Question

help please??
Assuming 100% dissociation, calculate the freezing point and boiling point of 2.64 m K3PO4(aq).
Tf=
Tb=

Answer 1

boiling point elevation and freezing point depression are related to the number of particles in solution

100% dissociation means 4 particles for each molecule of potassium phosphate, three K and one PO4

so the "particle" molarity is ... 4 * 2.64

Answer 2

so would that be Tf or Tb

Answer 3

You have to be kidding. You need to learn this material, just posting something in which you have no clue is revealing much about your effort.

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 <b>below</b> the normal freezing point, OC

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

I would encourage you to review your text materials on this.

Answer 4

Bob pursley is a loser