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anonymous

  • one year ago

PLEASE HELP! A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 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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    @abb0t

  2. anonymous
    • one year ago
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    @sweetburger @peachpi @mathway

  3. sweetburger
    • one year ago
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    so you need to find the molality which from what i remember is moles of solute/ kg of solvent

  4. sweetburger
    • one year ago
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    ΔT=-(Kf)m

  5. anonymous
    • one year ago
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    @sweetburger I got 735.33

  6. cuanchi
    • one year ago
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    "m" it is not mass, it is represent molality or moles of solute per kg of solvent, you have to calculate the molality of the solution to calculate the delta T. The delta T is the difference between the freezing point of the pure solvent and the freezing point of the solution. Then the T freezing point of the pure solvent in this case water I think you may know the value or you can find it in the internet. knowing that and the delta T you can calculate the freezing point of the solution.

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