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
2007-07-15 06:03:00 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
the Kf of water = 1.86
the Kb of water = 0.52
2007-07-15 06:24:38 · update #1
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
2007-07-19 04:19:23 · answer #1 · answered by Anonymous · 0⤊ 0⤋
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.
2007-07-15 13:38:13 · answer #2 · answered by reb1240 7 · 0⤊ 0⤋