A certain substance X has a normal freezing point of −3.1°C and a molal freezing point depression constant =Kf·6.23°C·kgmol−1. Calculate the freezing point of a solution made of 12.3g of urea NH22CO dissolved in 300.g of X. Be sure your answer has the correct number of significant digits.

Freezing T° of solution =  - 7.35 °C

Explanation:

This is about the freezing point depression, a colligative property which depends on solute.

The formula is: Freezing T° pure solvent - Freezing T° solution = m . Kf . i

Freezing T° of pure solvent is -3.1°C

At this case, i = 1. As an organic compound the urea does not ionize.

We determine the molality (mol/kg of solvent)

We convert mass to moles:

12.3 g . 1mol / 60.06 g = 0.205 moles

0.205 mol / 0.3 kg = 0.682 mol/kg

We replace data in the formula:

-3.1°C - Freezing T° of solution  = 0.682 mol/kg . 6.23 kg°C/mol . 1

Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1 + 3.1°C

Freezing T° of solution =  - 7.35 °C

e

explanation:

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.