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Chemistry 9 Online
OpenStudy (anonymous):

A solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 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.

OpenStudy (mertsj):

Hang on. I have to google this.

OpenStudy (mertsj):

21.5 g glucose = 21.5/180 moles = .1194 moles in 255 g =.468 moles per kg

OpenStudy (mertsj):

Guess you didn't care about this so much after all.

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