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brianna0905
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.
Calculate moles first: moles glucose = 15.5 g/ atomic mass = 15.5 g/ (6*12+12*1+6*16) = 15.5/180 = 0.086 Calculate molality next: (molality and molarity are the same when water is the solvent) 0.086 moles/ .245 kg water = 0.3515 m Calculate temp depression: -1.86 oC/M *.3515 m = 0.65 oC so the freezing point will drop from 0oC to -0.65oC.
i was so lost, but i get it more know