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Chemistry 24 Online
OpenStudy (anonymous):

Calculate the molarity of 1.0 liter of solution made with the following solutes... a.) 2.3g of NaCl b.) 1.2 grams of calcium carbonate c.) 0.09 grams of sodium sulfate

OpenStudy (anonymous):

@cuanchi

OpenStudy (anonymous):

Get moles of all compounds 1.) 2.3g NaCl * 1 moles Nacl 58.5Grams Nacl mole .039 Nacl .039moles Nacl/ 1L= .039Molarity 2.) 1.2g CaCO3 * 1mole CaCO3 100.869 Grams CaCO3 mole .012 CaCO3 .012/1L= .012 Molarity 3.) 0.09 NaSO4 * 1 mole NaSO4 100.87 Grams NaSo4 mole 6.34*10^-4 NaSO4 6.34*10^-4/1L= 6.34*10^-4 Molarity

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