Ways of Expressing
Concentrations of
Solutions
Mass Percentage
Mass % of A =
mass of A in solution
 100
total mass of solution
Parts per Million and
Parts per Billion
Parts per Million (ppm)
ppm =
mass of A in solution
 106
total mass of solution
Parts per Billion (ppb)
ppb =
mass of A in solution
 109
total mass of solution
Mole Fraction (X)
XA =
moles of A
total moles in solution
• In some applications, one needs the mole
fraction of solvent, not solute—make sure
you find the quantity you need!
Molarity (M)
M=
mol of solute
L of solution
• Because volume is temperature dependent,
molarity can change with temperature.
Molality (m)
m=
mol of solute
kg of solvent
Because neither moles nor mass change
with temperature, molality (unlike
molarity) is not temperature dependent.
SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations
(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg
of water. What is the mass percentage of solute in this solution? (b) A 2.5-g
sample of groundwater was found to contain 5.4 g of Zn2+ What is the
concentration of Zn2+ in parts per million?
PRACTICE EXERCISE
(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of
NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62
mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle
containing 2500 g of bleaching solution?
PRACTICE EXERCISE
A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate
(a) the molality and (b) the mole fraction of NaOCl in the solution.
Ideal solutions
Non-ideal solutions
Positive deviation from
Raoult’s law
1.Obey Raoult’s law at every 1.Do not obey Raoult’s
range of concentration.
law.
2.Hmix = 0; neither is evolved
nor absorbed during
2.Hmix>0. Endothermic
dissolution.
dissolution; heat is
absorbed.
3.Vmix = 0; total volume of
solution is equal to sum of
3.Vmix > 0. Volume is
volumes of the components. increased after dissolution.
4.P = pA + pB = pA0XA +
pB0XB
i.e., pA =
4.pA > pA0XA; pB > pB0XB
∴ pA + pB > pA0XA + pB0XB
Negative deviation from
Raoult’s law
1.Do not obey Raoult’s
law.
2.Hmix<0. Exothermic
dissolution; heat is evolved.
3.Vmix <0. Volume is
decreased during
dissolution.
4.pA < pA0XA; pB < pB0XB
∴ pA + pB < pA0XA +
pB0XB
Ideal solutions
Non-ideal solutions
Positive deviation from
Raoult’s law
Negative deviation from
Raoult’s law
5.A—B attractive force should
5.A—A, A—B, B—B
be weaker than A—A and B—B
interactions should be same, i.e., attractive forces. ‘A’ and ‘B’
‘A’ and ‘B’ are identical in
have different shape, size and
shape, size and character.
character.
5. A—B attractive force should
be greater than A—A and B—B
attractive forces. ‘A’ and ‘B’
have different shape, size and
character.
6. Escaping tendency of ‘A’
and ‘B’ should be same in pure
liquids and in the solution.
Examples: dilute solutions;
benzene + toluence: n-hexane +
n-heptane; chlorobenzene +
bromobenzene; n-butyl chloride
+ n-butyl bromide.
6. Escaping tendency of both
components ‘A’ and ‘B’ is
lowered showing lower vapour
pressure than expected
ideally. Examples: acetone +
aniline; acetone + chloroform;
CH3OH + CH3COOH; H2O +
HNO3; Choloroform + diethyl
ether, water + HCl; acetic acid
+ pyridine; chloroform +
benzene.
6. ‘A’ and B’ escape easily
showing higher vapour pressure
than the expected
value. Examples: acetone +
ethanol acetone + CS2; water +
methanol; water + ethanol;
CCl4 + toluene; CCl4 + CHCl3;
acetone + benzene; CCl4 +
CH3OH; Cyclohexane + ethanol