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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.

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Presentation on theme: "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."— Presentation transcript:

1 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!

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

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

4 Solutions Chapter 13 Properties of Solutions

5 Solutions Ways of Expressing Concentrations of Solutions

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

7 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

8 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%

9 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

10 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!

11 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.

12 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.

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

14 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

15 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

16 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

17 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

18 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

19 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

20 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

21 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

22 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

23 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

24 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 =

25 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

26 Solutions In-Class- pHet Molarity and concentration

27 Solutions HW: Ch. 13 sec. 5 reading notes


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