help please??
Assuming 100% dissociation, calculate the freezing point and boiling point of 2.64 m K3PO4(aq).
Tf=
Tb=
Bob pursley is a loser
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
so would that be Tf or Tb
Sure! To calculate the freezing point and boiling point of a solution, we need to use the formula:
ΔTf = -Kf * m
ΔTb = Kb * m
Where:
- ΔTf represents the change in freezing point
- ΔTb represents the change in boiling point
- Kf is the cryoscopic constant (freezing point depression constant) and depends on the solvent
- Kb is the ebullioscopic constant (boiling point elevation constant) and also depends on the solvent
- m is the molality of the solution, which is calculated by dividing the moles of solute by the mass of the solvent in kilograms (kg)
In this case, we are given the molality (m) of the K3PO4 solution as 2.64 m. However, we need additional information about the solvent in order to calculate the freezing point and boiling point accurately. Can you please provide the solvent used in the solution?
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.