colligative properties

Question

calculate the freezing point of solution perpared by dissolving 4.5g of gulcose ( m.mass = 180g per mol) in 250 g of bromo form . Given freezing point of bromo form = 7.8°C and Kf for bromo form = 14.4 K Kg per mol. i am getting 279.51 K does it's correct.

Answer 1

ΔT = imK

ΔT = change in freezing point = ?

i = van’t Hoff factor = 1 for glucose

m = molality = moles glucose/kg solvent = 4.5g/180 g/mol/0.25kg = 0.1 molal

K = freezing point constant for bromo form = 14.4°/m (note the units of K should be degrees per molal)

ΔT = (1)(0.10)(14.4) = 1.44 degrees

Freezing point of solution = 7.8 deg - 1.4 deg = 6.4 degrees C

Freezing point of solution = 273.15+ 6.4 = 279.55 = 279.6K

So, your answer is essentially correct.

Answer 2

What is freezing point for bromoform without the sugar? 273.15 C +7.8 C = 280.15 C.

Zero C is 273.15 K

If we had a 1 molal solution the freezing point would be dropped by 14.4 C and 280 - 14 is 266 C.

So now we ask what is the molality of our solution.

Number of moles sugar is easy 4.5 grams/180 grams/mole = 0.025 moles.

now the total mass of solution is not 250 grams rather 254.5 grams or 0.2545 Kg

Molality is 0.025/ 0.2545 moles/Kg solution and we have 2 sig figs for the mass of glucose making up this solution...m = 0.0982

14.4 x 0.0982 = 1.4 K

280.15 - 1.4 = 278.75 C