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  • one year ago

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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    1) you have to calculate the molality of the solution (moles of glucose per kg of solvent). 2) then apply the formula ΔTF = KF × b × i KF is the freezing point constant (-1.86 °C/m) b= molality i=i is the van 't Hoff factor (number of ion particles per individual molecule of solute, e.g. i = 2 for NaCl, 3 for BaCl2). in your case for glucose the value is =1

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