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

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Presentation on theme: "Notes 15.2 Describing Solution Composition. Mass Percent Mass percent= mass of solute X 100 mass of solution = grams of solute X 100 grams of solute +"— Presentation transcript:

1 Notes 15.2 Describing Solution Composition

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 Concentration can also be expressed in g/L

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 M 1 X V 1 = M 2 x V 2

9 example What volume of 16M sulfuric acid must be used to prepare 1.5 L of a 0.10M sulfuric acid solution? M 1 V 1 = M 2 V 2 16 M (V 1 ) = 0.10 M x 1.5 L V 1 = 0.0094 L or 9.4 mL


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