anonymous
  • anonymous
A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezing point depression of the solvent if the freezing point constant is -1.86 °C/m?
Chemistry
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
[delta]T=i Kf m?
aaronq
  • aaronq
Gotta use the freezing point equation: \(\sf \Delta T=i*m*K_f\) where \(\Delta T\) is the change in temp, \(i\) is the van't hoff constant, m is the molality, and Kf is the depression constant (given in the question). (we can ignore \(i\) because glucose doesnt dissociate into more particles in water). First find the molality though
aaronq
  • aaronq
yep

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anonymous
  • anonymous
M=number of moles/Volume
aaronq
  • aaronq
that's molaRity, this is molaLity
anonymous
  • anonymous
oh my b
aaronq
  • aaronq
\(\sf molality=\dfrac{moles~of~solute}{kg~of~solution}\)
aaronq
  • aaronq
lol no worries
anonymous
  • anonymous
I'm sorry how do i get my 25.5 g into moles?
anonymous
  • anonymous
25.5g x 1mol / 180g = 0.1417 mol
aaronq
  • aaronq
\(\sf moles=\dfrac{mass}{Molar~mass}\) yup thats it
anonymous
  • anonymous
m= 0.1417mol/ 0.398kg= 0.3359m
anonymous
  • anonymous
0.3559m x -1.86 C/m= 0.662C ??
aaronq
  • aaronq
the constant has a negative sign, so the answer should be negative
anonymous
  • anonymous
is that right?
aaronq
  • aaronq
you need a negative sign though
aaronq
  • aaronq
otherwise it's good
anonymous
  • anonymous
-0.662 C
aaronq
  • aaronq
yeah thats it
anonymous
  • anonymous
Thanks!
aaronq
  • aaronq
np!

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