Question

what is the freezing point of an aqueous 200m (NH4)PO4 salt solution? The Kf of water 1.86 C/m. Assume complete dissociation of the soluble salt.

Answer 1

Answer – Given, molality of (NH₄)₃PO₄ = 200 m , Kf = 1.86 ^oC/m

We know (NH₄)₃PO₄ has total 4 ion ( 3 for NH₄⁺ and 1 for PO₄^3-)

So, i = 4

We know the

∆Tf = i* Kf *m

       = 4 * 1.86 ^oC/m * 200 m

       = 1488 ^oC

We know the

Freezing point = Freezing point of pure water -∆Tf

                         = 0.0^oC -1488 ^oC

                         = -1488 ^oC

(It is too high, may be molality given is 0.200 m instate 200 m, if 0.200 m then answer is -1.49^oC)