Question

A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Answer 1

Assuming, the 5% glucose in water solution is also by mass, I am answering this question

Cane Sugar is Sucrose - C.12.H.22.O.11
(I have added dots for better understanding)

Now then molecular wt of Sucrose is 342 g/mole.

Glucose - C.6.H.12.O.6
Molecular wt of Glucose is 180

Lowering in freezing point in case of cane sugar is (273.15 - 271) = 2.15 K

Since the molecular wt. of Glucose is less than Sucrose, there will be more moles of Glucose in a 5% solution than moles of Sucrose.

Let assume that we have taken 3420 g of solute and corresponding amount of water for a 5% solution.

Hence no. of moles of

Sucrose = (3420/342) = 10
Glucose = (3420/180) = 19

Since molality values of both these solutions now only depends on the number of moles of the solute since we are taking same amount of water for both we can generalise it this way :

2.15 = K × 10
(where K is a proportionality constant)

Also

∆T = K × 19

∆T = (2.15×19)/10

∆T = 4.085 K

T(f) = 273.15 - 4.085 = 269.065 K

Hence the freezing point of 5% Glucose in Water solution is

269.065 K

Answer 2

Answer:

269.07 K

Explanation:

The depression in the freezing point of the solution is given by

ΔTf = freezing point of water − freezing point of solution

ΔTf  =273.15−271=2.15 K

Molar masses of glucose and sucrose are 180 g/mol and 342 g/mol respectively.

100 g of solution will contain 5 g of glucose or 5 g of sucrose.

Number of moles of glucose = 5/180 = 0.028 moles

Number of moles of sucrose =  5/342 = 0.0146 moles

Mass of solvent = total mass of solution - mass of solute =100−5=95 g or 0.095 kg

Molality of sucrose solution = 0.0146 /0.095 = 0.154 m

Kf = ΔTf/molarity = 2.15/0.154 = 13.97

For glucose solution, ΔTf = Kf ×m = 13.97×0.29 = 4.08

Freezing point of 5% glucose solution in water = freezing point of water - ΔTf = 273.15−4.08 = 269.07 K