Question

a) honey contains approximately 2.88 M fructose sugar (C6H12O6 MM=180.16 g/mol) and has a density of 1.36 g/mL. what is the boiling point of honey assuming it consists only of fuctose and water? kb=0.52 C/m for water

b) the freezing point of a solution prepared by dissolving 1.50 x 10^2 mg of caffeine in 10.0 g of camphor is 3.07 C lower than that of pure camphor (kf=39.7 C/m) what is the molar mass of caffeine?

Answer 1

Delta T = Kb*m
molality = Weight*1000/G.M.W *weight of solvent in gm
2.88M fructose solution contains 2.88 moles of fructose per liter of solution.
M = n/v in L
V = 1 liter
M = n ( no of moles)
mass of fructose = 2.88*180 = 518.4 gm
Density of honey = (mass of fructose + mass of water)/total volume
Density of honey = 1.36gm/ml
1360 = 518.4 + mass of water
mass of water = 1360-518.4 = 841.6 gm =0.8416Kg
Delta T = 0.52*2.88/0.8416 = 1.78C
The boliling point is approximately 101.78C

T = Kf *m

3.07 = 39.7*m

m= 0.0773 m

molality = W *1000/G.M.Wt*weight of solvent in gm

    0.0773 = 0.15*1000/G.M.Wt*10

G.M.Wt = 0.15*1000/0.0773*10

gram molar mass = 194