moles glycerin = grams/molar mass
solve for moles glycerin.
molality = mole/kg solvent
solve for molality
delta T = K*m
solve for delta T, then freezing point.
the freezing point of ethanol (C2H5OH) is -114.6C. the molal freezing point depression constant for ethanol is 2.00C/m. what is the freezing point (C) of a solution prepared by dissolving 50.0 g of glycerin (C3H8O3, A nonelectrolyte) in 200 g of ethanol.
any help would be appreciated im very confused.
3 answers
thank you looks like i double posted... sorry
i tried to solve it on my own but im confused about the last part, this i what i have so far..
moles of glycerin= 50.0/92.10=0.54
molality=0.54/0.200=2.7
is that right?
im a little confused about the numbers im supposed to use for step 3...
moles of glycerin= 50.0/92.10=0.54
molality=0.54/0.200=2.7
is that right?
im a little confused about the numbers im supposed to use for step 3...