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
help please??
Assuming 100% dissociation, calculate the freezing point and boiling point of 2.64 m K3PO4(aq).
Tf=
Tb=
4 answers
so would that be Tf or Tb
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 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
I would encourage you to review your text materials on this.
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
I would encourage you to review your text materials on this.
Bob pursley is a loser