Question

A 0.2436-g sample of an unknown substance was dissolved in 20.0 ml of cyclohexane. The density of cyclohexane is 0.779 g/ml. The freezing-point depression was 2.5ºc. Calculate the molar mass of the unknown substance

Answer 1

The depression in freezing point is related to molality of solution as follows:

$\Delta T_{f}=k_{f}m$

Here, $k_{f}$ is freezing point depression constant, for cyclohexane it is equal to 20°C kg/mol.

The value of freezing-point depression is 2.5 °C, molality can be calculated as follows:

$m=\frac{\Delta T_{f}}{k_{f}}=\frac{2.5 ^{o}C}{20 ^{o}C kg/mol}=0.125 mol/kg$

Molality is defined as number of moles of solute in 1 kg of solvent.

Here, solvent is cyclohexane, its volume is given 20.0 mL and density is 0.779 g/mol thus, mass of cyclohexane can be calculated as follows:

$m=d\times V=0.779 g/mL\times 20 mL=15.58 g$

Converting this into kg,

$1 g=10^{-3} kg$

Thus, 15.58 g will be 0.01558 kg.

Now, number of moles of unknown solute is related to its mass and molar mass as follows:

$n=\frac{m}{M}$

Putting the values of mass of solute which is 0.2436 or 0.0002436 kg

$n=\frac{ 0.0002436 kg}{M}$

Now, expression for molality of solution is:

$m=\frac{n_{solute}}{m_{solvent}}$

Putting all the values,

$0.125 mol/kg=\frac{0.0002436 kg}{M\times 0.01558 kg}$

Or,

$0.125 mol/kg=\frac{0.015635}{M}$

On rearranging,

$M=\frac{0.015635}{0.125 mol/kg}=0.1250 kg/mol$

Or,

$M=0.125 kg/mol(\frac{1000 g}{1 kg})=125 g/mol$

Therefore, molar mass of unknown sample is 125 g/mol