To calculate the freezing point of an aqueous 0.13 m MgI2 solution, we will use the freezing point depression formula:
ΔT_f = i imes K_f imes m
Where:
- ΔT_f = change in freezing point
- i = van ‘t Hoff factor (number of particles the solute breaks into)
- K_f = freezing point depression constant of the solvent (for water, K_f = 1.86 °C kg/mol)
- m = molality of the solution
In this case, MgI2 dissociates into three ions: 1 magnesium ion (Mg2+) and 2 iodide ions (I–). Therefore, the van ‘t Hoff factor (i) for MgI2 is 3.
Given:
- m = 0.13 m
- K_f = 1.86 °C kg/mol
- i = 3
Now, we can plug in the values:
ΔT_f = 3 imes 1.86 imes 0.13
ΔT_f = 0.7266 °C
This means the freezing point of the solution decreases by 0.7266 °C. Since the normal freezing point of water is 0 °C, we subtract the change:
Freezing Point = 0 °C – 0.7266 °C = -0.7266 °C
Therefore, the freezing point of the aqueous 0.13 m MgI2 solution is approximately -0.73 °C.