Question

The freezing point of water H2O is 0.00°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is urea. If 13.40 grams of urea, CH4N2O (60.10 g/mol), are dissolved in 153.2 grams of water...

The molality of the solution is _____ m.
The freezing point of the solution is _____ °C.

Answer 1

Molality = 1.46 molal

The freezing point of the solution = -2.72 °C

Step-by-step explanation:

Step 1: Data given

The freezing point of water H2O is 0.00°C at 1 atmosphere

urea = nonelectrolyte = van't Hoff factor = 1

Mass urea = 13.40 grams

Molar mass urea = 60.1 g/mol

Mass of water = 153.2 grams

Molar mass H2O = 18.02 g/mol

Kf = 1.86 °C/m

Step 2: Calculate moles urea

Moles urea = mass urea /molar mass urea

Moles urea = 13.40 grams / 60.1 g/mol

Moles urea = 0.223 moles

Step 3: Calculate the molality

Molality = moles urea / mass water

Molality = 0.223 moles / 0.1532 kg

Molality = 1.46 molal

Step 4: Calculate the freezing point of the solution

ΔT = i * Kf * m

ΔT = 1* 1.86 °C/m * 1.46 m

ΔT = 2.72 °C

The freezing point = -2.72 °C