Chemistry Question 4.15.20

Question

A 1.753 g sample of an unknown nonelectrolyte is dissolved in 25.00mL of cyclohexane. The resulting solution freezes at 2.46 ^oC, whereas the freezing point of pure cyclohexane is 6.47 ^o Cyclohexane has a density of 0.7739 g/mL and a K_f value of 20.2 ^oC/m. Determine the molar mass of the unknown compound.

Answer 1

Colligative properties:

∆T = imK

∆T = change in freezing point =6.47º - 2.46º = 4.01º

i = van't Hoff factor = 1 for a nonelectrolyte as it does not ionize or dissociate

m = molality = moles solute/kg solvent = ?

K = freezing point constant for cyclohexane = 20.2º/m

Solving for m we have:

m = ∆T/(i)(K) = (4.01º)/(1)20.2

m = 0.1985 molal = 0.1985 moles/kg cyclohexane

How many moles are actually present? Well, 25.00 ml x 0.7739 g/ml x 1 kg/1000 g = 0.01934 kg

0.1985 moles/kg x 0.01934 kg = 0.003840 moles

molar mass = grams/mole = 1.753 g/0.003840 moles

molar mass = 456.5 g/mole

(Check my math)