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

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each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i