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Chemistry 17 Online
OpenStudy (anonymous):

the molality of a calcium fluoride solution is 2.33 m. If the solution is prepared with 125 mL of water, how many grams of calcium fluoride were added to the solution?

OpenStudy (jfraser):

what are the units of molALity? (not molARity)

OpenStudy (anonymous):

molality= mol solute/ kg solvent

OpenStudy (jfraser):

so we know the molality, and we know the volume of solvent. What do we also know about the density of water?

OpenStudy (anonymous):

the density of water is 0.992g/mL

OpenStudy (jfraser):

so we have a way to convert the volume of water used into a mass:\[volume H_2O * (\frac{mass}{volume})\] that mass of 125mL of water is now the proper term to use in the molality equation: \[molality = \frac{mol solute}{kg solvent}\] the molality we know, and now we know the mass of solvent used. Find the moles of solute by rearranging, and then convert that moles into mass using the molar mass of calcium fluoride

OpenStudy (anonymous):

so the mass of 125mL is 124g?

OpenStudy (jfraser):

essentially, yes. we usually just call the density of water 1g/mL, for really easy conversions

OpenStudy (anonymous):

so the final answer would be .0289 grams?

OpenStudy (jfraser):

that's not what i get

OpenStudy (jfraser):

do it one step at a time

OpenStudy (anonymous):

I did. What did you get?

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