Question

Molar Mass Determination by Freezing Point Depression

Calculate and enter the freezing point depression of a solution of 74.2 g ethylene glycol (C₂H₆O₂) in 422 g H₂O.
K_f for H₂O is 1.86 °C kg/mol.

°C
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A solution which contains 57.1 g of an unknown molecular compound in 383 g of water freezes at -5.32°C.
What is the molar mass of the unknown?

g/mol

Answer 1

ans)

part a)

from above data that

Melecular weight = 62.068 g/mol for ethylene glycol.

now we can calculate the

Molality= 74.2 g *(1mol/62.068g) /(0.422 kg H20)

Molality= 2.832 m

DelT = Kf*molality

= (1.86 C*kg/mol)*(2.832 m)

= 5.267 C

Water has freezing point of 0 C,

Now it is 5.267 Celsius.

part b)

given that

del T = 5.32C

now we can calculate as

m = delT/Kf

= 5.32 C/(1.86 Ckg/mol)

= 2.860 mol/kg water.

mole = 2.860 mol/kg water *(0.383 kg water)

= 1.095 moles

Molecular Weight = 59.4 g/(1.095 moles)

Molecular Weight = 54.25 g/mol