freezing point of glycerine in water

Question

2-8. What mass of glycerin (C₃H₈O₃), a nonelectrolyte, must be dissolved in 200g H₂O to give a solution with a freezing point Tf=-1.50^oC？(For C₃H₈O₃, molar mass =92g/mol, and the K_f for water=1.86 K · kg·mol^-l)

Answer 1

∆T = imK

∆T = change in freezing point = 1.50ºC (change from 0º, normal freezing point of water)

i = van't Hoff factor for glycerine = 1 (glycerin is a non-electrolite)

m = molality of glycerine = moles / kg H2O = ?

K = freezing constant for water = 1.86º/m (note: deosn't matter if K or C as we are working with Changes)

Solving for m:

m = ∆T / (i)(K)

m = 1.50 / (1)(K) = 1.50 / 1.86

m = 0.806 mols glycerin / kg water

Since we have 200 g water, that is 0.200 kg water

m = 0.806 moles / kg x 0.200 kg = 0.161 moles glycerine

Mass of glycerin = 0.161 moles x 92.1 g / mol = 14.9 g glycerine

Answer 2

The equation for a nonelectrolyte in water is ΔT_FP = -(1.86 K/molal)* m where m is the molality or moles/kg water.

solving for m: |ΔT_FP/(k)| = (1.5/1.86) molal= .806 moles of nonelectrolyte / 1000 g of water

The amount of glycerine is 200g w (.806 moles/1000 g w)(92 g/mole glycerine) = g glycerine

Please consider a tutor. Take care.