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Solutions Entry Task: Nov 21 st Block 2 Question: What is the molality of a 200 g solution with 32 grams of NaCl? TURN IN ENTRY TASK SHEETS!! You have 5 minutes!

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Solutions Agenda: Discuss Ch. 13 sec. 4 pHet Concentration and Molarity Lab HW: Solutions ws

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Solutions I can … Express a concentration of a solution in different ways- %, ppm, mole fraction, Molarity, & Molality

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Solutions Chapter 13 Properties of Solutions

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Solutions Ways of Expressing Concentrations of Solutions

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Solutions Mass Percentage Mass % of A = mass of A in solution total mass of solution 100

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Solutions Parts per Million and Parts per Billion ppm = mass of A in solution total mass of solution 10 6 Parts per Million (ppm) Parts per Billion (ppb) ppb = mass of A in solution total mass of solution 10 9

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Solutions 13.3 problem (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water g solution X 100 = 2.91%

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Solutions 13.3 problem (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution? X 2500g X = X = 90.5g

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Solutions moles of A total moles in solution X A = Mole Fraction (X) In some applications, one needs the mole fraction of solvent, not solute— make sure you find the quantity you need!

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Solutions mol of solute L of solution M = Molarity (M) You will recall this concentration measure from Chapter 4. Because volume is temperature dependent, molarity can change with temperature.

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Solutions mol of solute kg of solvent m = Molality (m) Because both moles and mass do not change with temperature, molality (unlike molarity) is not temperature dependent.

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Solutions Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

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Solutions 13.4 problem What is the molality of a solution made by dissolving 36.5 g of naphthalene (C 10 H 8 ) in 425 g of toluene (C 7 H 8 )? 35g g = 0.28 mole kg 1 mole = kg

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Solutions 13.5 problem A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution. 3.62g 74 g = mole 1 mole 3.62 % in water means 3.62 g in water 3.62 g – 100 = g of water 96.38g 18 g = 5.35 mole 1 mole mole mole = mole mole mole 9.06 x 10 -3

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Solutions 13.5 problem A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution. 3.62g 74 g = mole 1 mole 3.62 % in water means 3.62 g in water 3.62 g – 100 = g of water kg = m mole

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Solutions a) Calculate the mass % of Na 2 SO 4 in solution containing 14.7 g Na 2 SO 4 in 345g H 2 O. b) Ore containing 7.35 g of silver per ton of ore. What is the concentration of silver in ppm? X 100 = 4.09% 14.7 g g 7.35 g 1 g = 8.09 ppm 1.10 x ton

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Solutions Calculate the mole fraction of methanol, CH 3 OH, in the following solutions: a) 7.5 g CH 3 OH in 245 gH 2 O; b)55.7 g CH 3 OH in 164g CCl g g = 0.23 mole of methanol 1 mole 245 g 18.0 g = 13.6 mole of water 1 mole 0.23 mole of methanol mole solution = mole fraction

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Solutions Calculate the mole fraction of methanol, CH 3 OH, in the following solutions: a) 7.5 g CH 3 OH in 245 gH 2 O; b)55.7 g CH 3 OH in 164g CCl g g = 1.74 mole of methanol 1 mole 164 g g = 1.07 mole of CCl 4 1 mole 1.74 mole of methanol mole solution = mole fraction

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Solutions Calculate the molarity of the following aqueous solutions: a) 10.5g KCl in ml of solution; b) 30.7g LiClO 4 3H 2 O in 125ml o solution; c) 25.0 ml 1.50M HNO 3 diluted to 0.500L 10.5 g 74 g = 0.14 mole 1 mole L = M 30.7 g g + 3(18g) = mole 1 mole L = 1.53 M (0.025L)(1.50) = (x)(0.500L) = M

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Solutions Calculate the molality of each the following solutions; a) 13.0 g benzene, C 6 H 6, dissolved in 17.0 g of CCl 4 ; b) 4.75 g NaCl dissolved in L of water whose density is 1.00g/ml g 78 g = mole 1 mole kg = 9.79 m 4.75 g = mole 1 mole kg = m

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Solutions A sulfuric acid solution containing 571.6g of H 2 SO 4 per liter of solution has a density of g/cm 3, Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of H 2 SO g 1329 g X 100 = 43.0% g 98 g = 5.83 moles of H 2 SO 4 1 mole g 18 g = 42.1 moles of H 2 O 1 mole 5.83 moles of H 2 SO = moles = 0.122

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Solutions A sulfuric acid solution containing 571.6g of H 2 SO 4 per liter of solution has a density of g/cm 3, Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of H 2 SO mol kg = 7.70 m 5.83 moles of H 2 SO 4 1 L = 5.83M

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Solutions A solution containing 80.5 g of ascorbic acid, C 9 H 8 O 6, dissolved in 210 g of water has a density of 1.22 g/cm 3, Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of C 9 H 8 O g 210 g g X 100 = 27.7% 80.5 g g = moles of C 9 H 8 O 6 1 mole 210 g 18 g = 11.7 moles of H 2 O 1 mole moles of C 9 H 8 O = moles =

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Solutions A solution containing 80.5 g of ascorbic acid, C 9 H 8 O 6, dissolved in 210 g of water has a density of 1.22 g/cm 3, Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of C 9 H 8 O mol kg = 2.18 m moles of C 9 H 8 O L = 1.92 M g 1.22 g = 0.238L 1 mL 1000 mL 1 L

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Solutions In-Class- pHet Molarity and concentration

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Solutions HW: Ch. 13 sec. 5 reading notes

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