What is the boiling & melting pt of .15 mol solute in 800 g H2O?

If 0.150 mole of a nonvolatile, nonelectrolyte solute is dissolved in 800 grams of water, what are the ideal melting and boiling points of the solution?
formulas to work with?
delta Tf = (Kf (mole solute)) / kg solvent
delta Tb = Kb of the same
freezing pt. of solution = 0 celcius - delta Tf
boiling pt. of solution = 100 celcius - delta Tb
thank you soooo much

Answer:
delta Tf = (Kf (mole solute)) / kg solvent
=1.86 X .15 X1000 /800
=.3487
m.pt= -0.3487degree

delta Tb = Kb of the same
=.52X.15X1000 /800=0.0975

b pt =100.0975degree

It s the same set up as in yur MgSO4 problem, except you only have .15 moles of solute because there are no ions here.

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What is the boiling & melting pt of .15 mol solute in 800 g H2O?

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