Final answer:
The freezing point of the solution with 21.6 g of NiSO4 in 100 g of H2O is calculated to be -7.8 °C after assuming 100% ionization for the solute and using the formula for freezing point depression.
Step-by-step explanation:
To compute the freezing point of a solution, we need to use the formula for freezing point depression, which is ΔT_f = i · K_f · m, where ΔT_f is the change in freezing point, i is the van't Hoff factor (number of particles the solute dissociates into), K_f is the freezing point depression constant for water (which is 1.86 °C·kg/mol), and m is the molality of the solution.
To start, we need to calculate the molality (m) of the solution. Molality is moles of solute per kilogram of solvent (water in this case).
The molar mass of NiSO4 (Nickel(II) sulfate) is approximately 154.75 g/mol. Therefore, the moles of NiSO4 are:
moles NiSO4 = 21.6 g NiSO4 ÷ 154.75 g/mol = 0.1396 mol
Next, we calculate the molality:
molality (m) = moles of solute ÷ kilograms of solvent = 0.1396 mol ÷ 0.100 kg = 1.396 m
Assuming 100% ionization, NiSO4 dissociates into 3 ions (Ni2+ and SO4^2-), so i = 3.
Now we can calculate the freezing point depression:
ΔT_f = i · K_f · m = 3 · 1.86 °C·kg/mol · 1.396 m = 7.8 °C
Water normally freezes at 0 °C, so the freezing point of this solution is:
freezing point = 0 °C - ΔT_f = 0 °C - 7.8 °C = -7.8 °C
Therefore, the freezing point of this solution, rounded to the nearest tenth, is -7.8 °C.