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
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Rushwr
  • Rushwr
tywower
  • tywower
tywower
  • tywower

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

tywower
  • tywower
tywower
  • tywower
tywower
  • tywower
tywower
  • tywower
tywower
  • tywower
tywower
  • tywower
@Ciarán95
tywower
  • tywower
tywower
  • tywower
NOOOOBODYYYY HELPED MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEeee
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.