Suppose you had a 1.00 m solution of AlCl3. Assuming complete dissociation, what is the theoretical change in the freezing point of this solution?
delta T = i*Kf*m
i = 4
DeltaT=-1.86x1.00mx4i=-7.44c
To determine the theoretical change in the freezing point of a solution, we need to use the concept of the van't Hoff factor (i). The van't Hoff factor represents the number of particles into which a compound dissociates when it dissolves in a solvent.
In this case, you mentioned that the AlCl3 solution is completely dissociated, so each AlCl3 molecule will dissociate into four particles: one Al3+ ion and three Cl- ions. Therefore, the van't Hoff factor (i) for AlCl3 is 4.
To calculate the theoretical change in the freezing point (∆Tf), we can use the equation:
∆Tf = i * Kf * m
Where:
∆Tf is the change in the freezing point of the solution,
i is the van't Hoff factor,
Kf is the cryoscopic constant (specific for each solvent), and
m is the molality of the solution.
In this case, since you mentioned that it is a 1.00 m (molal) solution of AlCl3, we can assume the molality is 1.00 mol AlCl3 / 1.00 kg of solvent.
The cryoscopic constant for water (the assumed solvent) is approximately 1.86 °C/m.
Plug in the values into the equation to calculate ∆Tf:
∆Tf = (4) * (1.86 °C/m) * (1.00 mol AlCl3 / 1.00 kg solvent)
∆Tf = 7.44 °C
Therefore, the theoretical change in the freezing point (∆Tf) of the 1.00 m solution of AlCl3, assuming complete dissociation, is 7.44 °C.