Assuming 100% dissociation, calculate the freezing point and boiling point of 2.64 m k3po4(aq).

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Freezing pointdepression= Kf*molality 
now Kf is freezing point depression, in this case water, which is 1.86deg*molality of particles 
= 1.86*4*2.64 
that the the depression below the normal freezing point, OC 

for boiling point elevation, Kb, the amount boiling point is elevated, kb=.52deg/mole 
boiling point elevation here 
= .52*4*2.64 degrees, and the boiling point then is 100C+ the elevation 

Ithink it’s 3 as the answer.

Ithink d hope that