Asked by Lauren
The freezing point of ethanol is -114.6 C and its Kf value is 2.00 C/m. What is the freezing point for the solution prepared by dissolving 50.0 g of glycerin (C3H8O3) in 200.0 g ethanol?
I used Change in temp=Kfm
kf=2.00
I calculated m to be 2.1736.
I ended up the answer of -118.4 C but the answer is apparently -120. I'm a few numbers off, can someone explain to me where I went wrong?
I used Change in temp=Kfm
kf=2.00
I calculated m to be 2.1736.
I ended up the answer of -118.4 C but the answer is apparently -120. I'm a few numbers off, can someone explain to me where I went wrong?
Answers
Answered by
DrBob222
mols glycerin = grams/molar mass = 50/92 = about 0.54 but you can clean that up (as well as the numbers following also).
m = mols/kg solvent
m = 0.54/0.200 = about 2.7
dT = Kf*m = 2.00*2.7 = about 5.4.
New freezing point = -114.6-5.4 = about 120.
m = mols/kg solvent
m = 0.54/0.200 = about 2.7
dT = Kf*m = 2.00*2.7 = about 5.4.
New freezing point = -114.6-5.4 = about 120.
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