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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 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.

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  1. anonymous
    • one year ago
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    Strategy: 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

  2. johnkim2330
    • one year ago
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    A 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.

  3. johnkim2330
    • one year ago
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    what about this one ^ @shadowhuntergaming im not sure what the molal boiling point constant is

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