100g of the solution contains 5g of urea and 10g of glucose. What will be freezing point of the solution? (kf of water is 1.86 k kg/mol)
To solve we use formula, ΔTf = Kf i m
where ΔTf is change in freezing temperature
i is van’t Hoff factor =1 (both glucose and urea does not dissociate )
Kf = molal freezing point depression constant = 1.86 Kkg/mol
m is molality of solute or
moles of glucose = 10/180 = 0.056mol
moles of urea = 5/60 = 0.083mol
total moles of solutes = 0.139mol
w1= 100g = 0.1 kg
Therefore, m = 0.139/0.1 = 1.39m
ΔTf = 1.86 1 1.39
ΔTf = 2.5854K or 2.5854°C
ΔTf = T°f - Tf
where T°f is freezing point of pure solvent, water = 273K
and Tf is the freezing point of solution.
2.5854 = 273 - Tf
Tf = 273- 2.5854 = 270.14K
where ΔTf is change in freezing temperature
i is van’t Hoff factor =1 (both glucose and urea does not dissociate )
Kf = molal freezing point depression constant = 1.86 Kkg/mol
m is molality of solute or
moles of glucose = 10/180 = 0.056mol
moles of urea = 5/60 = 0.083mol
total moles of solutes = 0.139mol
w1= 100g = 0.1 kg
Therefore, m = 0.139/0.1 = 1.39m
ΔTf = 1.86 1 1.39
ΔTf = 2.5854K or 2.5854°C
ΔTf = T°f - Tf
where T°f is freezing point of pure solvent, water = 273K
and Tf is the freezing point of solution.
2.5854 = 273 - Tf
Tf = 273- 2.5854 = 270.14K