Question

A certain substance X melts at a temperature of −9.9°C . But if a 350.g sample of X is prepared with 31.8g of urea NH22CO dissolved in it, the sample is found to have a melting point of −13.2°C instead. Calculate the molal freezing point depression constant Kf of X.

Answer 1

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

Answer 2

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