anonymous
  • anonymous
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? Show all of the work needed to solve this problem.
Chemistry
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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taramgrant0543664
  • taramgrant0543664
So for this one we have a couple extra steps so first we have to find the number of moles of glucose to do this: n=m/MM moles= mass / molar mass So mass is the 25.5g and the molar mass of glucose is 180.16 So 25.5/180.16=n n=?
anonymous
  • anonymous
n=0.142
taramgrant0543664
  • taramgrant0543664
Great! So next step we need to find moles per kg of water but first we have to convert 398g into kg so you divide by 1000 as that is the conversion

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anonymous
  • anonymous
0.398
taramgrant0543664
  • taramgrant0543664
Yes so now we can figure out the moles per kg m=0.142/0.398 m=?
anonymous
  • anonymous
0.357
taramgrant0543664
  • taramgrant0543664
Yes 0.357moles/kg, so the next step is the van't Hoff factor which again is represented by i, it is a dissociation factor, glucose doesn't dissociate in water so i=1 for this example
taramgrant0543664
  • taramgrant0543664
Second last step now is to find our change in temperature: T=iKfm i is 1, Kf is given so -1.86 and m is the 0.357 we found previously
anonymous
  • anonymous
T=(1)(-1.86)(0.357) = -0.664
taramgrant0543664
  • taramgrant0543664
Yes so that is the change in temperature when the glucose is mixed in so to find the freezing point we add this to the freezing point of water So 0+(-0.664)= -0.664 celcius is the freezing point of this reaction
anonymous
  • anonymous
why 0?
taramgrant0543664
  • taramgrant0543664
Because 0 is the freezing point of water
taramgrant0543664
  • taramgrant0543664
When we did the boiling point we used 100 as that is the boiling point of water
anonymous
  • anonymous
ok so the answer is -0.664?
taramgrant0543664
  • taramgrant0543664
Yes it would be
anonymous
  • anonymous
thx
taramgrant0543664
  • taramgrant0543664
No problem! Happy to help!!

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