Colligative properties

Question

If the boiling point constant of water is 0.51 C/M and the 400 g sample of aqueous solution whose boiling point is 101.1 C, how many grams of fructose (C5H10O5) are dissolved?

Answer 1

ΔT = imK

ΔT = change in temperature = 1.1^oC

i = vant Hoff factor = 1 for fructose

m = molarity = mols fructose/kg water = ?

K = constant = 0.51^oC/m

Solving for m…

1.1 = (1)(m)(0.51)

m = 1.1/0.51

m = 2.16 moles fructose/kg H2O

Next we find how many kg H2O we have:

2.16 mols x 180 g/mol = 388.8 g fructose

kg H2O = 400 g - 388.8 g = 11.2 g = 0.0112 kg H2O

Now we determine moles fructose in the sample:

2.16 moles fructose/kg x 0.0112 kg = 0.0242 moles fructose

And finally we calculate grams of fructose:

0.0242 moles fructose x 180 g/mol = 4.35 g fructose