Question

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?

Answer 1

The freezing point of the solution is -4.46 °C

Step-by-step 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

Answer 2

-
`T_(f,solution)=-4.5^oC`

- Equation:

`\Delta T_(freezing)=(T_(f,solution)-T_(f,water))=-imKf`

Step-by-step explanation:

Hello,

In this case, the freezing point depression of a solution when a solute is added is computed by the following equation:

`\Delta T_(freezing)=(T_(f,solution)-T_(f,water))=-imKf`

Whereas i accounts for the van't Hoff's factor that for magnesium bromide is 3 (since three ions are produced when it dissociates in water one Mg, and two Br), m for the solution's molality and Kf the freezing point depression constant. In such a way, we first compute the solutions molality as:

`m=(0.80mol)/(1.00kg)=0.8(mol)/(kg)`

Hence, as the freezing point of water is 0 °C, we obtain freezing point of the solution as shown below:

`T_(f,solution)=T_(f,water)-imKf=0^oC-3*0.8(mol)/(kg)*1.86(^oC*kg)/(mol) \\\\T_(f,solution)=-4.5^oC`

Best regards.