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How many grams of CaOH2 are needed to prepare 300.0 mL of a 0.115 M solution?
Wiki User
∙ 13y ago
Molarity = moles of solute/Liters of solution
( 300.0 ml = 0.300 liters ) ( Ca(OH)2 is correct formulation )
0.115 M Ca(OH)2 = moles Ca(OH)2/0.300 Liters
= 0.0345 moles Ca(OH)2 (74.096 grams/1 mole Ca(OH)2)
= 2.56 grams of Ca(OH)2 needed
Wiki User
∙ 13y ago
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