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

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  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

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  2. so would that be Tf or Tb

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  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 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.

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