Question

A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene

( C7H8 = 92 g/mol) and the freezing point is lowered by 1.33 oC. What is the molecular weight

of the organic compound? (Kf = 5.12 oC/m).

Answer 1

Molecular weight of the compound = 372.13 g/mol

Step-by-step explanation:

Depression in freezing point is related with molality of the solution as:

`\Delta T_f = K_f * m`

Where,

`\Delta T_f` = Depression in freezing point

`K_f` = Molal depression constant

m = Molality

`\Delta T_f = K_f * m`

`1.33 = 5.12 * m`

m = 0.26

Molality =
`(Moles\ of\ solute)/(Mass\ of\ solvent\ in\ kg)`

Mass of solvent (toluene) = 15.0 g = 0.015 kg

`0.26 = (Mole\ of\ compound)/(0.015)`

Moles of compound = 0.015 × 0.26 = 0.00389 mol

`Mol = (Mass\ in\ g)/(Molecular\ weight)`

Mass of the compound = 1.450 g

`Molecular\ weight = (Mass\ in\ g)/(Moles)`

Molecular weight =
`(1.450)/(0.00389) = 372.13\ g/mol`