Question

Consider a 3.08 m aqueous solution of glycerin, C₃H₈O₃? Assuming ideal solution behavior, what is the freezing point of this solution in degrees Celsius? Do not include units in your answer. Enter up to 4 decimal places, canvas does not recognize sig figs.

K_f for water is 1.86 ∘C/molal

normal freezing point of water is 0∘C at 1atm.

Answer 1

Ans:-

Since the depression in freezing point of any solution is given by

depression in freezing point = i X K_f X m

where i = Van't Hoff factor = 1 ( since glycerin is non electrolyte solute)

K_f = depression in freezing point constant = 1.86 ^∘C/m

m = molality of the solute = 3.08

Hence substituting the values we get

depression in freezing point = 1 X 1.86 X 3.08 = 5.7288 ^∘C

Since the freezing point of solution = freezing point of solvent i.e pure water - depression in freezing point

=> freezing point of solution = 0 - 5.7288 = 5.7288 ^∘C