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tywower
 one year ago
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.
tywower
 one year ago
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.

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tywower
 one year ago
Best ResponseYou've already chosen the best response.0NOOOOBODYYYY HELPED MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEeee

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0Hold on until tara or photon come online. I haven't still learned these stuff. Sorry I couldn't help u@@@@@

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.1convert 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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