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johnkim2330
 one year ago
HELP MEDAL WILL BE AWARDED
A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezingpoint depression of the solvent if the freezing point constant is 1.86 °C/m? Show all of the work needed to solve this problem.
johnkim2330
 one year ago
HELP MEDAL WILL BE AWARDED A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezingpoint depression of the solvent if the freezing point constant is 1.86 °C/m? Show all of the work needed to solve this problem.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Strategy: 1.Step 1: Calculate the freezing point depression of benzene. deltaTf = (Freezing point of pure solvent)  (Freezing point of solution) (5.5 oC)  (2.8 oC) = 2.7 oC 2.Step 2 : Calculate the molal concentration of the solution. molality = moles of solute / kg of solvent moles of naphthalene = (1.60 g) (1 mol / 128 g) = 0.0125 mol naphthalene molality of solution = (0.0125 mol) / (0.0200 kg) = 0.625 m 3.Step 3: Calculate Kf of the solution. deltaTf = (Kf) (m) (2.7 oC) = (Kf) (0.625 m) Kf = 4.3 oC/m

johnkim2330
 one year ago
Best ResponseYou've already chosen the best response.0A solution is made by dissolving 2.5 moles of sodium chloride (NaCl) in 198 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution? Show all of the work needed to solve this problem.

johnkim2330
 one year ago
Best ResponseYou've already chosen the best response.0what about this one ^ @shadowhuntergaming im not sure what the molal boiling point constant is
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