Question

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

Answer 1

Answer:

- $T_{f,solution}=-4.5^oC$

- Equation:

$\Delta T_{freezing}=(T_{f,solution}-T_{f,water})=-imKf$

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:

$m=\frac{0.80mol}{1.00kg}=0.8\frac{mol}{kg}$

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

$T_{f,solution}=T_{f,water}-imKf=0^oC-3*0.8\frac{mol}{kg}*1.86\frac{^oC*kg}{mol} \\\\T_{f,solution}=-4.5^oC$

Best regards.

Answer 2

Answer:

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