Asked by Laverne

Calculate the freezing point and boiling point of 12.0g of glucose (C6H12O6) in 50.0g H20
Answered by DrBob222
freezing point.
delta T = Kf*molality.
Convert 12.0 g glucose to moles. Moles = grams/molar mass.
Then use m = moles/kg solvent to calculate molality, substitute that into the first equation to calculate delta T. Subtract from 0<sup>o</sup> to find the freezing point of the solution.

boiling point. Done essentially the same way.
delta T = Kb*m
Post your work if you get stuck.
Answered by Laverne
Molar mass-180
12.0\180=
is the molar mass 6.67 or is it 6.67*10^-2
Answered by DrBob222
Neither. The molar mass = 180.
moles in 12.0 grams is
12.0/180 = 0.06667 and I would keep it like that until the end, then round the final answer.
Answered by Laverne
whats the kg solvent
Answered by DrBo222
The problem tell you 50.0 g H2O, which is 0.050 kg.
Answered by Victor
Freezing point DT=KF*M mol of glucose 12g/358=0.034mols kf=1.86c/mol SODT=1.86x0.034 =0.06
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