Chapter 8
Solutions
8.4
Percent Concentration
1
Concentration
The concentration of a solution
 is the amount of solute dissolved in a specific amount
of solution
concentration
=
amount of solute
amount of solution
2
Mass Percent Concentration
Mass percent (m/m) concentration is
 the percent by mass of solute in a solution
 calculated using the formula:
Mass =
mass of solute(g)
x 100%
percent mass of solute(g) + mass of solvent(g)
Mass percent = mass of solute(g) x 100%
mass of solution(g)
3
Mass of Solution
4
Calculating Mass Percent
The calculation of mass percent (m/m) requires the
 grams of solute (g of KCl) and
 grams of solution (g of KCl solution)
g of KCl
=
8.00 g
g of solvent (water)
=
42.00 g
g of KCl solution
=
50.00 g
8.00 g of KCl (solute)
x 100% = 16.0% (m/m)
50.00 g of KCl solution
5
Learning Check
A solution is prepared with 15.0 g of Na2CO3 and
235 g of H2O. What is the mass percent (m/m) of the
solution?
1) 15.0%(m/m) Na2CO3
2) 6.38%(m/m) Na2CO3
3) 6.00%(m/m) Na2CO3
6
Solution
3) 6.00% (m/m) Na2CO3
STEP 1 Given: 15.0 g of Na2CO3; 235 g of H2O
Need: mass percent (m/m) of Na2CO3 solution
STEP 2 Plan: mass percent(m/m) = g of solute x 100%
g of solution
Calculate mass of solution
= 15.0 g + 235 g
= 250 g of Na2CO3 solution
7
Solution (continued)
STEP 3 Write equalities and conversion factors:
15.0 g of Na2CO3 = 250 g of solution
15.0 g Na2CO3 and 250 g solution
250 g solution
15.0 g of Na2CO3
STEP 4 Set up problem:
mass
= 15.0 g Na2CO3 x 100 = 6.00%(m/m) Na2CO3
percent
250 g solution
(m/m)
8
Volume Percent
The volume percent (v/v) is
 percent by volume of solute (liquid) in a solution
 volume % (v/v) = volume of solute x 100%
volume of solution
 volume solute (mL) in 100 mL of solution.
volume % (v/v) =
mL of solute
100. mL of solution
9
Mass/Volume Percent
The mass/volume percent (m/v) is
 mass/volume % (m/v) = grams (g) of solute x 100%
volume(mL) of solution
 mass of solute(g) in 100 mL solution =
g of solute
100. mL of solution
10
Preparation of a Solution:
Mass/Volume Percent
11
Percent Conversion Factors
 Two conversion factors can be written for each
type of percent concentration.
12
Learning Check
Write two conversion factors for each solution:
A. 8.50% (m/m) NaOH
B. 5.75% (v/v) ethanol
C. 4.8% (m/v) HCl
13
Solution
A. 8.50% (m/m) NaOH
8.50 g NaOH
and
100 g solution
100 g solution
8.50 g NaOH
B. 5.75% (v/v) ethanol
5.75 mL ethanol and 100 mL solution
100 mL solution
5.75 mL ethanol
C. 4.8% (m/v) HCl
4.8 g HCl
and 100 mL HCl
100 mL solution
4.8 g HCl
14
Guide to Using Concentration to
Calculate Mass or Volume
15
Example of Using Percent
Concentration (m/m) Factors
How many grams of NaCl are needed to prepare
225 g of a 10.0% (m/m) NaCl solution?
STEP 1 Given: 225 g of solution; 10.0% (m/m) NaCl
Need: g of NaCl (solute)
STEP 2 Plan: g of solution
g of NaCl
16
Solution
STEP 3 Write equalities and conversion factors:
10.0 g of NaCl = 100 g of NaCl solution
10.0 g NaCl
and 100 g NaCl solution
100 g NaCl solution
10.0 g NaCl
STEP 4 Set up problem to cancel grams of solution:
225 g NaCl solution x 10.0 g NaCl
= 22.5 g of NaCl
100 g NaCl solution
17
Learning Check
How many grams of NaOH are needed to prepare
75.0 g of 14.0% (m/m) NaOH solution?
1) 10.5 g of NaOH
2) 75.0 g of NaOH
3) 536 g of NaOH
18
Solution
STEP 1 Given: 75.0 g of NaOH solution;
14.0% (m/m) NaOH solution
Need: g of NaOH (solute)
STEP 2 g of solution
g of NaOH
STEP 3 Write equalities and conversion factors:
14.0 g of NaOH = 100 of NaOH solution
14.0 g NaOH
and 100 g NaOH solution
100 g NaOH solution
14.0 g NaOH
STEP 4 Set up problem to cancel the grams of solution:
75.0 g NaOH solution x 14.0 g NaOH
= 10.5 g of NaOH
100 g NaOH solution
19
Learning Check
How many milliliters of a 5.75% (v/v) ethanol solution
can be prepared from 2.25 mL of ethanol?
1) 2.56 mL
2) 12.9 mL
3) 39.1 mL
20
Solution
STEP 1 Given: 2.25 mL of ethanol (solute)
5.75% (v/v) ethanol solution
Need: mL of ethanol solution
STEP 2 Plan: mL of ethanol
mL of solution
STEP 3 Write equalities and conversion factors:
100 mL of solution = 5.75 mL of ethanol
5.75 mL ethanol and 100 mL solution
100 mL solution
5.75 mL ethanol
STEP 4 Set up problem to cancel the mL of ethanol:
2.25 mL ethanol x 100 mL solution = 39.1 mL of solution
5.75. mL ethanol
21
Example Using Percent
Concentration (m/v) Factors
How many mL of a 4.20% (m/v) will contain 3.15 g KCl?
STEP 1 Given: 3.15 g of KCl (solute)
4.20% (m/v) KCl solution
Need: mL of KCl solution
STEP 2 Plan: g of KCl
mL of KCl solution
22
Using Percent Concentration(m/v)
Factors (continued)
STEP 3 Write equalities and conversion factors:
4.20 g KCl = 100 mL of KCl solution
4.20 g KCl
and 100 mL KCl solution
100 mL KCl solution
4.20 g KCl
STEP 4 Set up the problem:
3.15 g KCl x 100 mL KCl solution = 75.0 mL of KCl
4.20 g KCl
23
Learning Check
How many grams of NaOH are needed to prepare 125
mL of a 8.80% (m/v) NaOH solution?
1) 7.04 g of NaOH
2) 11.0 g of NaOH
3) 14.2 g of NaOH
24
Solution
2) 11.0 g of NaOH
STEP 1 Given: 125 mL of NaOH (solution)
8.80% (m/v) NaOH solution
Need: g of NaOH (solute)
STEP 2 Plan: mL of NaOH solution
g of NaOH
25
Solution (continued)
STEP 3 Write equality and conversion factors:
8.80 g of NaOH = 100 mL of NaOH solution
8.80 g NaOH
and 100 mL NaOH solution
100 mL NaOH solution
8.80 g NaOH
STEP 4 Set up the problem:
125 mL NaOH solution x
8.80 g NaOH
100 mL NaOH solution
= 11.0 g of NaOH
26
Chapter 8
Solutions
8.5
Molarity and Dilution
27
Molarity (M)
Molarity (M)
 is a concentration term for solutions
 gives the moles of solute in 1 L solution
molarity (M) = moles of solute
liter of solution
28
Preparing a 1.0 Molar NaCl
Solution
A 1.0 M NaCl solution is prepared
 by weighing out 58.5 g of NaCl (1.00 mole) and
 adding water to make 1.0 liter of solution
29
Guide to Calculating Molarity
30
Example of Calculating Molarity
What is the molarity of 0.500 L NaOH solution if it
contains 6.00 g of NaOH?
STEP 1 Given: 6.00 g of NaOH in 0.500 L solution
Need: molarity (mole/L) of NaOH solution
STEP 2 Plan:
g of NaOH
moles of NaOH
molarity
31
Example of Calculating Molarity
(continued)
STEP 3 Write equalitites and conversion factors:
1 mole of NaOH = 40.0 g of NaOH
1 mole NaOH
and 40.0 g NaOH
40.0 g NaOH
1 mole NaOH
STEP 4 Set up problem to calculate molarity:
6.00 g NaOH x 1 mole NaOH = 0.150 mole of NaOH
40.0 g NaOH
0.150 mole NaOH
= 0.300 mole NaOH
0.500 L NaOH solution
1 L NaOH solution
= 0.300 M NaOH
32
Learning Check
What is the molarity of a solution if 325 mL of the
solution contains 46.8 g of NaHCO3?
1) 0.557 M NaHCO3
2) 1.44 M NaHCO3
3) 1.71 M NaHCO3
33
Solution
3) 1.71 M
STEP 1 Given: 46.8 g of NaHCO3
325 mL (0.325 L) NaHCO3 solution
Need: molarity (mole/L) of NaHCO3 solution
STEP 2 Plan:
g of NaHCO3
moles of NaHCO3
molarity
STEP 3 Write equalities and conversion factors:
1 mole of NaHCO3 = 84.0 g of NaHCO3
1 mole NaHCO3 and 84.0 g NaHCO3
84.0 g NaHCO3
1 mole NaHCO3
34
Solution (continued)
STEP 4 Setup problem to calculate moles and molarity of
NaHCO3:
46.8 g NaHCO3 x 1 mole NaHCO3
84.0 g NaHCO3
= 0.557 mole of NaHCO3
0.557 mole NaHCO3
= 1.71 mole NaHCO3
0.325 L NaHCO3 solution
1 L NaHCO3 solution
= 1.71 M NaHCO3
35
Learning Check
What is the molarity of a KNO3 solution if 225 mL
of the solution contains 34.8 g of KNO3?
1) 0.344 M
2) 1.53 M
3) 15.5 M
36
Solution
2) 1.53 M KNO3
STEP 1 Given: 34.8 g of KNO3
225 mL (0.225 L) KNO3 solution
Need: molarity (mole/L) of KNO3 solution
STEP 2 Plan:
g of KNO3
moles of KNO3
molarity
STEP 3 Write equalities and conversion factors:
1 mole of KNO3 = 101.1 g of KNO3
1 mole KNO3 and 101.1 g KNO3
101.1 g KNO3
1 mole KNO3
37
Solution (continued)
STEP 4 Set up problem to calculate moles and molarity
of KNO3:
34.8 g KNO3 x 1 mole KNO3
101.1 g KNO3
= 0.344 mole of KNO3
0.344 mole KNO3
= 1.53 mole KNO3
0.225 L KNO3 solution
1 L KNO3 solution
= 1.53 M KNO3
In one setup:
34.8 g KNO3 x 1 mole KNO3 x
1
= 1.53 M
101.1 g KNO3 0.225 L
38
Molarity Conversion Factors
The units of molarity are used to write conversion
factors for calculations with solutions.
39
Example of Calculations Using
Molarity
How many grams of KCl are needed to prepare 125 mL
of a 0.720 M KCl solution?
STEP 1 Given: 125 mL (0.125 L) of 0.720 M KCl
Need: grams of KCl
STEP 2 Plan:
L of KCl
moles of KCl
g of KCl
40
Example of Calculations Using
Molarity (continued)
STEP 3 Write equalities and conversion factors:
1 mole of KCl = 74.6 g of KCl
1 mole KCl and 74.6 g KCl
74.6 g KCl
1 mole KCl
1 L of KCl = 0.720 mole of KCl
1L
and 0.720 mole KCl
0.720 mole KCl
1L
STEP 4 Set up problem to cancel mole KCl:
0.125 L x 0.720 mole KCl x 74.6 g KCl = 6.71 g of KCl
1L
1 mole KCl
41
Learning Check
How many grams of AlCl3 are needed to prepare
125 mL of a 0.150 M solution?
1) 20.0 g of AlCl3
2) 16.7 g of AlCl3
3) 2.50 g of AlCl3
42
Solution
3) 2.50 g of AlCl3
STEP 1 Given: 125 mL (0.125 L) of solution
0.150 M AlCl3 solution
Need: g of AlCl3
STEP 2 Plan:
L of solution
moles of AlCl3
g of AlCl3
STEP 3 Write equalities and conversion factors:
1 mole of AlCl3 = 133.5 g of AlCl3
1 mole AlCl3 and 133.5 g AlCl3
133.5 g AlCl3
1 mole AlCl3
43
Solution (continued)
STEP 3 (continued)
1 L of KCl = 0.150 mole of AlCl3
1L
and 0.150 mole AlCl3
0.150 mole AlCl3
1L
STEP 4 Set up problem:
0.125 L x 0.150 mole AlCl3 x 133.5 g = 2.50 g of AlCl3
1L
1 mole AlCl3
44
Learning Check
How many milliliters of 2.00 M HNO3 contain
24.0 g of HNO3?
1) 12.0 mL of 2.00 M HNO3
2) 83.3 mL of 2.00 M HNO3
3) 190 mL of 2.00 M HNO3
45
Solution
3) 190 mL of HNO3
STEP 1 Given: 24.0 g of HNO3
2.00 M HNO3 solution
Need: mL of HNO3 solution
STEP 2 Plan:
g of HNO3
moles of HNO3
L of HNO3 solution
STEP 3 Write equalities and conversion factors:
1 mole of = 63.0 g of HNO3
1 mole HNO3 and 63.0 g HNO3
63.0 g HNO3
1 mole HNO3
46
Solution (continued)
STEP 4 Set up problem to calculate volume, in mL, of
HNO3:
24.0 g HNO3 x 1 mole HNO3 x 1000 mL HNO3
63.0 g HNO3 2.00 moles HNO3
= 190 mL of a 2.00 M HNO3 solution
47
Dilution
In a dilution,
 water is added
 volume increases
 concentration of solute decreases
48
Initial and Diluted Solutions
In the initial and diluted solution,
 the moles of solute are the same
 the concentrations and volumes are related
by the following equations:
For percent concentration
C1V1
=
C2V2
Concentrated
solution
For molarity
M1V1
Concentrated
solution
Diluted
solution
=
M2V2
Diluted
solution
49
Guide to Calculating Dilution
Quantities
50
Example of Dilution Calculations
Using Percent Concentration
What volume of a 2.00% (m/v) HCl solution can be
prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution?
STEP 1 Prepare a table:
C1 = 14.0% (m/v)
V1 = 25.0 mL
C2 = 2.00% (m/v)
V2 = ?
STEP 2 Solve dilution expression for the unknown
C1V1 = C2V2
V2 = V1C1
C2
STEP 3 Set up the problem using known quantities:
V2 = V1C1 = (25.0 mL)(14.0%) = 175 mL
C2
2.00%
51
Learning Check
What is the percent (m/v) of a solution prepared by
diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
52
Solution
What is the percent (m/v) of a solution prepared
by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
STEP 1 Prepare a table:
C1 = 9.00% (m/v)
V1 = 10.0 mL
C2 = ?
V2 = 60.0 mL
STEP 2 Solve the dilution expression for the unknown:
C1V1 = C2V2
C2 = C1V1
V2
STEP 3 Set up the problem using known quantities:
C2 = C1V1 = (10.0 mL)(9.00%) = 1.50% (m/v)
V2
60.0 mL
53
Example of Dilution Calculations
Using Molarity
What is the molarity (M) of a solution prepared
by diluting 0.180 L of 0.600 M HNO3 to 0.540 L?
STEP 1 Prepare a table:
M1 = 0.600 M
V1 = 0.180 L
M2 = ?
V2 = 0.540 L
STEP 2 Solve the dilution expression for the unknown:
M1V1 = M2V2
STEP 3 Set up the problem using known quantities:
M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M
V2
0.540 L
54
Learning Check
What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a 0.300 M solution?
1) 27.0 mL of a 1.80 M KOH
2) 60.0 mL of a 1.80 M KOH
3) 90.0 mL of a 1.80 M KOH
55
Solution
What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a 0.300 M solution?
STEP 1 Prepare a table:
M1 = 1.80 M
V1 = 15.0 mL
M2 = 0.300 M
V2 = ?
STEP 2 Solve the dilution expression for the unknown:
M1V1 = M2V2
STEP 3 Set up the problem using known quantities:
V2 = M1V1 = (1.80 M)(15.0 mL) = 90.0 mL
M2
0.300 M
56
Molarity in Chemical Reactions
In a chemical reaction,
 the volume and molarity of a solution are used to
determine the moles of a reactant or product
molarity ( mole ) x volume (L) = moles
1L
 if molarity (mole/L) and moles are given, the volume
(L) can be determined
moles x
1 L = volume (L)
moles
57
Guide to Calculations Involving
Solutions in Chemical Reactions
58
Using Molarity of Reactants
How many mL of 3.00 M HCl are needed to completely
react with 4.85 g of CaCO3?
2HCl(aq) + CaCO3(s)
CaCl2(aq) + CO2(g) + H2O(l)
STEP 1 Given: 3.00 M HCl; 4.85 g of CaCO3
Need: volume of HCl in mL
STEP 2 Write a plan:
g of CaCO3
moles of CaCO3
moles of HCl
mL of HCl
STEP 3 Write equalities and conversion factors:
1 mole of CaCO3 = 100.1 g
1 mole CaCO3 and 100.1 g CaCO3
100.1 g CaCO3
1 mole CaCO3
59
Using Molarity of Reactants
(continued)
STEP 3 (continued)
1 mole of CaCO3 = 2 moles of HCl
1 mole of CaCO3 and 2 moles of HCl
2 moles of HCl
1 mole of CaCO3
1000 mL of HCl = 3.00 moles of HCl
1000 mL of HCl and 3.00 moles of HCl
3.00 moles of HCl
1000 mL of HCl
STEP 4 Set up problem to calculate mL of HCl:
4.85 g CaCO3 x 1 mole CaCO3 x 2 moles HCl x 1000 mL HCl
100.1 g CaCO3 1 mole CaCO3 3.00 moles HCl
= 32.3 mL of HCl required
60
Learning Check
If 22.8 mL of 0.100 M MgCl2 is needed to completely
react 15.0 mL of AgNO3 solution, what is the molarity of
the AgNO3 solution?
MgCl2(aq) + 2AgNO3(aq)
2AgCl(s) + Mg(NO3)2(aq)
1) 0.0760 M
2) 0.152 M
3) 0.304 M
61
Solution
3) 0.304 M AgNO3
STEP 1 Given: 22.8 mL (0.228 L) of 0.100 M MgCl2
Need: molarity of AgNO3
STEP 2 Write a plan to calculate molarity:
mL of MgCl2
moles of MgCl2
moles of AgNO3
molarity of AgNO3
STEP 3 Write equalities and conversion factors:
0.100 mole of MgCl2 = 1 L of MgCl2
0.100 mole MgCl2 and 1 L MgCl2
1 L MgCl2
0.100 mole MgCl2
62
Solution (continued)
STEP 3 (continued)
1 mole of MgCl2 = 2 moles AgNO3
2 moles AgNO3
1 mole MgCl2
and 1 mole MgCl2
2 moles AgNO3
STEP 4 Set up problem to calculate molarity of AgNO3:
0.0228 L x 0.100 mole MgCl2 x 2 moles AgNO3 x
1____
1L
1 mole MgCl2
0.0150 L
= 0.304 mole/L = 0.304 M AgNO3
63
Learning Check
How many liters of H2 gas at STP are produced when
125 mL of 6.00 M HCl reacts with sufficient Zn?
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
1) 4.20 L of H2
2) 8.40 L of H2
3) 16.8 L of H2
64
Solution
2) 8.40 L of H2 gas
STEP 1 Given: 125 mL (0.125 L) of 6.00 M HCl
Need: liters of H2 at STP
STEP 2 Write a plan to calculate liters of H2:
L of HCl
moles of HCl
moles of H2
liters of H2
STEP 3 Write equalities and conversion factors:
6.00 moles of HCl = 1 L of HCl
6.00 moles HCl and
1 L HCl
1 L HCl
6.00 moles HCl
22.4 L of H2 = 1 mole of H2 at STP
22.4 L H2 and
1 mole H2
1 mole H2
22.4 L H2
65
Solution (continued)
STEP 3 (continued)
2 moles HCl = 1 mole H2
2 moles HCl and
1 mole H2
1 mole H2
2 moles HCl
STEP 4 Set up problem to calculate liters of H2:
0.125 L x 6.00 moles HCl x 1 mole H2 x 22.4 L
1L
2 moles HCl 1 mole H2
= 8.40 L of H2 gas
66
Chapter 8
Solutions
8.6
Properties of Solutions
67
Solutions
Solutions
 contain small particles (ions or molecules)
 are transparent
 do not separate, even by filtration or through a
semipermeable membrane
 do not scatter light
68
Colligative Properties
Colligative properties
 are changes in the properties of a solvent when
solute particles are added
 depend on the number of solute particles in solution
 involve the lowering of freezing point
 involve an increase in the boiling point
69
Freezing Point Lowering and
Boiling Point Elevation
When 1 mole of solute particles is added to 1000 g of
water, the
 freezing point of water decreases by 1.86 °C (from 0 °C
to –1.86 °C)
 boiling point of water increases by 0.52 °C (from 100 °C
to 100.52 °C)
70
Moles of Particles
The number of moles of particles depends on the type of
solute:
 nonelectrolytes dissolve as molecules
1 mole of nonelectrolyte = 1 mole of particles in water
 strong electrolytes dissolve as ions
1 mole of electrolyte = 2 to 4 moles of particles in water
71
Effect of Solute Particles
72
Learning Check
A 1000-g sample of water contains 0.50 mole of Na3PO4.
What are the new freezing point (FP) and boiling point
(BP) of the solution?
1) FP 1.86 °C; BP 100.52 °C
2) FP 3.72 °C; BP 101.04 °C
3) FP 7.44 °C; BP 102.08 °C
73
Solution
A 1000-g sample of water contains 0.50 mole of Na3PO4.
What are the new freezing point (FP) and boiling point
(BP) of the solution?
2) FP 3.72 °C; BP 101.04 °C
Na3PO4
3Na+ +
PO430.50 mole
1.5 moles
0.5 mole
Total moles of solute = 2.0 moles
FP = 0 °C – 2(1.86 °C) = 0 °C – 3.72 °C = –3.72 °C
BP = 0 °C + 2(0.52 °C) = 0 °C + 1.04 °C = 101.04 °C
74
Osmosis
In osmosis,
 water (solvent) flows from
the lower solute
concentration into the
higher solute concentration
 the level of the solution with
the higher solute
concentration rises
 the concentrations of the
two solutions become equal
with time
75
Example of Osmosis
A semipermeable membrane separates a 4% starch
solution from a 10% starch solution. Starch is a colloid
and cannot pass through the membrane, but water can.
What happens?
4% starch
10% starch
H2O
semipermeable membrane
76
Example of Osmosis (continued)
 The 10% starch solution is diluted by the flow of water
out of the 4% solution, and its volume increases.
 The 4% solution loses water, and its volume decreases.
 Eventually, the water flow between the two becomes
equal.
7% starch
7% starch
H2O
77
Osmotic Pressure
Osmotic pressure
 is produced by the solute particles dissolved in a
solution
 is the pressure that prevents the flow of additional
water into the more concentrated solution
 increases as the number of dissolved particles in the
solution increases
78
Learning Check
A semipermeable membrane separates a 10% (m/v)
starch solution (A) from a 5% (m/v) starch solution (B).
If starch is a colloid, fill in the blanks in the statements
below.
1. Solution ____ has the greater osmotic pressure.
2. Water initially flows from ___ into ___.
3. The level of solution ____ will be lower.
79
Solution
A semipermeable membrane separates a 10% (m/v)
starch solution (A) from a 5% (m/v) starch solution (B).
If starch is a colloid, fill in the blanks in the statements
below.
1. Solution A has the greater osmotic pressure.
2. Water initially flows from B into A.
3. The level of solution B will be lower.
80
Osmotic Pressure of the Blood
Red blood cells
 have cell walls that are semipermeable membranes
 maintain an osmotic pressure that cannot change
without damage occuring
 must maintain an equal flow of water between the
red blood cell and its surrounding environment
81
Isotonic Solutions
An isotonic solution
 exerts the same osmotic
pressure as red blood cells
 is known as a “physiological
solution”
 of 5.0% (m/v) glucose or
0.9% (m/v) NaCl is used
medically because each has
a solute concentration equal
to the osmotic pressure equal
to red blood cells
82
Hypotonic Solutions
A hypotonic solution
 has a lower osmotic
pressure than red blood
cells (RBCs)
 has a lower concentration
than physiological solutions
 causes water to flow into
RBCs
 causes hemolysis (RBCs
swell and may burst)
83
Hypertonic Solutions
A hypertonic solution
 has a higher osmotic
pressure than RBCs
 has a higher concentration
than physiological solutions
 causes water to flow out of
RBCs
 causes crenation (RBCs
shrink in size)
84
Dialysis
In dialysis,
 solvent and small solute
particles pass through an
artificial membrane
 large particles are retained
inside
 waste particles such as
urea from blood are
removed using
hemodialysis (artificial
kidney)
85
Learning Check
Indicate if each of the following solutions is
1) isotonic 2) hypotonic 3) hypertonic
A. ____ 2% (m/v) NaCl solution
B. ____ 1% (m/v) glucose solution
C. ____ 0.5% (m/v) NaCl solution
D. ____ 5% (m/v) glucose solution
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Solution
Indicate if each of the following solutions is
1) isotonic
2) hypotonic
3) hypertonic
A. _3_ 2% (m/v) NaCl solution
B. _2_ 1% (m/v) glucose solution
C. _2_ 0.5% (m/v) NaCl solution
D. _1_ 5% (m/v) glucose solution
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Learning Check
When placed in each of the following, indicate if a red
blood cell will
1) not change 2) hemolyze 3) crenate
A. ____ 5% (m/v) glucose solution
B. ____ 1% (m/v) glucose solution
C. ____ 0.5% (m/v) NaCl solution
D. ____ 2% (m/v) NaCl solution
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Solution
When placed in each of the following, indicate if a red
blood cell will
1) not change 2) hemolyze 3) crenate
A. _1_ 5%(m/v) glucose solution
B. _2_ 1%(m/v) glucose solution
C. _2_ 0.5%(m/v) NaCl solution
D. _3_ 2%(m/v) NaCl solution
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Hemodialysis
 When the kidneys fail, an artificial kidney uses
hemodialysis to remove waste particles such as
urea from blood.
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Learning Check
Each of the following mixtures is placed in a dialyzing
bag and immersed in pure water. Which substance, if
any, will be found in the water outside the bag?
A. 10% (m/v) KCl solution
B. 5% (m/v) starch solution
C. 5% (m/v) NaCl and 5% (m/v) starch solutions
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Solution
Each of the following mixtures is placed in a dialyzing
bag and immersed in pure water. Which substance, if
any, will be found in the water outside the bag?
A. 10%(m/v) KCl solution: KCl ( K+, Cl−)
B. 5%(m/v) starch solution: None, starch is retained.
C. 5%(m/v) NaCl and 5%(m/v) starch solutions:
NaCl (Na+, Cl−) will be found in the water outside
the bag, but the starch is retained inside the bag.
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