ $=$ $0.0156 $
$M_1 $=$\frac{2.40}{0.0156}$ $ \approx$ $ 154 \frac{g}{mol}$
This corresponds to the molar mass of $C_{12}H_{10}$ !