What will be the freezing point and boiling point of an aqueous solution containing 55.0 g of glycerol, C3H5(OH)3, and 250 g of water?
Kb(H2O) = 0.51 oC/m
Kf = 1.86o C/m.
To determine the freezing point and boiling point of the solution, we need to calculate the molality of the glycerol in water.
Step 1: Calculate the molality (m) of the glycerol:
Molar mass of glycerol (C3H5(OH)3) = 92.09 g/mol
moles of glycerol = 55.0 g / 92.09 g/mol = 0.5982 mol
mass of water = 250 g
molality (m) = moles of solute / mass of solvent (in kg)
mass of solvent (in kg) = mass of water / 1000 = 250 g / 1000 = 0.250 kg
molality (m) = 0.5982 mol / 0.250 kg = 2.3932 m
Step 2: Calculate the freezing point depression (ΔTf):
ΔTf = Kf * m
ΔTf = 1.86 oC/m * 2.3932 m = 4.454 oC
Step 3: Calculate the freezing point of the solution:
Freezing point of water = 0 oC
Freezing point of solution = Freezing point of water - ΔTf
Freezing point of solution = 0 oC - 4.454 oC = -4.454 oC
Therefore, the freezing point of the aqueous solution is -4.454 oC.
Step 4: Calculate the boiling point elevation (ΔTb):
ΔTb = Kb * m
ΔTb = 0.51 oC/m * 2.3932 m = 1.221 oC
Step 5: Calculate the boiling point of the solution:
Boiling point of water = 100 oC
Boiling point of solution = Boiling point of water + ΔTb
Boiling point of solution = 100 oC + 1.221 oC = 101.221 oC
Therefore, the boiling point of the aqueous solution is 101.221 oC.