Question

The freezing point of water, H2O, is 0.000 °C at 1 atmosphere. Kf(water) = 1.86 °C/m In a laboratory experiment, students synthesized a new compound and found that when 12.86 grams of the compound were dissolved in 242.1 grams of water, the solution began to freeze at -1.591 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound?

Answer 1

According to depression in freezing point concept:

T_s - T₀ = -i * K_f * (w/M) * 1/W

Where T_s = the freezing point of solution = -1.591 °C

T₀ = the freezing point of water = 0 °C

i = van't Hoff factor = no. of ions in aqueous solution = 1 (for non-electrolyte)

Kf = molar freezing point depression constant = 1.86 °C/m = 1.86 °C/(mol/kg) = 1.86 °C.kg/mol (since 1 m = 1 mol/Kg)

w = weight of compound dissolved in water = 12.86 g

M = molecular weight of the compound, which has to be determined

W = Weight of water = 242.1 g = 0.2421 Kg (since 1 Kg = 1000 g)

Therefore, -1.591 °C - 0 °C = -1 * 1.86 °C.kg/mol * (12.86 g/M) * 1/0.2421 Kg

i.e. M = {1.86*12.86/(1.591 * 0.2421)} g/mol

i.e. M = 62.1 g/mol