Question

calculate the freezing point of water (Kf= 1.86 C/m) for an ideal solution of sodium phosphate which has a concetration of 3.25 m .

Answer 1

Answer – We are given, molality of Na₃PO₄ = 3.25 m , Kf= 1.86 ^oC/m,

We know pure solvent, means water freezing point = 0.00^oC,

First we need to calculate the freezing point depression and we know formula   

∆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

             = 24.18^oC

So, Freezing point of ideal solution of Na₃PO₄ = 0.00^oC - ∆Tf

                                                                                = 0.00^oC – 24.18^oC

                                                                                = – 24.18^oC