Question

Estimate the freezing point of 150cm3 of water sweetened with 75 g of sucrose.

Answer 1

we have formula dT = i x kf x m

dT = depression in freezing point = water FP - solution FP = 0 -T

i = vatoff factor = 1 for non dissociative solids like sucrose

Kf = molal depression constanat for water 1.86 kg/mol

m = molality = moles of surcose / water mass in kg

water vol = 150 cm3 = 150 ml = 0.15 L

water mass = 0.15 kg ( mass = vol x density = 0.15 x 1 = 0.15 kg)

sucrose moles = mass/molar mass of sucrose = 75/ 342.2965 = 0.219

molality m = moles of solute / solvent mass in kg = 0.219 /0.15 =1.46

now dT = 1 x 1.86 x 1.46

dT = 2.7 = 0 -T

T = solution FP = -2.7 C