How to mix a Standard Solution
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480
Process of Making a Standard
Solution from Liquids
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 483
Markings on Glassware
Beaker
500 mL + 5%
Range = 500 mL + 25 mL
475 – 525 mL
Graduated Cylinder
Volumetric Flask 500 mL + 0.2 mL
Range = 499.8 – 500.2 mL
TC 20oC “to contain at a temperature of 20 oC”
TD “to deliver”
22
s
T
“time in seconds”
Concentration…a measure of solute-to-solvent ratio
concentrated
“lots of solute”
dilute
“not much solute”
“watery”
Add water to dilute a solution; boil water off to concentrate it.
A. mass % = mass of solute
mass of sol’n
B. parts per million (ppm)  also, ppb and ppt
-- commonly used for minerals or
contaminants in water supplies
C. molarity (M) = moles of solute
L of sol’n
mol
-- used most often in this class
M 
D.
mol
L
molality (m) = moles of solute
kg of solvent
M
L
WRONG
7.85 kg KCl are dissolved in 2.38 L of solution.
Find molality.
m
kg solute 7.85 kg


L sol' n
2.38 L
3.30 m KCl
24.8 g table sugar (i.e., sucrose, C12H22O11) are mixed into 450 g water.
Find molality.
m
kg solute 0.0248 kg


L sol' n
0.450 L
0.055 m C12H22 O11
What mass of CaF2 must be added to 1,000 L of water
so that fluoride atoms are present at a conc. of 1.5 ppm?
23
 1000 mL  1 g  1 mol  6.02 x 10 m' cule 
X m' cule H2O  1000 L 

= 3.34 x 1028 m’cules H2O



1 mol
 1L  1 mL  18 g 

1.5 atom F
X atoms F

1,000,000 m' cule H2O 3.34 x 1028 m' cule H2O
 1 m' c CaF2 
22
X  5.01 x 1022 at. F  times 
  2.505 x 10 m' c CaF2
 2 at. F 
1 mol

  78.1 g 
X g CaF2  2.505 x 1022 m' c 

  3.25 g CaF2
23
 6.02 x 10 m' c   1 mol 
mol
M
L
How many moles solute are required to make 1.35 L of 2.50 M solution?
mol = M L = 2.50 M (1.35 L) =
3.38 mol
A. What mass sodium hydroxide is this?
 40.0 g 
X g NaOH  3.38 mol 
  135 g NaOH
 1 mol 
B. What mass magnesium phosphate is this?
 262.9 g 
X g Mg 3 (PO 4 )2  3.38 mol 
  889 g Mg 3 (PO 4 )2
1
mol


Find molarity if 58.6 g barium hydroxide are in 5.65 L solution.
 1mol Ba(OH) 2 

58.6 g Ba(OH) 2 
 171.3 g Ba(OH) 2  
XM
5.65 L
0.061 M Ba(OH) 2
You have 10.8 g potassium nitrate.
How many mL of solution will make this a 0.14 M solution?
 1 mol 
10.8 g KNO 3 

101.1
g

  0.763 L  1000 mL  
XL 


0.140 M
1
L


76.3 mL
convert to mL
Concentration
 The
amount of solute in a solution.
 Describing
Concentration
• % by mass - medicated creams
• % by volume - rubbing alcohol
• ppm, ppb - water contaminants
• molarity - used by chemists
• molality - used by chemists
Molality
moles of solute
molality (m) 
kg of solvent
0.25 mol
0.25m 
1 kg
mass of solvent only
1 kg water = 1 L water
Molarity of Solutions
Molarity of Solutions
Molarity of Solutions
http://www.unit5.org/christjs/tempT
27dFields-Jeff/Solutions1.htm