Question

A 0.1539 molal aqueous solution of cane sugar (mol mass = 342 g mol⁻¹) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (mol. mass = 180 g mol⁻¹) per 100 g of solution?

Answer 1

Given: Molality of cane sugar solution = 0.1539 mol/kg

Freezing point of cane sugar solution = 271 K

Freezing point of pure water = 273.15 K

ΔT_f = 273.15 − 271

ΔT_f = 2.15 K

ΔT_f = K_f . m

`K_f = 2.15/0.1539`

K_f = 13.97 K

Mass of glucose = 5 g

Molar mass = 180 g/mol

Total solution mass = 100 g

So, mass of solvent (water) = 100 − 5 = 95 g = 0.095 kg

Moles of glucose = `5/180`

= 0.02778 mol

Molality = `0.02778/0.095`

= 0.2924 mol/kg

ΔT_f = K_f . m

= 13.97 × 0.2924

= 4.08 K

Freezing point of solution = 273.15 − 4.08

= 269.07 K

∴ The freezing point of the glucose solution is 269.07 K.