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Describing Solution Composition

Describing Solution Composition. Notes 15.2. Mass Percent. Mass percent= mass of solute X 100 mass of solution = grams of solute X 100 grams of solute + grams of solvent. Example.

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Describing Solution Composition

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  1. Describing Solution Composition Notes 15.2

  2. Mass Percent • Mass percent= mass of solute X 100 mass of solution = grams of solute X 100 grams of solute + grams of solvent

  3. Example • A solution is prepared by mixing 1.00g of ethanol with 100.0g of water. Calculate the mass percent of ethanol in this solution. • Mass% = 1.00 g of ethanol x 100 100.0 g of water + 1.00 g ethanol • Mass % of ethanol = 0.99 %

  4. Molarity • Molarity is the number of moles of solute per volume of solution in liters • M= molarity= moles of solute = mol Liters of solution L

  5. example • Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. • 11.5 g NaOH x 1 mol NaOH= .288 mol NaOH 40 g NaOH • Molarity= moles liters • Molarity = .288 mol NaOH = 0.192 M NaOH 1.50 L NaOH

  6. Molarity • Liters of solution x Molarity = moles of solute • L x M = mols

  7. example • Formalin is an aqueous solution of formaldehyde, HCHO, used as a preservative for biological specimens. How many grams of formaldehyde must be used to prepare 2.5 L of 12.3M formalin? • Moles = Liters x Molarity • Moles= 2.5 L x 12.3 M = 31 mols 31 mols HCHO x 30 g HCHO = 930 g HCHO 1 mol HCHO

  8. molarity • A standard solution is a solution whose concentrations is accurately known. • The process of adding more solvent to a solution is call dilution • M1 X V1 = M2 x V2

  9. example • What volume of 16M sulfuric acid must be used to prepare 1.5 L of a 0.10M sulfuric acid solution? • M1 V1 = M2 V2 16 M (V1) = 0.10 M x 1.5 L V1 = 0.0094 L or 9.4 mL

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