# chemistry

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Calculate the freezing point and boiling point of 12.0g of glucose (C6H12O6) in 50.0g H20

• chemistry -

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 0o 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.

• chemistry -

Molar mass-180
12.0\180=
is the molar mass 6.67 or is it 6.67*10^-2

• chemistry -

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.

• chemistry -

whats the kg solvent

• chemistry -

The problem tell you 50.0 g H2O, which is 0.050 kg.

• chemistry -

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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