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**Entry Task: Nov 21st 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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**Agenda: Discuss Ch. 13 sec. 4 pHet Concentration and Molarity Lab**

HW: Solutions ws

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

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**Chapter 13 Properties of Solutions**

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**Ways of Expressing Concentrations of Solutions**

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**Mass Percentage mass of A in solution Mass % of A =**

total mass of solution Mass % of A = 100

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**Parts per Million and Parts per Billion**

Parts per Million (ppm) mass of A in solution total mass of solution ppm = 106 Parts per Billion (ppb) mass of A in solution total mass of solution ppb = 109

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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. 1.50 51.5g solution X = 2.91%

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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 36.2 100 100 X = 90500 X = 90.5g

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**total moles in solution**

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

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**Molarity (M) mol of solute M = L of solution**

You will recall this concentration measure from Chapter 4. Because volume is temperature dependent, molarity can change with temperature.

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**Molality (m) mol of solute m = kg of solvent**

Because both moles and mass do not change with temperature, molality (unlike molarity) is not temperature dependent.

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**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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13.4 problem What is the molality of a solution made by dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8)? 35g 1 mole = 0.28 mole = kg 128.1 g 0.425 kg

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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. 3.62 % in water means 3.62 g in water 3.62 g – 100 = g of water 3.62g 1 mole = mole mole 5.398 mole 74 g 96.38g 1 mole 9.06 x 10-3 = 5.35 mole 18 g mole mole = mole

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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. 3.62 % in water means 3.62 g in water 3.62 g – 100 = g of water 3.62g 1 mole = mole 74 g mole = m kg

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13.21 13.21 a) Calculate the mass % of Na2SO4 in solution containing 14.7 g Na2SO4 in 345g H2O. b) Ore containing 7.35 g of silver per ton of ore. What is the concentration of silver in ppm? 14.7 g X 100 = 4.09% g 7.35 g 1.10 x 10-6 ton = 8.09 ppm 1 g

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13.23 Calculate the mole fraction of methanol, CH3OH, in the following solutions: a) 7.5 g CH3OH in 245 gH2O; b)55.7 g CH3OH in 164g CCl4. 7.5 g 1 mole = 0.23 mole of methanol 32.01 g 245 g 1 mole = 13.6 mole of water 18.0 g 0.23 mole of methanol = mole fraction mole solution

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13.23 Calculate the mole fraction of methanol, CH3OH, in the following solutions: a) 7.5 g CH3OH in 245 gH2O; b)55.7 g CH3OH in 164g CCl4. 55.7 g 1 mole = 1.74 mole of methanol 32.01 g 164 g 1 mole = 1.07 mole of CCl4 g 1.74 mole of methanol = mole fraction mole solution

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13.25 Calculate the molarity of the following aqueous solutions: a) 10.5g KCl in ml of solution; b) 30.7g LiClO4•3H2O in 125ml o solution; c) 25.0 ml 1.50M HNO3 diluted to 0.500L 10.5 g 1 mole = 0.14 mole = M 74 g 0.250 L 30.7 g 1 mole = mole = 1.53 M g + 3(18g) 0.125 L (0.025L) (1.50) = (x)(0.500L) = M

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13.27 13.27 Calculate the molality of each the following solutions; a) 13.0 g benzene, C6H6, dissolved in 17.0 g of CCl4; b) 4.75 g NaCl dissolved in L of water whose density is 1.00g/ml. 13.5 g 1 mole = mole = 9.79 m 78 g 0.017 kg 4.75 g 1 mole = mole = m 58.45 0.250 kg

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13.29 13.29 A sulfuric acid solution containing 571.6g of H2SO4 per liter of solution has a density of g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of H2SO4 571.6 g X = % 1329 g 571.6 g 1 mole = 5.83 moles of H2SO4 98 g 757.4 g 1 mole = 42.1 moles of H2O 18 g 5.83 moles of H2SO4 = 0.122 = moles

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13.29 13.29 A sulfuric acid solution containing 571.6g of H2SO4 per liter of solution has a density of g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of H2SO4 5.83 mol = m kg 5.83 moles of H2SO4 = 5.83M 1 L

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13.30 13.30 A solution containing 80.5 g of ascorbic acid, C9H8O6, dissolved in 210 g of water has a density of 1.22 g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of C9H8O6 80.5g X = % 210 g g 80.5 g 1 mole = moles of C9H8O6 176.1 g 210 g 1 mole = 11.7 moles of H2O 18 g 0.457 moles of C9H8O6 = = moles

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13.30 13.30 A solution containing 80.5 g of ascorbic acid, C9H8O6, dissolved in 210 g of water has a density of 1.22 g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of C9H8O6 0.457 mol = m 0.210 kg 290.5 g 1 mL 1 L = 0.238L 1.22 g 1000 mL 0.457 moles of C9H8O6 = 1.92 M 0.238 L

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**In-Class- pHet Molarity and concentration**

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**HW: Ch. 13 sec. 5 reading notes**

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