chemistry
posted by jessica .
How many grams of C6H12O6 need to be dissolved in 350g of water vto have a solution which freezes at 2.50 degrees celcius?
delta T = K_{f}*m
You know K_{f}, m = molality = mols/kg and mols = grams/molar mass, delta T is 2.50. Solve for grams. Post your work if you get stuck.
what is delta t i never learned that
Delta T is T_{normal f.p.}  T_{new f.p.}. Delta T in this problem is 2.50 degrees C or since the normal freezing point is 0 and you want it to be 2.50, that is 0  (2.50) = 2.50.
i really don't understand can you help me throught the first part... what do you make moles? i really need help i have a big test tommorrow!!
delta T = i*K_{f} *m
delta T = 2.50
i = 1
K_{f} = 1.86
solve for m
2.50 = 1*1.86*m
m = 2.50/(1*1.86) = 1.34 molal.
The definition of molality = #mols/kg.
m=mols/kg
1.34 molal = mols/0.350 kg
1.34*0.350 = mols = 0.469
mols = grams/molar mass
0.469 = g/180
g = 0.469*180 = 84.4 g C6H12O6.
Let me know if this isn't clear. Check my thinking. Check my arithmetic.
okay so do i do the exact same thing for the other one but i=3
For the CaCl2 problem, i = 3, Kf is the same, I think kg solvent is not 0.350 kg but something else. Otherwise, yes, the same procedure.
did you get 13.92 g
No, I obtained an answer of 84.4 g C6H12O6.
molality = mols/kg.
1.34 molal = mols/0.350 kg
1.34*0.350 = mols = 0.469
mols = grams/molar mass
0.469 = g/180
g = 0.469*180 = 84.4 g C6H12O6.
no for the other one sorry!! 13.92!!
You're mixing yourself up AND me by not posting the answers with the question. Let's keep it simple. Answer the question with THAT post and not some other post. I didn't get 13.92.
If you will show your work I can find your error.
2.0=1.86(3)M
TIMES 1.86 TIMES 3=5.58 THEN DIVIDE
2.0/5.58 =.358
.358 TIMES .350 =.1254
THEN .1254 TIMES 110.98 (THAT IS THE MOLAR MASS OF CACL2) = 13.92
If you will go back and read the original problem, I believe it is in 100 g water, not 350 g water.
So 0.358*0.100*110.98 = ?? and I have 3.97 g CaCl2.
thank you so much!!
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