Answer :
Answer:
Kf = 2.18 °C/m
Explanation:
We must apply the colligative property of freezing point depression:
Freezing T° of pure solvent - Freezing T° of solution = Kf . m
Let's determine data.
Freezing T° of pure solvent = -9.9°C
Freezing T° of solution = -13.2°C
ΔT = -9.9°C - (-13.2°C) = 3.3°C
Kf is the unknown
Let's determine m (molality), moles of solute in 1kg of solvent
X is the solvent. We convert the mass from g to kg
350 g . 1kg / 1000 g = 0.350 kg
We convert the mass of solute (urea) to moles: 31.8 g / 60g/mol = 0.53 moles
Molality = 0.53 mol / 0.350 kg = 1.51 m
We replace data in the main formula: 3.3°C = Kf . 1.51 m
Kf = 3.3°C / 1.51 m = 2.18 °C/m
Answer:
The molal freezing point depression constant Kf of X is 2.185 °C / m
Explanation:
Step 1: Data given
Substance X melts at a temperature of −9.9°C
Mass of sample X = 350 grams
Mass of urea = 31.8 grams
Molar mass = 60.06 g/mol
the sample is found to have a melting point of −13.2°C
Step 2: Calculate moles urea
Moles urea = mass urea / molar mass urea
Moles urea = 31.8 grams / 60.06 g/mol
Moles urea = 0.529 moles
Step 3: Calculate molality
Molality = moles urea / mass X
Molality = 0.529 moles / 0.350 kg
Molality = 1.51 molal
Step 4: Calculate the molal freezing point depression constant Kf of X
ΔT =i* Kf * m
⇒with ΔT = the freezing point depression = 3.3 °C
⇒with i = the van't Hoff factor of urea = 1
⇒with Kf = the molal freezing point depression constant Kf of X. = TO BE DETERMINED
⇒with m = the molality = 1.51 molal
3.3 °C = Kf * 1.51 m
Kf = 3.3 °C / 1.51 m
Kf = 2.185 °C / m
The molal freezing point depression constant Kf of X is 2.185 °C / m