Question

The molal freezing point depression constant =Kf6.19·°C·kgmol−1 for a certain substance X . When 10.g of urea NH22CO are dissolved in 500.g of X , the solution freezes at −6.8°C . Calculate the freezing point of pure X . Be sure your answer has the correct number of significant digits.

Answer 1

-4.7°C

Step-by-step explanation:

The freezing point depression is a colligative property that can be calculated using the following expression.

ΔTf = Kf . b

where

ΔTf: depression in the freezing point

Kf: molal freezing point depression constant

b: molality

The moles of urea (solute) are:

`(10g)/(60.06g/mol) =0.17mol`

The mass of solvent (X) is 500 g (0.500 kg).

The molality of the urea is:

`b=(0.17mol)/(0.500kg) =0.34mol/kg`

ΔTf = Kf . b = (6.19 °C.kg/mol) × 0.34 mol/kg = 2.1°C

The freezing point of pure X is:

-6.8°C + 2.1°C = -4.7°C