Answer – We are given, molality of
Na3PO4 = 3.25 m , Kf= 1.86
oC/m,
We know pure solvent, means water freezing point =
0.00oC,
First we need to calculate the freezing point depression and we
know formula  Â
∆Tf = i * Kf * m
We know, I is the van't Hoff factor and for the
Na3PO4 it is 4
So, ∆Tf = 4 * 1.86 oC/m * 3.25 m
           Â
= 24.18oC
So, Freezing point of ideal solution of
Na3PO4 = 0.00oC -
∆Tf
                                           Â
                                   =
0.00oC – 24.18oC
                                            Â
                                  =
– 24.18oC
