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

We know that :

`$\Delta T_f = k_f.m$` and
`$m=(w_2)/(m_2 * w_1)$`

Then,
`$\Delta T_f = k_f.(w_2)/(m_2.w_1)$` ..................(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),

`$3.14 = (5.12 * 24.3)/(m_2 * 0.250)$`

`$m_2 = (5.12 * 24.3)/(3.14 * 0.250)$`

`m_2= 158.49` g/mole

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