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

the correct answer is boyle

the answer is na (sodium)