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
Answer:
- [tex]T_{f,solution}=-4.5^oC[/tex]
- Equation:
[tex]\Delta T_{freezing}=(T_{f,solution}-T_{f,water})=-imKf[/tex]
Explanation:
Hello,
In this case, the freezing point depression of a solution when a solute is added is computed by the following equation:
[tex]\Delta T_{freezing}=(T_{f,solution}-T_{f,water})=-imKf[/tex]
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:
[tex]m=\frac{0.80mol}{1.00kg}=0.8\frac{mol}{kg}[/tex]
Hence, as the freezing point of water is 0 °C, we obtain freezing point of the solution as shown below:
[tex]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[/tex]
Best regards.
