boiling/freezing points.. help plz!!?

Question

Calculate the boiling points and freezing points of each of the following: 3m sugar water, 3m lithium nitrate (aq), 3m lead (II) chlorate (aq), and 3m ammonium phosphate (aq).

Answer 1

For water Kf=1.858 and Tf0= 0 deg C, Kb= 0.512 and Tb0=100 deg C

generally ΔT =i*K*m
where m= molality
K the respective constant for elevation/depression
i the van't Hoff coefficient

For non-electrolytes i=1
For strong electrolytes i=n, where n is the number of ions produced by its dissociation according to the molecular formula
For weak electrolytes i=1-a+na where n the same as above and a the degree of dissociation.

Sugar is a non electrolyte thus i=1 and
ΔTf= 1*1.858*3 =5.574= 5.57 deg C
Tf= Tf0-ΔTf= 0-5.57= -5.57 deg C
ΔTb = 1*0.512*3 = 1.536 =1.54 C
Tb= Tb0+ ΔTb= 100+1.54 =101.54 C

LiNO3 is a strong electrolyte which dissociates into 1 Li+ and 1 NO3-, thus 2 ions in total, so i=2 and
ΔTf= 2*1.858*3 =11.148= 11.15 => Tf= -11.15 C
ΔTb = 2*0.512*3 = 3.07 => Tb= 103.07 C

Pb(ClO3)2 dissociated into 1 Pb(+2) and 2 ClO3(-), thus 3 ions in total, thus i=3
ΔTf= 3*1.858*3 = 16.72=> Tf= -16.72 C
ΔTb = 3*0.512*3 = 4.608=4.61 => Tb= 104.61 C

(NH4)3PO4 is a special case. When it dissociates it gives 3 NH4+ and 1 PO4(-3) so you would expect that i=3+1=4. In reality the ions will hydrolyze and we should treat it as a weak electrolyte and try to find the true number of particles in solution. However this is a bit complicated in this case so I will assume we are talking about a not so advanced case and treat it like a strong electrolyte, without considering the hydrolysis of the ions. If you need the latter tell me.
So
ΔTf= 4*1.858*3 = 22.296= 22.30 > Tf= -22.30 C
ΔTb = 4*0.512*3 = 6.144=6.14 => Tb= 106.14 C