Question

A 5.00-gram sample of glucose, C6H12O6, [molar mass = 180.] is dissolved 10.0 grams of water

What is the molality of the glucose solution?

Calculate the boiling(BP) and freezing points(FP) of this solution to the nearest 0.1 C°.


The vapor pressure of pure water at 30ºC is 25.0 torr. What is the vapor pressure of the solution described above at 30°C?


moles glucose = grams/molar mass
Solve for moles.

m = #moles/L soln
You know #moles from above and you know L soln. Solve for m.

f.p.
delta T = Kf*m, then solve for f.p. knowing normal f.p. is 0C.

b.p.
delta T = Kb*m, then solve for b.p. knowing normal b.p. is 100 C.

mole fraction water = moles H2O/total # moles (that is, moles water + moles glucose), then
P<sub>soln</sub> = X<sub>H2O</sub>*25.0 torr.

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