Solutions / Concentrations Study Guide
Chemistry Ch. 15 Dr. Hazlett
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The best way of explaining the methodologies involved in solutions and concentrations is
to do so via examples.
I. Percent by Mass = Mass of Solute x 100
Mass of Solution
A. Example: How many grams of C6H12O6 and of water are in 500 g of a 5.3%
by mass glucose solution?
A 5.3% by mass soln means there is 5.3 g of glucose per 100 g
 5.3 g C6H12O6 x 500 g soln = 26.5 g glucose
100 g soln
The reminder of the soln is water, so 500 – 26.5 = 437.5 g H2O
II. Percent by Volume = Volume of solute x 100
Volume of soln
A. Example: Rubbing alcohol is an aqueous solution that has 70% isopropyl
alcohol by volume. How would you prepare a 250 mL rubbing
alcohol soln from pure isopropyl alcohol?
 70 % isopropyl alcohol = 70 mL iso. Alcohol x 250 mL soln
100 mL soln
= 175 mL isopropyl alcohol
So, add water to the 175 mL alcohol until reach 250 mL
III. Molarity (M) = mols of a solute
Volume soln (L)
A.  a 3M soln of HCl contains 3 mols of HCl acid per 1 liter of solution
B. Example: How many mols of HCl acid are in 200 mL of 0.15 M HCl?
 0.125 mol HCl is the conversion factor
1 L soln
And, 200 mL of 0.15 M HCl x 1L
x
0.15 mol HCl
1000 mL
1 L of .15 M HCl
= 0.3 mol HCl
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Solutions / Concentrations Study Guide
Chemistry Ch. 15 Dr. Hazlett
C. Example: What mass of NaOH is needed to make 100 mL of a 0 .125 M soln?
 0.125 mol NaOH and 1 mol NaOH =  40 g NaOH
1 L soln
? g NaOH = 100 mL of 0.125 M NaOH x
and
1L
x
1000 mL
0.125 mol NaOH x 40 g NaOH =
1 L soln
1 mol NaOH
0.5 g NaOH
D. Example: What volume of 3.25 M H2SO4 is needed to prepare 0.5 L of 0.13
M H2SO4?
 ? mol H2SO4 in 0.5L 0f .13 M H2SO4
0.5 L of 0.13 M H2SO4 x 0.13 mol H2SO4 = 0.065 mol H2SO4
1 L soln
And, ? L of 3.25 M H2SO4 = 0.065 mol H2SO4
x 1.0 L of 3.25 M H2SO4
3.25 mol H2SO4
=
0.2 L of 3.25 M
This gives the volume of concentrated sulfuric acid that provides the mols
needed for a dilute solution. The volume of 3.25 M H2SO4 would be
dissolved in 480 mL of water (500 mL – 20 mL = 480) to prepare 0.5 L
(500 mL) of a 0.13 M H2SO4.
+
20 mL of
3.25 M
H2SO4
(.065 mol acid)
=
0.480L
H2O
500 mL of
.13 M H2SO4
(.065 mol acid)
IV. Molality (m) = Moles of solute
Kg of solvent
A. Example: What is the molality of a soln w/ 16.3 g KCl dissolved in 845 g of
water?
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Solutions / Concentrations Study Guide
Chemistry Ch. 15 Dr. Hazlett
Convert mass of solute to moles:
16.3 g KCl x 1 mol KCl = 0.218 mols KCl
74.6 g KCl
 Molar mass of KCl
And, 845 g = 0.845 Kg
 m = 0.218 mol KCl = 0.258 KCl = 0.258
0.845 Kg H2O
1 Kg H2O
V. Mole Fraction
A. The ratio of mols of solute or solvent to the total number of moles of solute
plus solvent in a solution
1. XA = mole fraction solvent =
nA
nA + nB
XB = mole fraction solute =
n
nB
nA + nB
= # moles
2. Example: XA + XB of a solution w/ 215 g H2O and 44 g NaOH?
 nH2O = 215 g H2O x 1 mol H2O = 11.9 mol H2O
 18 g/mol
nNaOH = 44 g NaOH x 1 mol NaOH = 1.1 mol NaOH
 40 g/mol
XH2O =
n H2O
=
11.9
= 0.915
nH2O + nNaOH
11.9 + 1.1
XNaOH =
nNaOH
=
1.1
= 0.0846
nH2O + nNaOH
11.9 + 1.1
and 0.915 + 0.0846 =  1; fractions are correct!
VI. Diluting Solutions
A. In dilution, are adding only solvent (in mols or g)
B. Moles or Grams before dilution = Moles or Grams after dilution
C. Since M = mols solute / vol. soln (L); moles of solute = (M)(Volume)
1. And:
M1V1 = M2V2
D. Example: How would you prepare a 500 mL solution of 3 M HCl using a
6 M HCl as a supply source. In other words, how much 6 M HCl
would you mix to get the right dilution?
1st – determine the volume of 6 M HCl:
6 M (vol. of 6M) = 3 M HCl (500 mL)
Volume 6 M HCl = 3 M HCl (500 mL) = 250 mL 6 M HCl
6 M HCl
Solutions / Concentrations Study Guide
Chemistry Ch. 15 Dr. Hazlett
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2nd – determine the amount of H2O to add to 6 M HCl.
Since total volume = 500 mL, and volume of 6 M is 250 mL,
VH2O = 500 - 250 = 250 mL H2O
VII. Osmotic Pressure
A.  = iMRT
1. P = (n / V)RT ; w/ n = # moles, and V is in liters
2. i is the van’t Hoff constant and is an ionization factor for the substance
and its degree of dissociation
a. i is approximately equal to the number of species that a
molecule will ionize into; and equals 1 if it does not ionize
3. Example: What is the  at 25oC of seawater. It has a mass of
approximately 27 g NaCl per liter.
Convert g to mols: 27 g / 55.85 g/mol = 0.483 mol NaCl
 = i (0.483)(0.08206)(298)
= NaCl  Na+ + Cl= 1 mol  1mol + 1 mol ;  i = 2
= 2 (0.483)(0.08206)(298)
= 23.622
VIII. Boiling Point Elevation / Freezing Point Depression
A. ΔTb = i Kb m (boiling)
ΔTf = i Kf m (freezing)
B. Example: Seawater is  3.5% by weight dissolved salt. What is its boiling
point?
3.5 % means 3.5 g NaCl per 100 g of seawater soln, so this means
3.5 g NaCl in 96.5 g water.
Mols NaCl = 3.5 g / 58.5 g/mol = 0.0598 mols
Molality (m) NaCl = 0.0598 mol / 0.0965 kg = 0.612 m
 Δ Tf = (2)(0.52oC/m)(0.612m) = 0.64oC
 Constant from sourcebook
And, 100oC + 0.64oC = 100.64oC
C. Example of freezing point depression – subtract the calculated number from
the freezing point of the solvent
What is the freezing point depression of a 0.1 m soln of HF (aq)
which is -0.201oC. What is the percent dissociation of the HF?
ΔTf = i Kf m
= 0.201oC = i (1.86oC/m)(0.1m)
i = 1.08;  8% Do
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Solutions / Concentrations Study Guide
Chemistry Ch. 15 Dr. Hazlett
IX. Henry’s Law
S1 = S2
P1
P2
and
p = kHc
where p is partial pressure,
c is solute concentration, and kH is
a constant
A. At a constant temperature, the amount of a given gas dissolved in a given type
and volume of liquid is directly proportional to the partial pressure of that gas in
equilibrium with that liquid.
1. The solubility of a gas in a liquid is proportional to the pressure of that
gas above the liquid.
a. Example is carbonated drinks – they go flat (reach equilibrium)
if pressure of gas in liquid allowed to equalize w/ outside pressure
Some forms of Henry's law and constants (gases in water at 298 K), derived from
equation:
units:
dimensionless
O2
769.23
1.3×10−3
4.259×104
3.180×10−2
H2
1282.05
7.8×10−4
7.099×104
1.907×10−2
CO2
29.41
3.4×10−2
0.163×104
0.8317
N2
1639.34
6.1×10−4
9.077×104
1.492×10−2
He
2702.7
3.7×10−4
14.97×104
9.051×10−3
Ne
2222.22
4.5×10−4
12.30×104
1.101×10−2
Ar
714.28
1.4×10−3
3.955×104
3.425×10−2
CO
1052.63
9.5×10−4
5.828×104
2.324×10−2
c = amount concentration of gas in solution (in mol/L)
p = partial pressure of gas above the solution (in atm)
x = mole fraction of gas in solution
Solutions / Concentrations Study Guide
Chemistry Ch. 15 Dr. Hazlett
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X. Beer-Lambert Law
A. A=bc
1. Where A is absorbance (no units, since A = log10 P0 / P )
 is the molar absorbtivity with units of L mol-1 cm-1; b is the path length
of the sample - that is, the path length of the curvette in which the sample
is contained expressed in centimeters; and c is the concentration of the
compound in solution, expressed in mol L-1
2. Absorbance is directly proportional to the other parameters, as long as
the law is obeyed.
a. The absorption of UV light by a compound, A = log10 P0 / P
XI. Raoult’s Law
A. P1 = i x1p1o
1. Where P1 is the vapor pressure; i is the # mole of the soln; x1 is the
mole fraction; and p1o is vapor pressure of the pure solvent
2. As number of components in a soln increases, the individual vapor
pressures decrease since mole fraction of each decreases as components
added
3. Law only works with ideal solns and is used to examine bonding
between liquids in solns. If predicted vapor pressure is less than expected,
then fewer molecules have left soln and bonding is greater between
liquids. If vp higher than predicted, less bonding, lower energy and more
molecules leaving soln.