Question

Assuming 100% dissociation, calculate the freezing point (T_f) and boiling point (T_b) of 2.32 m K₃PO₄(aq). Colligative constants can be found in the chempendix.

Question 16 of 19> Assuming 100% dissociation, calculate the freezing point (T) and boiling point (Th) of 2.32 m K3PO4(aq). Colligative constants can be found in the chempendix "С Ть — "С

Answer 1

Kf   = 1.86⁰C/m

m    = 2.32m

K3PO4(aq) ---------------> 3K^+ (aq) + PO4^3- (aq)

i = 4

$\Delta$ Tf   = i*m*mKf

           = 4*2.32*1.86

           = 17.261⁰C

$\Delta$ Tf     = Tf of solvent - Tf solution

17.261   = 0-Tf solution

Tf solution = -17.261⁰C

The freezing point of solution = -17.261⁰C

Kb   = 0.512⁰C/m

$\Delta$ Tb = i*m*mKb

           = 4*2.32*0.512

          = 4.7514⁰C

The boiling point of solution = 100+4.7514   = 104.7514⁰C >>>>answer