Aim # 18: What are some other methods of expressing the concentration of a solution? H.W. # 18 Study pp. 474-481 (sec. 15.4-15.5) STUDY class notes Complete.

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Aim # 18: What are some other methods of expressing the concentration of a solution? H.W. # 18 Study pp. 474-481 (sec. 15.4-15.5) STUDY class notes Complete handout sheet

I Solution concentration- the quantity of solute dissolved in a specific quantity of solvent or solution. II Methods of calculating solution concentration A. percent by mass (also called mass percent) % = mass of solute x 100 mass of solution problem: What is the concentration, in percent by mass, of a solution of 30 g of sucrose in 300 g of water? answer: 30g sucrose x 100 = 9% (30g sucrose + 300g water)

B. parts per million (Ref. Table T) parts per million = grams solute x 106 grams solution problem: If 100g of river water contains 5 x 10-4 g of Hg, what is the concentration of the mercury expressed in ppm? ppm = g solute x 106 g solution = 5 x 10-4 g Hg x 106 = 5 ppm 102 g solution

answer: 6.0 x 10-3 g Pb x 106 = 3 x 101 2 x 102 g solution = 30 ppm problem: Express the concentration in parts per million (ppm) of a solution that contains 6.0 x 10-3 g Pb dissolved in every 200 g of solution. answer: 6.0 x 10-3 g Pb x 106 = 3 x 101 2 x 102 g solution = 30 ppm C. Molarity (M) – the number of moles of solute dissolved in 1L of solution. Molarity = moles of solute liters of solution e.g. A 1.0 molar solution (1.0 M) has 1.0 mole of solute in every liter of solution.

problem: What is the molarity of a solution that contains 6 problem: What is the molarity of a solution that contains 6.0 moles of AgNO3 dissolved in 3.0 L of solution? answer: M = 6.0 mol = 2.0 mol/L 3.0 L problem: How many moles of NaCl are there in 200. mL of a solution whose concentration is 0.300 M? How many grams of NaCl are there? 0.300 mol x .200 L = 0.0600 mol NaCl L 0.0600 mol NaCl x 58.5 g NaCl = 3.51 g NaCl mol NaCl

Problem: If 54. 75 g HCl is dissolved in enough water to prepare a Problem: If 54.75 g HCl is dissolved in enough water to prepare a .500 M solution, what is the volume of the solution? 54.75 g HCl x 1 mol HCl = 1.500 mol HCl 36.45 g HCl 1.500 mol HCl = .500 mol HCl V L L V = 1.500 mol HCl = 3.00 L .500 mol HCl L

How many grams of potassium nitrate are needed to prepare 1.5 L of a 0.250 M solution?