tywower
  • tywower
A solution is made by dissolving 23.5 grams of glucose (C6H12O6) in .245 kilograms of water. If the molal freezing point constant for water (Kf) is -1.86 °C/m, what is the resulting Δtf of the solution? Show all the steps taken to solve this problem.
Chemistry
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SOLVED
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  • chestercat
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Rushwr
  • Rushwr
@Photon336
tywower
  • tywower
@dan815
tywower
  • tywower
@abb0t

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Rushwr
  • Rushwr
Hold on until tara or photon come online. I haven't still learned these stuff. Sorry I couldn't help u@@@@@
Photon336
  • Photon336
@tywower do you know this formula? \[\Delta T = ikbm \] delta T is the temperature changed caused by adding a certain amount of solute in the solvent. m = molality moles of solute per kilogram of solvent. i = Van Hoff factor Kb = our constant.
Photon336
  • Photon336
convert grams of glucose to moles of glucose \[23.5g x (1mol/180g) = 23.5/180 = 0.130 \] they gave us the amount of kilograms of water which is = 0.245 kg now we calculate the total molality: \[molality = moles/kilogram of solvent = 0.130moles/0.245kg = 0.53 m \] we put all of this together: kf = -1.86 C/m \[-1.86 C/m *0.53 m*1 = -0.99 \] we realize that the temperature changed by a factor of -0.99 or 1 and the freezing point of water we know is 0 degrees celsius so our new temperature = -1 degree celsius. so what this means is that by adding a certain amount of solute the freezing point went down by 1 degree celsius.

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