Question

Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.14 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound

Answer 1

Solution :

We know that :

$\Delta T_f = k_f.m$  and   $m=\frac{w_2}{m_2 \times w_1}$

Then, $\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$   ..................(1)

Where,

$w_1$ = amount of solvent (in kg)

$w_2$ = amount of solute (in kg)

$m_2$ = molar mass of solute (g/mole)

$m$ = molality of solution (mole/kg)

Given :

$\Delta T_f$ = $3.14\ ^\circ C$,   $k_f= 5.12\ ^\circ C/m$

                              $=5.12 \ ^\circ C/mole/kg$

                              $=5.12 \ ^\circ C \ kg/mole$

$w_1$ = 0.250 kg,  $w_2$ = 24.3 g

Then putting this values in the equation is (1),

$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$

$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$

$m_2= 158.49$  g/mole

So, the molar mass of the unknown compound is 158.49 g/mole.