 # II III I II. Concentration Solutions. A. Concentration  The amount of solute in a solution.  Describing Concentration % by mass - medicated creams %

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II III I II. Concentration Solutions

A. Concentration  The amount of solute in a solution.  Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

B. % by Mass  Remember …  % = part x 100 whole  % by mass = mass solute x 100 mass solution

Example  What is the % by mass of a solution with 3.6 g of NaCl dissolved in 100.0 g of water?  % = (3.6 / 103.6) x 100 = 3.5% NaCl

C. % by Volume  Remember …  % = part x 100 whole  % by volume = volume solute x 100 volume solution

Example  What is the % by volume of 75.0 ml of ethanol dissolved in 200.0 ml of water?  % = (75.0 / 275.0) x 100 = 27.3%

D. Molarity  Molarity = moles of solute/liter of solution  Note: it’s liters of solution, not liters of solvent

Molarity Examples  Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate in 125 ml of solution  23.4 g Na 2 SO 4  0.165 mol  125 ml  0.125 L  M = mol / L  M = 0.165mol / 0.125 L  M = 1.32 M

Molarity Examples  Calculate the molarity of a solution made by dissolving 5.00 g of C 6 H 12 O 6 in enough water to make 100.0 ml of solution  0.280 M

Molarity Examples  How many grams of Na 2 SO 4 are required to make 0.350 L of a 0.500 M solution of Na 2 SO 4 ? (hint: find moles first, then convert to grams)  24.9 g Na 2 SO 4

E. Dilution  When chemists purchase solutions, they generally purchase “stock solutions” which are extremely concentrated solutions  This way a chemist can dilute the strong solution to any concentration that they wish. This stops the chemist from having to buy several concentrations

E. Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

E. Dilution  M 1 V 1 = M 2 V 2  M 1 = initial molarity  V 1 = initial volume  M 2 = final molarity  V 2 = final volume  The units for V 1 & V 2 do not matter as long as they are the same  M 1 & M 2 MUST be in molarity

E. Dilution Problems  Suppose we want to make 250 ml of a 0.10 M solution of CuSO4 and we have a stock solution of 1.0 M CuSO4. How many mL of the stock solution do we need?  First do the math  M 1 V 1 = M 2 V 2  (0.10M)(250ml) = (1.0)(V 2 )  V 2 = 25 ml

E. Dilution Problems  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

F. Preparing Solutions  500 mL of 1.54M NaCl mass 45.0 g of NaCl add water until total volume is 500 mL 500 mL mark 500 mL volumetric flask

F. Preparing Solutions  250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!”

Solution Preparation Lab  Turn in one paper per team.  Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution.  For each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25m NaCl in 100.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.1M concentrate.

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