solutions entry task: nov 21 st block 2 question: what is the molality of a 200 g solution with 32...
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Solutions
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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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 13Properties of Solutions
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Solutions
Ways of Expressing
Concentrations of Solutions
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Mass Percentage
Mass % of A =mass of A in solutiontotal mass of solution
100
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Solutions
Parts per Million andParts per Billion
ppm =mass of A in solutiontotal mass of solution
106
Parts per Million (ppm)
Parts per Billion (ppb)
ppb =mass of A in solutiontotal mass of solution
109
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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.
1.50 51.5g 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
36.2100
100 X = 90500
X = 90.5g
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Solutions
moles of Atotal moles in solution
XA =
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 soluteL 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 solutekg 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 (C10H8) in 425 g of toluene (C7H8)?
35g
128.1 g = 0.28 mole
0.425 kg
1 mole= 0.670 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 = 0.0489 mole1 mole
3.62 % in water means 3.62 g in water3.62 g – 100 = 96.38 g of water
96.38g
18 g = 5.35 mole1 mole
0.0489 mole + 5.35 mole = 5.398 mole
0.0489 mole 5.398 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 = 0.0489 mole1 mole
3.62 % in water means 3.62 g in water3.62 g – 100 = 96.38 g of water
0.09638 kg= 0.507 m
0.0489 mole
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Solutions
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?
X 100 = 4.09%14.7 g14.7 + 345 g
7.35 g
1 g = 8.09 ppm
1.10 x 10-6 ton
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13.23 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
32.01 g = 0.23 mole of methanol
1 mole
245 g
18.0 g = 13.6 mole of water
1 mole
0.23 mole of methanol13.6 + 0.23 mole solution
= 0.017 mole fraction
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Solutions
13.23 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
32.01 g = 1.74 mole of methanol
1 mole
164 g
153.81 g = 1.07 mole of CCl4
1 mole
1.74 mole of methanol1.74 + 1.07 mole solution
= 0.620 mole fraction
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13.25 13.25 Calculate the molarity of the following aqueous solutions: a) 10.5g KCl in 250.0 ml of solution; b) 30.7g LiClO4•3H2O in 125ml o solution; c) 25.0 ml 1.50M HNO3 diluted to 0.500L10.5 g
74 g = 0.14 mole
1 mole
0.250 L = 0.560 M
30.7 g
106.397 g + 3(18g) = 0.191 mole
1 mole
0.125 L = 1.53 M
(0.025L)(1.50) = (x)(0.500L) = 0.075 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 0.250 L of water whose density is 1.00g/ml.
13.5 g
78 g = 0.167 mole
1 mole
0.017 kg = 9.79 m
4.75 g
58.45 = 0.081 mole
1 mole
0.250 kg = 0.325 m
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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 1.329 g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of H2SO4
571.6 g
1329 gX 100 = 43.0%
571.6 g
98 g = 5.83 moles of H2SO4
1 mole
757.4 g
18 g = 42.1 moles of H2O
1 mole
5.83 moles of H2SO4
5.83 + 42.1 = 47.93 moles = 0.122
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Solutions
13.29 13.29 A sulfuric acid solution containing 571.6g of H2SO4 per liter of solution has a density of 1.329 g/cm3 , Calculate a) the mass percentage; b) the mole fraction; c) molality; d) molarity of H2SO4
5.83 mol
0.7574 kg= 7.70 m
5.83 moles of H2SO4
1 L = 5.83M
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Solutions
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
210 g + 80.5gX 100 = 27.7%
80.5 g
176.1 g = 0.457 moles of C9H8O6
1 mole
210 g
18 g = 11.7 moles of H2O
1 mole
0.457 moles of C9H8O6
0.457+ 11.7 = 12.16 moles = 0.0376
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Solutions
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
0.210 kg= 2.18 m
0.457 moles of C9H8O6
0.238 L = 1.92 M
290.5 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