Transcript Chapter 15 - Solutions

Solution Concentration
• Concentration – how much solute
dissolved in amount of solvent
what is difference between concentrated
and diluted?
Concentration
• 3 different units of concentration
a) percent by mass
b) molarity (M)
c) molality (m)
Percent by mass
• Formula
mass of solute
x 100
mass of solution
OR
mass solute
x 100
mass solute  mass solvent
Percent by mass
• Ex prob: If 3.6 g NaCl is dissolved in 100 g
H2O, what is the percent by mass?
What is the solute?
NaCl
What is the solvent?
H2O
Percent by mass
• Mass of solute (NaCl) =
3.6 g
• Mass solvent (H2O) =
100 g
• Mass solution =
3.6 + 100 = 103.6 g
Percent by mass
Percent by mass
=
=
3.6 g x 100
103. 6 g
3.5 % NaCl
Molarity
• Defn - # of moles per liter of solution
• Formula
• unit
mol solute
L solution
mol = M (capital M)
L
Molarity Ex prob #1
• A solution has a volume of 250 mL and
has 0.70 mol NaCl. What is the molarity?
0.70 mol
0.250 L
=
2.8 mol/L
or
2.8 M
Molarity ex prob #2
• What is the molarity of a solution made of
47.3 g NaOH in 500 mL water?
step 1: convert grams to moles
47.3 g NaOH 1 mol NaOH
40 g NaOH
= 1.1825 mol NaOH
Molarity ex prob #2
Step 2: divide moles by volume (L)
1.1825 mol
= 2.37 mol/L NaOH
0.500 L
or
2.37 M NaOH
Molarity ex prob #3
• How many moles of solute are present in
1.5 L of 2.4 M NaCl? How many grams?
# moles = volume x molarity
1.5 L
x
2.4 mol NaCl
L
= 3.6 mol NaCl
Molarity ex prob #3
• moles to grams
3.6 mol NaCl
58.5 g NaCl
1 mol NaCl
= 210.6 g
NaCl
Molality
• Defn - # moles of solute in one kg solvent
• Formula
mol solute
kg solvent
• Units
mol = m (lower case m)
kg
Molality ex problem
• What is the molality of a solution with 8.4
g NaCl in 255 g of water?
Step 1: convert grams to moles
8.4 g NaCl
1 mol NaCl
58.5 g NaCl
=
0.14 mol NaCl
Molality ex problem
Step 2: divide by mass (kg)
0.14 mol NaCl
0.255 kg
0.55
mol/kg
NaCl
=
or
0.55 m NaCl
Diluting Solutions
• Defn – add more solvent to original
solution
• Formula
M1V1 = M2V2
M1 is more concentrated than M2
Diluting Solutions
• What volume of a 2.0 M stock solution is
needed to make 0.50 L of a 0.300 M
solution?
M1= 2.0 M
V1= ?
M2= 0.300 M
V2= 0.50 L
(2.0 M) V1 = (0.300 M)(0.50 L)
V1 = 0.075 L
Diluting Solutions
• If you dilute 20.0 mL of a 3.5 M solution
to make 100.0 mL of solution, what is the
molarity of the dilute solution?
M1= 3.5 M
V1= 20.0 mL
M2= ?
V2= 100 mL
(3.5 M)(20.0 mL) = M2 (100.0 mL)
M2 = 0.7 M