CONCENTRATION OF
SOLUTIONS
GENERAL CHEMISTRY 2
QUARTER 3 – LESSON 7
Use different ways of expressing concentration of solutions:
• percent by mass,
• mole fraction,
• molarity,
• molality,
• percent by volume, and
• ppm
PRETEST
ITEM NO. 1
• There are ways on how to A. The moles of solute present in the
express the concentration of solution.
a solution, one of which is B. The volume of solution (in liters)
mole fraction, which of the containing the solute.
following are needed to C. The volume of solution (in liters)
containing the solvent.
calculate the mole fraction?
D. The number of moles of each
component present in the solution.
D. The number of moles of each component
present in the solution.
3
PRETEST
ITEM NO. 2 A. Percent by Mass
•The following are units of
B. Mole Fraction
concentration, except:
C. Parts per Million
D. Mole
D. Mole
4
PRETEST
ITEM NO. 3
•What is the unit of A. mol/kg
molarity? B. mol
C. mol/L
D. none of the above
C. mol/L
5
PRETEST
ITEM NO. 4 A. Sum of moles of all
•The mass of solution is components
equal to: B. Mass of solute + mass of
solvent
C. Volume of solute + volume of
solvent
D. All of the above
B. Mass of solute + mass of solvent
6
PRETEST
ITEM NO. 5
•What is the molarity of a A. 0.182
solution with 5.0 g CaCl2 B. 0.00182
in 250 mL solution? C. 20
D. 0.0022
A. 0.182
7
What is a Solution?
A solution is a homogeneous mixture of
one or more solutes dissolved in a solvent.
SOLUTE + SOLVENT = SOLUTION
SOLUTE – substance dissolving
SOLVENT – substance the solute dissolves in 8
SOLUTIONS
A solute can be a gas, a
liquid or a solid.
Solvents can also be
gases, liquids or solids
HOW DO WE EXPRESS THE
CONCENTRATION OF A
SOLUTION? 9
Concentration of a Solution
MOLARITY
is a useful way to describe solution
concentrations for reactions that are
carried out in solution.
MOLE FRACTIONS
are used not only to describe gas
concentrations
but also, to determine the vapor pressures
of mixtures of similar liquids.
10
Concentration
of a Solution
• is a macroscopic property,
• represents the amount of solute
dissolved in a unit amount of
solvent or of solution, and
• can be expressed in a variety of
ways (qualitatively and
quantitatively). 11
Qualitative Expressions of Concentration
A solution can be qualitatively
described as :
DILUTE → CONCENTRATED
DILUTE: a solution that contains a small proportion of solute
relative to solvent
CONCENTRATED: a solution that contains a large proportion 12
of solute relative to solvent.
Semi-Quantitative
Expressions of
Concentration
A solution can be semi-
quantitatively described as:
• UNSATURATED: a solution in which more solute will dissolve,
• SATURATED: a solution in which no more solute will dissolve,
• SUPERSATURATED: a solution in which solute crystallizes at an
elevated temperature.
13
QUANTITATIVE
EXPRESSIONS OF
CONCENTRATION
There are several ways to express
the relative amounts of solute and
solvent in a solution.
I. II. III.
Percent by Mass Mole Fraction Molarity
IV. V. VI.
Molality Percent by Volume Parts per million
14
I. PERCENT BY MASS
This expresses the mass of solute per 100g of solution.
Mass of solution is equal to the mass of solute plus the
mass of solvent.
*+,, ./ ,.01&"
30g !"#$"%& () *+,, =
*+,, ./ ,.01&2.%
×455
*+,, ./ ,.01&2.% = *+,, ./ ,.01&" + *+,, ./ ,.07"%&
70g 30 % by mass sugar
30 g sugar per
100 g solution
15
Given:
mass of solute = 30 g NaCl ; mass of solvent = 105 g water
Required: % by mass NaCl
SAMPLE Solution:
PROBLEM 1 %&'' 06 '0-12304 = 30 7 +&,- + 105 7 06 9&2:;
%&'' 06 '0-12304 = 135 7
A sample of 30
grams (g) of sodium
chloride (NaCl) is 30 " +&,-
dissolved in 105 % #$ %&'' = ×100
135 "'0-12304
grams of water.
What is the percent
by mass of NaCl in % %& '()) = ++. ++ %
the solution?
16
II. MOLE FRACTION
Ratio of the number of mole of one component to the total
number of moles in a solution.
It is represented by a capital letter X.
!"#$% %"#'($
<0-: =;&>2304 ? @0-12: A =
!"#$% %"#'()"*
%0-:' '0-B:42
<0-: =;&>2304 ? @0-B:42 C =
%0-:' '0-12304
41%#:; 06 %0-:' '0-12304 = 41%#:; 06 %0-:' '0-12: + 41%#:; 06 %0-:' '0-B:42
<0-: =;&>2304 06 '0-12: + <0-: =;&>2304 06 '0-B:42 = 1
?+ + ?, = 1
17
SAMPLE PROBLEM 2
A solution is made by dissolving 1.25 g of Na2SO4 in 65.0 g water. Calculate
the mole fraction of the solute and the solvent.
Given: mass of solute=1.25 g Na2SO4 ; mass of solvent = 65.0 g water
Required: mole fraction
Solution:
)#** ,- "#2 %&4 1.25 3 "#2 %&4
!"#2 %&4 = = 3 = 0.0088 ),.
),.#/ )#** ,- "#2 %&4 142
),. "#2 %&4
&'(( *+ "2 $ 65.0 2 "2 $
!"2 $ = = 2 = 3.61 &*,
&*,'- &'(( *+ "2 $ 18
&*, "2 $
! "#$%& = !(#&)$* + !(#&,*!$
!(#&)$- #! = !.%2014 + !321
!(#&)$-#! = 0.0018 + 3.61 = 3.61 ;#&
18
SAMPLE PROBLEM 2
A solution is made by dissolving 1.25 g of Na2SO4 in 65.0 g water. Calculate
the mole fraction of the solute and the solvent.
! "#$%& =g!Na
Given: mass of solute=1.25 2SO4+
(#&)$* !(#&,*!$
; mass of solvent = 65.0 g water
Required: mole fraction
!(#&)$- #! = !.%2014 + !321
Solution:
!(#&)$-#! = 0.0018 + 3.61 = 3.61 ;#&
,1"#2*$ "' 1"#3$,*
!"#$ &'()*+", (.) =
,1"#2*+",
0.0088 7"#
.1"#2*$ = = ;. ;;<=
3.62 7"#
3.61 7"#
.1"#3$,* = = ;. ??@
3.62 7"#
Check your answer: !"#$%&' + !"#$*'+& = 1
0.0024 + 0.997 = 0.999 ≈ 1
19
III. MOLARITY (M)
Ratio of the number of moles of solute per liter of solution.
The unit of molarity is mol/L
%0-:' 06 '0-12:
<0-&;32$(<) =
-32:; 06 '0-12304
A 3M (3molar) NaOH means that
there are three moles NaOH in 3 mol
one liter of solution or three moles NaOH
NaOH/liter solution.
20
Given:
mole of solute = 0.2 mol NaOH ;
liter of solution (volume) = 100 mL
SAMPLE
PROBLEM 3
Required: molarity, M
Determine the
molarity of the Solution:
solution with
0.2 mol of 0.2 ."# 1000.0
NaOH in 100 !"#$%&'( = × = 2345/7
mL of solution. 100.0 10
21
Given:
Molarity, M = 0.5 M H2SO4 ;
liter of solution (volume) = 100 mL
SAMPLE
PROBLEM 4 Required: moles of solute, n
Calculate the
Solution:
number of
moles of solute *.0", (%) ./ ,.01&" = 02&"#, ./ ,.01&2.% ×:.0+#2&)(:)
in each of
100mL of 0.5 M 1, 0.5 +#$
of H2SO4. !"#$ %&' = 100+, × × = 1. 12 345
1000+, ,
22
IV. MOLALITY (m)
The number of moles of solute per kilogram of solvent.
The unit of molality is mol/kg.
%0-:' 06 '0-12:
<0-&-32$(%) =
%&'' 06 '0-B:42 (F7)
A 3m solution also means G HIJ KIJLMN
3 moles solute/kilogram GH =
OP
solvent
23
Given:
molality, m = 4.0 mol Mg(OH)2 ;
kg solvent = 550 g water
SAMPLE
PROBLEM 5 Required: moles of solute, m
How many Solution:
moles of solute
is present in HIJNK (Q) KIJLMN = HRKK IS KIJTNQM U VIJRJWMX(H)
4.0m Mg(OH)2
in 550 g water? 12+ 44#$ 5+(/- )2
!"#$%&' = 550+ -2 / × × = 8. 8 :;<
1000+ 2+
24
SAMPLE PROBLEM 6
The density of a 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g/mL.
What is the molality of the solution? The molar mass of methanol is 32.04 g.
Given:
Density (CH3OH) = 0.976 g/mL ; Molarity = 2.45 M ; m, mass (CH3OH) =32.04 g
Required: molality, m
Solution: assume 1L soln, so the number of moles (CH3OH) is 2.45 mol
1000 81 2345 0.976 =
masstotal (CH3 OH)soln = 11 2345 × × = 976 =
11 2345 181 2345
!"## %& '2 ) = !"## %& #%+, − !"## %& #%+./0
32.04 4 9'3 )'
!"## %& '2 ) = 976 4 − 52.45 !%+ 9'3 )' × > = 898 4
1 !%+ 9'3 )'
25
SAMPLE PROBLEM 6
The density of a 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g/mL.
What is the molality of the solution? The molar mass of methanol is 32.04
g/mol.
Given:
Density (CH3OH) = 0.976 g/mL ; Molarity = 2.45 M ;
molar mass (CH3OH) =32.04 g/mol
Required: molality, m
Solution: assume 1L soln, so the number of moles (CH3OH) is 2.45 mol
!"## %& '2 ) = !"## %& #%+, − !"## %& #%+./0
32.04 4 9'3 )'
!"## %& '2 ) = 976 4 − 52.45 !%+ 9'3 )' × > = 898 4
1 !%+ 9'3 )'
2.45 !"# ./3 1/ :;<
!"#$#%&' = = 7. 89
0.898 56 /2 1 =>
26
V. PERCENT BY VOLUME
It is used to express the concentration of a solution when
the volume of a solute and the volume of a solution are
given, multiplied by 100.
B0-1%: 06 '0-12:
Y:;>:42 #$ B0-1%: = ×100
B0-1%: 06 '0-12304
B0-1%: 06 '0-12304 = B0-1%: 06 '0-12: + B0-1%: 06 '0-B:42
The percent by volume is a unitless number
because it is a ratio of two similar quantities.
27
Given:
Volume of solvent = 120.6 mL;
Volume of solute = 4.94 mL
SAMPLE Required: percent by volume
PROBLEM 7
Solution:
In a solution, there is B0-1%: 06 '0-12:
Z:;>:42 #$ B0-1%: = ×100
120.6 mL solvent B0-1%: 06 '0-12304
and 4.94 mL solute
present. Find the 4.94 37
percent by volume. % -. /01234 = ×100
120.6 37 + 4.94 37
% -. /01234 = >. ?@ %
28
VI. PARTS PER MILLION (ppm)
The number of units of mass of a chemical or contaminant
per million units of total mass.
Parts per million is mathematically expressed as:
*+,, ./ ,.01&" 7.01*" ./ ,.01&"
;;* (*⁄*) = ×45! ;;* (7⁄7) = ×45!
*+,, ./ ,.01&2.% 7.01*" ./ ,.01&2.%
or
*+,, ./ ,.01&"(=)
;;* (*⁄7) = ×45!
7.01*" ./ ,.01&2.% (*>)
A 'C )DEFGH
A BB' =
A I )DEJ
29
Given:
Concentration of solution = 1.24 g/L
SAMPLE
Required: ppm
PROBLEM 8
A solution has a Solution:
concentration of
1.24 g/L. What is 1.24 ) 1000 ")
its concentration !!" = × = -./0 112
in ppm? 1* 1)
30
Given:
% by mass solute = 0.000077%
Required: ppm
SAMPLE
PROBLEM 9 Solution:
<&'' 06 '0-12: = % #$ %&'' '0-12:
What is the <&'' 06 '0-4 100
concentration
in ppm of a 0.000077 % 0.00000077 7 '0-12:
solution that is = =
100 % 7 '0-4
0.000077% by
mass solute? 0.00000077 7 '0-12:
ZZ% = ×1000000 = ]. ^^ __H
7 '0-4
31
BIG IMAGE
32
LET’S DO THIS
1. A solution is prepared by
dissolving 5.02 g of KNO3 in
95.0 g of water. Calculate the
percent by mass of the solute.
% by mass of KNO3 = 5.02 %
33
LET’S DO THIS
2. A solution is prepared by
dissolving 5.02 g of KNO3 in
95.0 g of water. Calculate the
Mole Fraction:
mole fraction of the solute and X solute(KNO3) = 0.0094
X solvent(H2O) = 0.9906
the solvent.
34
LET’S DO THIS
3. Calculate the molarity of a
45.3g NaNO3 dissolved in
water to produce 225 ml of
Molarity :
solution. 2.37 mol/L
or
2.37 M
35
LET’S DO THIS
4. Calculate the molality
of 0.40 g of Na2CO3
(molar mass = 106)
Molality :
dissolved in 150 g of 0.025 mol/kg
or
water 0.025 m
36
LET’S DO THIS
5. A solution is made by
adding 50 mL of ethanol
to 180 mL of water.
Calculate the percent by
% by mass of ethanol = 21.74 %
volume of ethanol.
37
LET’S DO THIS
6. A solution is prepared
by dissolving 6.08 g of
NaCl in 115 g of water.
Calculate the
Parts per million=50,215 ppm
concentration of the
solution in ppm.
38
SUMMARY
Concentration refers to the amount of solute present in a given amount of solvent.
DEFINITION FORMULA
1. Percent by %&'' )* ')+,-.
mass of solute per 100 g of solution % #$ %&'' = ×233
mass %&'' )* ')+,-/)0
ratio of the number of mole of one
%)+.' )* 9
2. Mole Fraction component to the total number of moles in a 4)+. 56&7-/)0 8 )* 9 =
%)+.' ')+,-/)0
solution.
ratio of the number of moles of solute per %)+.' )* ')+,-.
3. Molarity 4)+&6/-$(4) =
liter of solution +/-.6 )* ')+,-/)0
the number of moles of solute per kilogram %)+.' )* ')+,-.
4. Molality 4)+&+/-$(%) =
of solvent %&'' )* ')+<.0- (=>)
concentration of a solution when the volume <)+,%. )* ')+,-.
5. Percent by ?.67.0- #$ <)+,%. = ×233
of a solute and the volume of a solution are <)+,%. )* ')+,-/)0
volume
given, multiplied by 100.
The number of units of mass of a chemical %&'' )* ')+,-.
6. Parts per @@% (%⁄%) = ×23!
or contaminant per million units of total %&'' )* ')+,-/)0
million
mass.
39
