3/11/2013 Standards: 6 (Solutions) Objectives: ● Students will be able to calculate problems involving Molarity (M)
and molality (m)DO NOW: 1. Calculate M (molarity) of a solution with 4.7 L of solution in which
there are 2.2 mol of NaClO4HOMEWORK: 1. Ch 15.2 PP's p. 466 17-20 and Ch 15 Assess. p. 485 76 (a, b, & c)
5 pts (Std 6) 2. Two PRINTS (see website) 3. HANDOUTS TODAY - Tutorial Ch 15.1 - What Are Solutions,
Ch 15.2 - Molarity - Molality - Colligative Properties Std 6===========================================================STAMPS: ● Journal 3/8 1. App A, p. 880 1-2 (1 pt) Std 6 2. Ch 15.2 p. 464-465 PP's 11-16 (4 pts) Std 6
Units of measure in Henry's Law
S (solubility)g/L g/mL mol/L mol/mL (or any unit of volume besides L or mL)
P (pressure)atm, mmHg, Torr, kPa, Pa, psi
(any unit of pressure is OK)
http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html
SOLVATION ANIMATION
1. agitation
2. increase surface area
3. temperature
Demo - see videomade in classyesterday
Units of measure in Henry's Law
S (solubility)g/L g/mL mol/L mol/mL (or any unit of volume besides L or mL)
P (pressure)atm, mmHg, Torr, kPa, Pa, psi
(any unit of pressure is OK)
M
Molarity
Molarity
Molarity
Molarity
Molarity
Molarity
Molarity
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Molarity
Molarity
Molarity
Molarity
Molarity
Molarity
Molarity
Molarity
Molarity
MolarityMolarity
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Mola
rity
Mola
rity
M =
mol L
molarity
molarity
solute
solution
Calculate the molarity of a solution having 15.0 g NaCl in 2.80 L of water (volume of solution is still 2.80 L).
Molar mass Cl = 35.5Molar mass Na = 23.0
In your answer, use 3 sig. figs and do not use any units of measure. NUMBER ONLY
1
m = (milli-) = 1
1000
How to derive equality with prefixes
What is 1 mL equal to in terms of L ?
1
1000mL = L x 10001000 x
1
1000mL = L x 10001000 x
1000 mL = 1 L
1
1000mL = L
k = (kilo-) = 1000
1 kg = 1000 g
How to derive equality with prefixes
What is 1 kg equal to in terms of g ?
Summary
Is it 1000 kg = 1 g or 1000 g = 1 kg ?
prefix letter x its mass unit (or volume)
1 kg = 1000 g
1
10001 mL = L
1 mg = g1
1000
1 kL = 1000 L
prefix value x its mass unit
Make --
=
PREFIXES
1 kg = 1000 g 1 mL = L1
1000
Everything in blue will be a fraction.
Everything in green will be a fraction.
http://mrwiggersci.com/chem/Tutorials/conv-units-all-prefixes.htm
Big Bigger
M
Molarity
2 What is the value of S if 25.0 g of gas is dissolved in 1.00 L of water?
Number only3 sig figs onlyno units of
measure
3 What is the solubility of a gas if 25.0 g of the gas will dissolve in 1.00 L of water.
Number only3 sig figs onlyno units of
measure
p. 463
possible solutes: a. salt (NaCl)b. sugar (sucrose)c. MgCl2 d. C (smaller portion of steel)e. paint pigments
possible solvents:f. waterg. iron (Fe major portion of steel)h. laquer thinner
`
3/13/2013 Standards: 6 (Solutions) (same as yesterday) Objectives: ● Practice Calculations of molarity and mass percentDO NOW: 1. Ch 15 Assess - p. 484 69-71 and 77 (all) & 78 (all) will be stamped as
a 5 pt HW stamp for Th. (5 pts) Std 6HOMEWORK:
None===========================================================STAMPS: ● Journal 3/11 1. Ch 15.2 PP's p. 466 17-20 and Ch 15 Assess. p. 485 76 (a, b, & c)
5 pts (Std 6)
AT YOUR DESK: 1. Two PRINTS (see website) (Bigger - Units of Measure, and
Mass% & Molarity Lab Sheet) (1 pts) Std 6
For Quiz 3.3 number correct x 2.4 = points earned
60 = total possible
Summary
Is it 1000 kg = 1 g or 1000 g = 1 kg ?
prefix letter x its mass unit (or volume)
1 kg = 1000 g1
10001 mL = L
1 mg = g1
10001 kL = 1000 L
prefix value x its mass unit =
Make --
PREFIXES
answers:
69. 3.9%70. 33.6%71. 25%77. a. 0.02 M (mol/L)
b. 0.27 M (mol/L) c. 0.0847 M (mol/L)78. a. 0.02 M mol
b. 0.00625 molc. 0.648 mol
3/13/2013 Standards: 6 (Solutions) (same as yesterday) Objectives: ● Practice Calculations of molarity and mass percentDO NOW: 1. Ch 15 Assess - p. 484 69-71 and 77 (all) & 78 (all) will be stamped as
a 5 pt HW stamp for Th. (5 pts) Std 6HOMEWORK:
None===========================================================STAMPS: ● Journal 3/11 1. Ch 15.2 PP's p. 466 17-20 and Ch 15 Assess. p. 485 76 (a, b, & c)
5 pts (Std 6)
AT YOUR DESK: 1. Two PRINTS (see website) (Bigger - Units of Measure, and
Mass% & Molarity Lab Sheet) (1 pts) Std 6
3/14/2013 Standards: 6 (Solutions) Objectives: ● Students will be able to calculate boiling point INcreases, and freezing
point DEcreases. ● Demonstrate conducive solutions. Students will be able to recognize
compounds that ionize (make 2 or more particles, salts, acids & bases) as different from those who don't ionize (sugar).
DO NOW: 1. Name 3 kinds of substances ionize in water, and give one example
of each?HOMEWORK: 1. Ch 15 Assess p. 485 79 (a,b,c), Ch 15.4 p. 475 33, 34,
Ch 15 Assess p. 485 87 (a,b,c) (5 pts ) Std 6 (total problems = 8)===========================================================STAMPS: ● Journal 3/13 (for periods 1,3,5) or 3/14 (for periods 2,6) 1. Ch 15 Assess - p. 484 69-71 and 77 (all) & 78 (all) will be stamped as
a 5 pt HW stamp for Th. (5 pts) Std 6
period 2 was in assembly, which I photographed. It was on driving safely.
Answers: 78 a = 0.0320 mol; b = 0.00625 mol ; c = 0.648 mol 79 a = 0.0859 g; b = 34.6 g; c = 5.91 g 87 a = fp = 10.2 deg C, bp = 103.8 deg C
b = fp = -115.8 deg C, bp = 79.1 deg C c = fp = -1.0 deg C, bp = 83.3 deg C
m = molality =
mole solute
kg solvent
M = molarity =
mole solute
L solution
Boiling & freezing points of water can be changed by dissolving:
BP
100°C
FP 0°C
ionic compounds (salts, acids, bases) non-ionic compounds (like sugar that don't form ions)
p. 474p. 472
More in the changes shown in charts
BIGGER
particles
changes
BP
FP
100°C
0°C
Kb °C/m
Kf °C/m
p. 474p. 472
More in the changes shown in charts
BIGGER
particles
changes
molaltiymol (solute)
kg (kg solvent)=
molaltiymol (solute)
kg (kg solvent)=
molaltiymol (solute)
kg (kg solvent)=
p. 474p. 472
More in the changes shown in charts
BIGGER
particles
changes
3/15/2013 Standards: 6 (Solutions) Objectives:● Polish skills learned in this chapterDO NOW: 1. Due end of period (8 points):
Directions: All problems require calculations. So, show given, find, and calculations. Each problem is worth one point.
p. 484 67 1 x 101 atm p. 487 3 a71 25% 4
cp. 485 78 (b) 0.00625 mol 5 b
89 fp = -6.46 deg C bp = 101.8 deg
HOMEWORK: Enjoy your week off and Think about Easter===========================================================STAMPED: ● Journal 3/14 1. Ch 15 Assess p. 485 79 (a,b,c), Ch 15.4 p. 475 33, 34,
Ch 15 Assess p. 485 87 (a,b,c) (5 pts ) Std 6
3/14/2012 Standards: 6 (Solutions) Objectives:● Complete data gathering and calculations of your mass% & molarity and turn in to basket before end of the periodDO NOW: READ LAB SILENTLY AND try to figure out your procedure.HOMEWORK: Enjoy your week off and Think about Easter===========================================================TURNED IN TO BASKET: ● Journal NONE TODAY
(for this lab, assume vol of solution =
20.00mL)
1.000 g = 1.000 mL (only for water)
mass % =
g soluteg solution x 100
M = (molarity)
mol (solute) L (solution)
When you come to me to get your salt, you will need to show me these masses:
1. filter paper2. graduated cylinder3. 50 mL beaker
All masses must have 3 decimals Examples: 0.l56 g
0.033 g3.427 g
Gp #
g NaCl g H20 g soln Mass % M
1
2
3
4
5
6
7
8
9
Mass % & Molarity LAB - RESULTS BY GROUPPer 1
Gp #
gNaCl g H20 g soln Mass % M
1
2
3
4
5
6
7
8
9
Mass % & Molarity LAB - RESULTS BY GROUPPer 2
Gp #
gNaCl g H20 g soln Mass % M
1
2
3
4
5
6
7
8
9
Mass % & Molarity LAB - RESULTS BY GROUPPer 3
Mass % & Molarity LAB - RESULTS BY GROUPPer 5
Gp #
gNaCl g H20 g soln Mass % M
1
2
3
4
5
6
7
8
9
Gp #
gNaCl g H20 g soln Mass % M
1
2
3
4
5
6
7
8
9
Mass % & Molarity LAB - RESULTS BY GROUPPer 6
Gp #
gNaCl g H20 g soln Mass % M
1
2
3
4
5
6
7
8
9
Mass % & Molarity LAB - RESULTS BY GROUPPer
Work problems p. 475 33-36