Question

a 0.1539 molal aqueous solution of cane sugar ( molecular mass 342 g/mol) has freezing point of 271 k while the freezing point of water is 273.15k. what will bw the freezing point of an aqueous solution containing 5g of glucose ( molecular mass 180 b / mol ) for 100 g of solution.

Answer 1

The molality of the cane sugar solution, m = 0.1539 m

and the depression in freezing point (ΔT_f) = 273.15 − 271 = 2.15 K

Since ​ΔT_f = K_f m

or K_f = ​ΔT_f / m = 2.15 K / 0.1539 m = 13.97 K/m

Now, weight of glucose, W₂ = 5 g
Molecular mass of glucose, M₂ = 180 g mol⁻¹
Then, ​ΔT_f = K_f m
Or, ∆ T f = K f × W 2 × 1000 W 1 × M 2 = 13 . 97 × 5 × 1000 100 × 180 = 3 . 88 K

ΔT_f = 3.88 K
Then, Freezing point of solution = 273.15 - 3.88 K = 269.27 K