0.80 mol MgBr2 is added to 1.00 kg water. Determine the freezing point of the solution. Water has a

freezing point depression constant of 1.86°C. kg/mol.

Which equation should you use?

The freezing point of the solution is -4.46 °C

Explanation:

Step 1: Data given

Number of moles MgBr2 = 0.80 moles

Mass of water = 1.00 kg

Water has a  freezing point depression constant of 1.86°C.kg/mol

Freezing point of water = 0°C

Step 2: Calculate the freezing point of the solution

ΔT = i * Kf * m

⇒ΔT = the freezing point depression = TO BE DETERMINED

⇒with i = the can't Hoff factor of MgBr2 = 3

⇒with Kf = the freezing point depression constant of 1.86°C/molal

⇒with m = the molality = 0.80 moles / / 1.00 kg = 0.80 molal

ΔT = 3 * 1.86 °C/molal * 0.80 molal

ΔT = 4.46 °C

Step 3: Calculate the freezing point of the solution

0°C - 4.46 °C = -4.46 °C

The freezing point of the solution is -4.46 °C

h let me think

heat or electricity in the atmosphere- c.