SOLUTIONS II

SOLUTIONS
Solutions are homogeneous mixtures of two or more co mponents. If a solution which consists of only two
components, then it is said to be a binary solution. The co mponent that is present in the largest quantity is known as
solvent. And the components present in the solution other than solvent are called solutes.
Concentration of Solutions
Co mposition of a solution can be described in terms of concentration. We can describe the concentration of the
solution quantitatively by;
(i) Mass percentage (w/w):
Mass percentage of a component = x 100

(ii) Volume percentage (V/V):

Volume percentage of a component = x 100
(iii) Parts per million (ppm):
Parts per million = x 106
(iv) Mole fraction ( x):
Mole fraction of a component, x =
Consider a binary solution, having co mponents A and B. n A and n B represents their number of mo les. Then, mole
fraction of A will be;
XA = and that of component B will be ; XB =
In a given solution sum of all the mole fractions is unity, i.e.
XA + XB +……. + Xi = 1
(v) Molarity (M):
Molarity (M) is defined as number of moles of solute dissolved in one litre of solution.
Molarity = or
=

(vi) Molality (m):

Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent.
Molality (m) = or
=
(vii) Normality (N):
Normality (N) is defined as the number of gram equivalent of solute dissolved in one litre of solution.
Normality (N) =
(viii) Formality (F): It is the number of formula weights of solute present per litre of the solution.
Formality = moles of substance added to solution / volume of solution (in L)

Mass %, ppm, mole fraction and molality are independent of temperature, whereas molarity is a function of
temperature. This is because volume depends on temperature and the mass does not.
Solubility
Solubility of a substance is its maximu m amount that can be dissolved in a specified amount of solvent at a specified
temperature.
Solubility of a Solid in a Liquid:
A solute dissolves in a solvent if the intermo lecular interactions are similar in the two. When a solid solute is added
to the solvent, some solute dissolves and its concentration increases in solution. This process is known a s
dissolution. So me solute particles in solution collide with the solid solute particles and get separated out of solution.
This process is known as crystallisation. A stage is reached when the two processes occur at the same rate.
Solute + Solvent ↔ Solution
At this stage the concentration of solute in solution will remain constant under the given temperature and pressure.

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The concentration of solute in such a solution is its solubility . According to Le Chateliers Princip le, if the
dissolution process is endothermic, the solubility should increase with rise in temperature and if it is exothermic the
solubility should decrease.
But pressure does not have any significant effect on solubility of solids in liquids.
Solubility of a Gas in a Liquid
Solubility of gases in liquids is greatly affected by pressure and temperature. The solubility of gases increase with
increase of pressure. A quantitative relation between pressure and solubility of a gas in a solvent is given by Henry
and is known as Henry‟s law. The law states that at a constant temperature, the solubility of a gas in a liquid is
directly proportional to the partial pressure of the gas present above the surface of liquid or solution.
The most commonly used form of Henry‟s law states that “the partial pressure of the gas in vapour phase (p) is
proportional to the mole fraction of the gas (x) in the solution” and is expressed as:
p = KH x
Here KH is the Henry‟s law constant and is a function of the nature of the gas. Higher the value of KH at a given
pressure, the lower is the solubility of the gas in the liquid.
The solubility of gases increases with decrease of temperature. It is due to this reason that aquatic species are more
comfortable in cold waters rather than in warm waters.
Henry‟s law finds several applications:
• To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.
• Scuba divers have high concentrations of dissolved gases in blood while breathing air at high pressure
underwater. When they come towards surface, the pressure gradually decreases. This releases the dissolved gases
and leads to a medical condition known as bends. To avoid bends, as well as, the toxic effects of high concentrations
of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium.
• At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low
concentrations of oxygen in the blood and tissues of climbers and leads to a medical condition known as anoxia.

Vapour Pressure of Liquid- Liquid Solutions and Raoult’s law

Consider a binary solution having components A and B taken in a closed vessel. Both the components are volatile
and equilibriu m would be established between vapour phase and the liquid phase. Let the total vapour pressure at
this stage be Ptotal and PA and PB be the partial vapour pressures of the two components A and B respectively. XA
and XB represent the mole fractions of the two components A and B respectively.
Francois Marte Raoult gave the quantitative relationship between mole fraction and partial vapour pressure. The
relationship is known as the Raoult‟s law and it states that „the partial vapour pressure of an volatile component in a
solution is directly proportional to its mole fraction‟.
Thus, for component A
PA ∝ XA or PA = KA XA
When A is in pure state XA = 1; then the proportionality constant KA = P0 A , the vapour pressure of the pure
component A
Thus, PA = P0 A XA ………………………. (i)
Similarly, for component B
PB = P0 B XB `````````````````````````````````````````` (ii)
0
Where P B represent the vapour pressure of the pure component B
Thus partial vapour pressure of an volatile component in a solution is equal to the product of vapour pressure of the
component in pure state and its mole fraction. This is another way of stating Raoult‟s law.
According to Dalton‟s law of partial pressures, the total pressure (Ptotal ) is given as:
Ptotal = PA + PB
Thus Ptotal = P0 A XA + P0 BXB
= (1-XB) P0 A + P0 BXB
= P0 A - P0 A XB + P0 BXB
= P0 A + ( P0 B - P0 A )XB ……………….. (iii)

The composition of vapour phase in equilibriu m with the solution is determined by the partial pressures of the
components. If YA and YB are the mole fractions of the components A and B respectively in the vapour phase then,
using Dalton‟s law of partial pressures:
PA = YA P total
PB = YB P total
In general, Pi = Yi P total ……………………………… (iv)

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Vapour Pressure of Solid- Liquid Solutions and Raoult’s law

Consider a binary solution having components A and B taken in a closed vessel. Here co mponent A is volatile wh ile
B is non-volatile. XA and XB represents their mole fractions and PA vapour pressure of component A and PS that of
solution.
In a pure liqu id the entire surface is occupied by the molecules of the liquid. While in the solution,
the surface has both solute and solvent molecules. If a non-volatile solute is added to a solvent to give a solution the
vapour pressure of the solution is solely fro m the solvent alone. Thus vapour pressure of the solution found to be
lower than the vapour pressure of the pure solvent.
In the solution, the surface has both solute and solvent molecules; thereby the fraction of the
surface covered by the solvent molecules gets reduced. Consequently, the number of solvent molecu les escaping
from the surface is reduced, thus, the vapour pressure is also reduced.
Thus vapour pressure of solution, Ps = PA only.
But PA = P0 A XA or Ps = P0 A XA or Ps = P0 A (1-XB ) or = 1 - XB

or = XB ………..………………….. (i)

Fro m this it is clear that, the relative lowering of vapour pressure is equal to the mole fraction of non - volatile
component present in the solution. This is another way of stating Raoult‟s law.
Ideal and Non- ideal Solutions
Ideal Solutions:
The solutions which obey Raoult‟s law over the
entire range of concentration are known as ideal
solutions. Here enthalpy of mixing (ΔH mix ) and the
volume of mixing (ΔV mix ) is zero,
i.e., ΔH mix = 0, ΔV mix = 0
In the case of ideal solution the force of attraction
exist between A & A and B & B is same as that exist
between A & B. So the escaping tendency of A and B
fro m solution found to be independent of each other.
[A and B components of solution]
Eg : solution of n-hexane and n-heptane,
bromoethane and chloroethane,
benzene and toluene. The plot of vapour pressure and mole fraction of an ideal
solution at constant temperature.
Non-ideal Solutions:
When a solution does not obey Raoult‟s law over the entire range of concentration, then it is called non-ideal
solution. The vapour pressure of such a solution is
either higher or lower than that predicted by Raoult‟s
law. If it is higher, the solution exhibits positive
deviation and if it is lower, it exhibits negative
deviation from Raoult‟s law.
In case of positive deviation from Raoult‟s
law, the force of attraction exist between A & B
found to be weaker than those between A & A or B
& B. So the escaping tendency of A and B fro m
solution increases. This will increase the vapour
pressure and result in positive deviation. Here
enthalpy of mixing (ΔH mix ) and the volu me of
mixing (ΔV mix ) is positive. Solution that shows positive deviation from Raoult's law
i.e., ΔH mix = +ve , ΔV mix = +ve
Eg: Mixtures of ethanol and acetone,
carbon disulphide and acetone.
In case of negative deviation fro m Raoult‟s law, the force of attraction exist between A & B found to be greater than
those between A & A or B & B. So the escaping tendency of A and B fro m solution decreases. This will decrease
the vapour pressure and result in negative deviation. Here enthalpy of mixing (ΔH mix ) and the volu me of mixing
(ΔV mix ) is negative.
i.e., ΔH mix = -ve , ΔV mix = -ve
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Eg: mixture of phenol and aniline, a mixture of chloroform and acetone.

Azeotropes are binary mixtures having the same co mposition in liquid and vapour phas e and boil at a constant
temperature. There are two types of azeotropes called min imu m boiling azeotrope and maximu m bo iling azeotrope.
The solutions which show a large positive deviation fro m Raoult‟s law form minimum boiling azeotrope at a specific
composition.
Eg : ethanol-water mixture (95% by volume of ethanol.)
The solutions that show large negative deviation fro m Raoult ‟s law form maximum boiling azeotrope at a specific
composition. Eg : Nitric acid and water (68% nitric acid and 32% water by mass )
Colligative Properties and Determination of Molar Mass
Properties of solution wh ich depend on the number of solute particles and are independent of their nature are called
colligative properties. They are: (1) relat ive lowering of vapour pressure of the solvent (2) depression of freezing
point of the solvent (3) elevation of boiling point of the solvent and (4) osmotic pressure of the solution.
1. Relative Lowering of Vapour Pressure:
The vapour pressure of the solvent in solution is less than that of t he pure solvent. The relation between vapour
pressure of the solution, mole fraction and vapour pressure of the solvent is p 1 = x1 p 1 0
p 1 = vapour pressure of the solution; x1 = mole fraction of the solute; p 1 0 = vapour pressure of the solvent.

is called relat ive lowering of vapour pressure and is equal to the

mole fraction of the solute .For dilute solutions n2 << n1 hence neglecting n2 in the denominator we have

=
Ostwald and Walker method is used to determine the relative lowering of vapo ur pressure.
2. Elevation in Boiling Point (ΔT b )
Boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the at mospheric pressure.
As the vapour pressure of a solution containing a non-volatile solute is lo wer than that of the pure solvent, it boiling
point will be higher than that of the pure solvent.
Let T0 b be the boiling point of pure solvent and Tb be
the boiling point of solution. The increase in the
boiling point
Tb -T0 b =△Tb is the elevation in boiling point,
which is proportional to the lowering of vapour
pressure, △p.
It is found that equimolal solutions of d ifferent
solutes in the same solvent boil at the same
temperature. Hence elevation in boiling point is
directly proportional to the molality of the solution.
i.e. △Tb α m or .△Tb=Kb m.
Where Kb is the proportionality constant.
When m =1, .△Tb = Kb.
So Kb is the elevation in boiling point fo r a solution
of unit molality. Hence it is called molal elevation
constant (Ebullioscopic Constant) and its unit is
kelvin/molality.

If W2 g of solute of molecular weight M2 , dissolved in W1 g of solvent; then the molality of the solution;
m= W2 1000/W1 M2 .

Therefore .△Tb= Kb Using this relation it is possible to calculate the molecular mass of solute.
The boiling point elevation of a solution is determined by
(i) Landsberger‟s method (ii) Cottrell‟s method

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3. Depression in freezing point (ΔTf ) :

This is another consequence of the lowering of vapour pressure. When a pure liquid freezes, the pure solid has an
equilibriu m vapour pressure, which is equal to the vapour pressure of the liqu id at its freezing point. Since the
vapour pressure of a solution is less than the vapour pressure of pure solvent at the freezing point of the pure
solvent, the solution will not freeze at this temperature. It will freeze at a lower temperature at which the vapour
pressure of pure solid solvent is equal to the vapour pressure of the solution.
Fro m the graph Tf0 is the freezing point of pure solvent. At Tf0 the vapour pressure of solution is less than the
equilibriu m vapour pressure of solid solvent. Then the solution frezes at a lo wer temperature, T f when its vapour
pressure becomes equal to the equilibrium vapour pressure of pure solid solvent.
So Tf0 –Tf = △Tf, the depression in freezing point. And is proportional to the lowering of vapour pressure △p.
It is found that equimolal solutions of different solutes in the same solvent freezes at the same temperature. Hence

depression in freezing point is directly p roportional to

the molality of the solution.
i.e. △Tf α m or .△Tf=Kf m.
Where Kf is the proportionality constant.
When m =1, .△Tf = Kb.
So Kf is the depression in freezing point for a
solution of unit molality. Hence it is called mo lal
depression constant or Cryoscopic Constant and its
unit is kelvin/molality.
If W2 g of solute of mo lecular weight M2 , dissolved in
W1 g of solvent; then the molality of the solution;
m= W2 1000/W1 M2 .

Therefore .△Tf= Kf
Using this relation it is possible to calcu late the
mo lecular mass of solute. The freezing point
depression is determined by Beckmann method or
Rast method.

Calculations of molal elevation constant (Kb ) and molal depression constant (Kf ):

and
Here R and M 1 are the gas constant and molar mass of the solvent, respectively and Tf and Tb denote the freezing
point and the boiling point of the pure solvent respectively in Kelvin. ΔHfus and ΔHvap are the enthalpies for the
fusion and vapourisation of the solvent, respectively.

4. Osmotic pressure (π) :

Osmosis is the spontaneous flow of solvent from solvent to solution or from dilute solution to concentrated solution,
when the two liquids are separated by a semipermeable membrane. A semipermeab le membarene is a membrane ,
which allows only solvent molecules to pass through it. Eg : cellophane, egg membrane.
The passage of solute molecules from concentrated solution to dilute solution or from solution to solvent is called
diffusion.
As a result of osmosis, a hydrostatic pressure will be developed on the solution. The maximu m hydrostatic pressure
developed on a solution when it is separated from its solvent by a semipermeable membrane is called osmotic
pressure.
It can also be defined as the minimum pressure to be applied on the solution so that it is just su fficient to prevent the
entry of solvent to solution when the two liquids are separated by a semipermeable membrane.
Vant Ho lf suggested that solute mo lecules in solution behave like mo lecules in a gas. So the osmotic pressure of the
solution is given by the ideal equation of state.
Thus πv = nRT or Π = C R T
Where π- osmot ic pressure, v – volu me of solution containing „n‟moles of a solute, R- gas constant, T- kelv in
temperature c-concentration (molarity) of solution.

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Since n =
πv = RT
Fro m this relation mo lecular mas of solute can be calculated. It would provide a good method for the determination
of molecular masses of polymers.
Osmotic pressure can be determined by
(i) Pfeffer‟s method (ii) Berkeley and Hartley‟s method (iii) Morse and Frazer‟s method

*[Osmosis may be
(i) Exosmosis : It is outward flow of water or solvent from a cell through semipermeable membrane.
(ii) Endosmosis: It is inward flow of water or solvent from a cell through a semipermeable membrane.

On the basis of osmotic pressure, -the solution can be

(i) Hypertonic solution: A solution is called hypertonic if its osmotic pressure is higher than that of the solution
from which it is separated by a semipermeable membrane. When a plant cell is placed in a hypertonic solution,
the fluid from the plant cell comes out and cell shrinks, this phenomenon is called plasmolysis.
(ii) Hypotonic solution: A solution is called hypotonic if its osmotic pressure is lower than that of the solution from
which it is separated by a semipermeable membrane.
(iii) Isotonic solution: Two solutions are called isotonic if they exert the same osmotic pressure and they have the
same molality. These solutions have same molar concentration. For isotonic solutions in the same solvent,
W1/M1 = W2/M2 ,
where W1 is the mass of solute of molecular mass M1 and W2 is the mass of another solute of molecular mass
of M2.
0.91% solution of pure NaCl is isotonic with human RBC‟s. When placed in water containing less than 0.9% salt,
blood cells collapse due to loss of water by osmosis.
Reverse osmosis:
If the pressure applied on the solution is greater than the osmotic pressure, then solvent starts passing from solution
into solvent. This is called reverse osmosis, generally used for purification of sea water or hard water.]

Abnormal Molecular Masses:

Colligative properties, which depend only on the number of particles of solute, are measured to determine mo lecular
masses of solute. Any factor that affects the nu mber of part icles in solution will therefore give abnormal v alue for a
colligative property. Hence the molecular mass calculated also will be abnormal. The two factors that affects are;
A. Association:
Some solutes in solution undergo association. For example in ben zene, benzo ic acid undergoes dimerization through
hydrogen bonding. Therefore the number o f particles is reduced to half. The observed colligative property will be
half the expected value and molecular mass will be twice the theoretical value.
B. Dissociation :
Electrolytes like acids, bases, salts etc, undergoes dissociation in solu tion. For example, NaCl dissociate into Na+
and Cl- ions. This increases the number of particles in solution. Consequently the measured colligative property will
be higher and molecular mass will be less.
van’t Hoff Factor (i):
van‟t Hoff introduced a factor i, known as the van‟t Hoff factor, to account for the extent of dissociation or
association of solute in solution. It is the ratio of observed value of colligative property to the calculated value of
colligative property.
i = observed value of colligative property / calculated value of colligative property
or i = normal molecular mass / observed molecular mass
or i = number of particles after association or dissociation / number of particles initially
So to correct the observed value of molar mass, van‟t Hoff factor (i) must be included in different expressions for
colligative properties.
Relative lowering of vapour pressure of solvent,

Elevation of Boiling point, ΔTb = i Kb m

Depression of Freezing point, ΔTf = i Kf m

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Osmotic pressure of solution, π = i RT

van‟t Hoff factor (i) > 1 for solutes undergoing dissociation and it is < 1 for solutes undergoing association.
For electrolytes the degree of dissociation α is related to the van t Holf factor by the expression,
α = where „r‟is the number of ions produced from 1 mole of the electro lyte. Eg : for NaCl ; r = 2 and for
Na2 CO3 , „r‟is 3.

SOLVED EXAMPLES:

1. The depression in freezing point of water observed for the same amount of acetic acid trichloroacetic acid and
trifluoroacetic acid increases in the order given above. Explain briefly.
Sol. The depression in freezing points are in the order : acetic acid < trichloroacetic acid < trifluoroacetic acid
Fluorine, being most electronegative, has the highest electron withdrawing inductive effect. Consequently,
Trifluoroacetic acid is the strongest acid while acetic acid is the weakest acid. Hence, trifluoroacetic acid ionizes to
the largest extent while acetic acid ionizes to the min imu m extent to give ions in their sloutions in water. Greater the
ions produced, greater is the depression in freezing point. Hence, the depression in freezing point is maximu m for
the fluroacetic acid amd minimum for acetic acid.
2. Give an example of solid solution in which the solute is a gas.
In case a solid solution is formed between two substances, an interstitial solid solution will be formed. For examp le,
a solution of hydrogen in palladium is a solid solution in which the solute is a gas.
3. What role does the molecular interaction play in a solution of alcohol and water?
In pure alcohol and water, the mo lecules are held t ightly by a strong hydrogen bonding. The interaction between the
mo lecules of alcohol and water is weaker than alcohol−alcohol and water−water interactions. As a result, when
alcohol and water are mixed, the intermolecular interactions become weaker and the mo lecules can easily escape.
This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.
4. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in
liquids is an exothermic process.
Gas + Liquid →Solution + Heat
Therefore, when the temperature is increased, heat is supplied and the equilibriu m shifts backwards, thereby
decreasing the solubility of gases.
5. Based on solute-solvent interactions, arrange the following in order of i ncreasing solubili ty in n-octane and
explain. Cyclohexane, KCl, CH3 OH, CH3 CN.
n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in
the n-octane. The order of increasing polarity is: Cyclohexane < CH3 CN < CH3 OH < KCl
Therefore, the order of increasing solubility is: KCl < CH3 OH < CH3 CN < Cyclohexane
6. Henry‟s law constant for CO2 in water is 1.67 × 10 8 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of
soda water when packed under 2.5 atm CO2 pressure at 298 K.
It is given that:
KH = 1.67 × 108 Pa = 0.00152
PCO2 = 2.5 atm = 2.5 × 1.01325 × 105 Pa We can write, x = =
= 2.533125 × 105 Pa
[Since, n CO2 is negligible as compared to n H2O ]
According to Henry‟s law:
In 500 mL of soda water,
the volume of water = 500 mL
[Neglecting the amount of soda present]
We can write: 500 mL of water = 500 g of water
= mol of water
= 27.78 mol of water
Now, =x
= 0.00152
n CO2 = 0.042 mol

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Hence, quantity of CO2 in 500 mL of soda water

= (0.042 × 44)g = 1.848 g
7.Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH 2 CONH2 ) is dissolved in 850 g of
water.Calculate the vapour pressure of water for this solution and its relative lowering.

It is given that vapour pressure of water, = 23.8 mm of Hg ; Weight of water taken, w1 = 850 g

Weight of urea taken, w2 = 50 g

Molecular weight of water, M1 = 18 g mol−1

Molecular weight of urea, M2 = 60 g mol−1

Now, we have to calculate vapour pressure of water

in the solution. We take vapour pressure as p 1 .

Now, from Raoult‟s law, we have:

Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.
8. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that
it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol −1 .
Here, elevation of boiling point
ΔTb = (100 + 273) − (99.63 + 273)
= 0.37 K
Mass of water, wl = 500 g
Molar mass of sucrose (C12 H22 O11 ), M2
= 11 × 12 + 22 × 1 + 11 × 16
= 342 g mol−1
Molal elevation constant, Kb = 0.52 K kg mol−1

We know that:
= 121.67 g (approximately)
Hence, 121.67 g of sucrose is to be added.
9. Calculate the mass of ascorbic aci d (Vitamin C, C6 H8 O6 ) to be dissolved in 75 g of acetic aci d to lower its
melting point by 1.5°C. Kf = 3.9 K kg mol −1 .

Mass of acetic acid, w1 = 75 g

Molar mass of ascorbic acid (C6 H8 O6 ), M2
= 6 × 12 + 8 × 1 + 6 × 16
= 176 g mol−1
Lowering of melting point, ΔTf = 1.5 K
We know that:

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= 5.08 g (approx)
Hence, 5.08 g of ascorbic acid is needed to be dissolved.
10. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of
molar mass 185,000 in 450 mL of water at 37°C.
It is given that: Volume of water, V = 450 mL = 0.45 L Temperature, T = (37 + 273)K = 310 K
Number of moles of the polymer, n = mol , We know that:
Osmotic pressure, π = RT = mol x x 8.314 x 103 Pa L K-1 mol-1 x 310 K
= 30.98 Pa = 31 Pa (approximately)
11. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each
component in the solution? If the density of solution is 1.2 g mL −1 , then what shall be the molarity of the
solution?
10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of
glucose is present in (100 − 10) g = 90 g of water.

Molar mass of glucose (C6 H12 O6 )

= 6 × 12 + 12 × 1 + 6 × 16 And, mole fraction of water, Xw = 1- Xg
= 180 g mol−1 = 1 − 0.011
Then, number of moles of glucose = = 0.989
If the density of the solution is 1.2 g mL −1 , then the
= 0.056 mol
volume of the 100 g solution can be given as:
Molality of solution =
=
= 0.62 m
= 83.33mL
Number of moles of water = = 83.33 x 10-3 L
= 5 mol Molarity of the solution =
Hence mole fraction of glucose Xg = = 0.67 M
= 0.011
12. Calculate the amount of benzoic acid (C 6 H5 COOH) required for preparing 250 mL of 0.15 M solution in
methanol.
0.15 M solution of benzoic acid in methanol means, Molar mass of benzoic acid (C6 H5 COOH)
1000 mL of solution contains 0.15 mo l of benzoic = 7 × 12 + 6 × 1 + 2 × 16
acid = 122 g mol−1
Therefore, 250 mL of solution contains Hence, required benzoic acid
= 0.0375 mol × 122 g mol−1
mol of benzoic acid.
= 4.575 g
= 0.0375 mol of benzoic acid

13. Determine the amount of CaCl 2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
We know that,

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Here, R = 0.0821 L atm K-1 mol-1

M = 1 × 40 + 2 × 35.5
= 111g mol-1

Therefore, w =
= 3.42 g
Hence, the required amount of CaCl 2 is 3.42 g.
14. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K 2 SO4 in 2 liter of water at
25° C, assuming that it is completely dissociated.
When K2 SO4 is dissolved in water, K+ and SO4 2- ions
are produced.

Total number of ions produced = 3

i =3
Given, w = 25 mg = 0.025 g
V =2 L
T = 250 C = (25 + 273) K = 298 K
Also, we know that: R = 0.0821 L atm K-1 mol-1
M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol-1
Appling the relation,

6. Low concentration of o xygen in the blood and

tissues of people living at high altitude is due to _
1. Which of the following units is useful in relating (i) low temperature (ii) low atmospheric pressure
concentration of solution with its vapour pressure? (iii) high atmospheric pressure
(iv) both low temperature and high atmospheric pressure
(i) mole fraction (ii) parts per million
(iii) mass percentage (iv) molality
7. Considering the formation, breaking and strength
2. On dissolving sugar in water at room temperature of hydrogen bond, predict which of the following
solution feels cool to touch. Under which of the mixtures will show a positive deviation from Raoult‟s law?
following cases dissolution of sugar will be most rapid? (i) M ethanol and acetone. (ii) Chloroform and acetone.
(i) Sugar crystals in cold water. (iii) Nitric acid and water. (iv) Phenol and aniline.
(ii) Sugar crystals in hot water.
(iii) Powdered sugar in cold water. 8. Colligative properties depend on ____________.
(iv) Powdered sugar in hot water. (i) the nature of the solute particles dissolved in solution.
(ii) the number of solute particles in solution.
3. At equilibriu m the rate of dissolution of a solid
(iii) the physical properties of the solute particles
solute in a volatile liquid solvent is ___. dissolved in solution. (iv) the nature of solvent particles.
(i) less than the rate of crystallisation
(ii) greater than the rate of crystallisation 9. Which of the following aqueous solutions should
(iii) equal to the rate of crystallisation (iv) zero have the highest boiling point?
4. A beaker contains a solution of substance „A‟. (i) 1.0 M NaOH (ii) 1.0 M Na2 SO4
Precip itation of substance „A‟ takes place when s mall (iii) 1.0 M NH4 NO3 (iv) 1.0 M KNO3
amount of „A‟ is added to the solution. The solution is __.
(i) saturated (ii) supersaturated 10. The unit of ebulioscopic constant is __________.
(iii) unsaturated (iv) concentrated (i) K kg mol –1 or K (molality) –1
(ii) mol kg K –1 or K –1 (molality)
5. Maximum amount of a solid solute that can be (iii) kg mol –1 K –1 or K –1 (molality) –1
dissolved in a specified amount of a given liquid (iv) K mol kg –1 or K (molality)
solvent does not depend upon _____
(i) Temperature (ii) Nature of solute
(iii) Pressure (iv) Nature of solvent
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11. In co mparison to a 0.01 M solution of glucose, 19. We have three aqueous solutions of NaCl labelled
the depression in freezing point of a 0.01 M MgCl2 as „A‟, „B‟ and „C‟ with concentrations 0.1M, 0.01M
solution is and 0.001M , respectively. The value of van‟t Hoff
(i) the same (ii) about twice factor for these solutions will be in the order______.
(iii) about three times (iv) about six times (i) i A < i B < i C (ii) i A > i B > i C
(iii) i A = i B = i C (iv) i A < i B > i C
12. An unripe mango placed in a concentrated salt
solution to prepare pickle, shrivels because ____ 20. On the basis of information given below mark the
(i) it gains water due to osmosis. correct option.
(ii) it loses water due to reverse osmosis. Information:
(iii) it gains water due to reverse osmosis. (A) In bromoethane and chloroethane mixture
(iv) it loses water due to osmosis. intermolecular interactions of A–A and B–B type
are nearly same as A–B type interactions.
13. At a given temperature, osmotic pressure of a
(B) In ethanol and acetone mixture A–A or B–B type
concentrated solution of a substance _____
intermolecular interactions are stronger than A–B
(i) is higher than that at a dilute solution.
type interactions.
(ii) is lower than that of a dilute solution.
(C) In chlorofo rm and acetone mixture A–A or B–B
(iii) is same as that of a dilute solution. (iv) cannot
be compared with osmotic pressure of dilute solution. type intermolecular interactions are weaker than A–
B type interactions.
14. Which of the following statements is false? (i) Solution (B) and (C) will follow Raoult‟s law.
(i) Two different solutions of sucrose of same molality (ii) Solution (A) will follow Raoult‟s law.
prepared in different solvents will have the same (iii) Solution (B) will show negative deviation from
depression in freezing point. Raoult‟s law.
(ii) The osmotic pressure of a solution is given by the
(iv) Solution (C) will show positive deviation from
equation Π = CRT ( where C is the molarity of the
solution). Raoult‟s law.
(iii) Decreasing order of osmotic pressure for 0.01 M 21. If two liquids A and B form minimum boiling
aqueous solutions of barium chloride, potassium azeotrope at some specific composition then ___
chloride, acetic acid and sucrose is
BaCl2 > KCl > CH 3COOH > sucrose. (i) A– B interactions are stronger than those
(iv) According to Raoult‟s law, the vapour pressure between A–A or B–B.
exerted by a volatile component of a solution is directly (ii) vapour pressure of solution increases because
proportional to its mole fraction in the solution. more nu mber of molecules of liquids A and B can
escape from the solution.
15. The values of Van‟t Hoff factors for KCl, NaCl (iii) vapour pressure of solution decreases because
and K2 SO4 , respectively, are ___. less number of molecu les of only one of the liquids
(i) 2, 2 and 2 (ii) 2, 2 and 3 escape from the solution.
(iii) 1, 1 and 2 (iv) 1, 1 and 1 (iv) A–B interactions are weaker than those
16. Which of the following statements is false? between A–A or B–B.
(i) Un its of atmospheric pressure and osmotic 22. 4L of 0.02 M aqueous solution of NaCl was
pressure are the same. (ii) In reverse osmosis, diluted by adding one lit re of water. The mo lality of
solvent molecules move through a semipermeable the resultant solution is-
memb rane fro m a region o f lower concentration of (i) 0.004 (ii) 0.008 (iii) 0.012 (iv) 0.016
solute to a region of higher concentration. 23. KH value for Ar(g), CO2 (g), HCHO (g) and CH4 (g) are
(iii) The value of molal depression constant 40.39, 1.67, 1.83×10 –5and 0.413 respectively. Arrange
depends on nature of solvent. (iv) Relative lo wering these gases in the order of their increasing solubility .
of vapour pressure, is a dimensionless quantity. (i) HCHO < CH4 < CO2 < Ar
17. Value of Henry‟s constant KH _____ (ii) HCHO < CO2 < CH4 < Ar
(i) increases with increase in temperature. (iii) Ar < CO2 < CH4 < HCHO
(ii) decreases with increase in temperature. (iv) Ar < CH4 < CO2 < HCHO
(iii) remains constant. 24. Which of the following factor (s) affect the
(iv) first increases then decreases. solubility of a gaseous solute in the fixed volu me of
18. The value of Henry‟s constant KH is ____ liquid solvent?
(i) greater for gases with higher solubility. (a) nature of solute (b) temperature (c) pressure
(ii) greater for gases with lower solubility. (i) (a) and (c) at constant T
(iii) constant for all gases. (ii) (a) and (b) at constant P
(iv) not related to the solubility of gases. (iii) (b) and (c) only (iv) (c) only

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25. Intermolecular forces between two benzene 27. Isotonic solutions must have the same ___
mo lecules are nearly of same strength as those (i) solute (ii) density
between two toluene molecules. For a mixture of (iii) elevation in boiling point
benzene and toluene, wh ich of the following are not (iv) depression in freezing point
true?
(i) Δ mix H = zero (ii) Δ mix V = zero 28. Which of the following binary mixtures will have
(iii) These will form minimum boiling azeotrope. same composition in liquid and vapour phase?
(iv) These will not form ideal solution. (i) Benzene – Toluene (ii) Water-Nitric acid
(iii) Water-Ethanol (iv) n-Hexane - n-Heptane
26. Relative lo wering of vapour pressure is a
colligative property because ___.
29. In isotonic solutions ___
(i) It depends on the concentration of a non
(i) solute and solvent both are same.
electrolyte solute in solution and does not depend on
(ii) osmotic pressure is same.
the nature of the solute molecules.
(iii) solute and solvent may or may not be same.
(ii) It depends on number of particles of electrolyte
(iv) solute is always same solvent may be different.
solute in solution and does not depend on the nature
of the solute particles.
30. Colligative properties are observed when __
(iii) It depends on the concentration of a non (i) a non-volatile solid is dissolved in a volatile liquid.
electrolyte solute in solution as well as on the nature (ii) a non-volatile liquid is dissolved in another volatile
of the solute molecules. liquid.
(iv ) It depends on the concentration of an electrolyte (iii) a gas is dissolved in non-volatile liquid.
or nonelectrolyte solute in solution as well as on the (iv) a volatile liquid is dissolved in another volatile liquid
nature of solute molecules.

15. Components of a binary mixture of two liquids A and B were being separated by distillation. After some time
separation of components stopped and composition of vapour phase became same as t hat of liquid phase. Both
the components started coming in the distillate. Explain why this happened.
Since both the components are appearing in the distillate and co mposition of liquid and vapour is same, this shows
that liquids have formed azeotropic mixture and hence cannot be separated at this stage by distillation.

16. Explain why on addition of 1 mol of NaCl to 1 litre of water, the boiling point of water increases, while
addition of 1 mol of methyl alcohol to one litre of water decreases its boiling point.
NaCl is a non-volatile solute, therefore, addition of NaCl to water lowers the vapour pressure of water. As a result
boiling point of water increases. Methyl alcohol on the other hand is more volatile than water, therefore its addition
increases, the total vapour pressure over the solution and a decrease in boiling point of water results.

17. Explain the solubility rule “like dissolves like” in terms of intermolecular forces that exist in solutions.
A substance (solute) dissolves in a solvent if the intermo lecular interactions are similar in both the components; for
example, polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents thus we can say “like
dissolves like”.

18. Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of
temperature, however molarity is a function of temperature. Explain.
Molarity of a solution is defined as the number of moles of solute dissolved in one litre of solution. Since volume
depends on temperature and undergoes a change with change in temperature, the mo larity will also change with
change in temperature. On the other hand, mass does not change with change in temperature, as a result other
concentration terms given in the question remain unchanged by changing temperature. According to the definition of
all these terms, mass of the solvent used for making the solution is related to the mass of solute.

19. What is the significance of Henry‟s Law constant K H ?

Higher the value of Henry‟s law constant KH , the lower is the solubility of the gas in the liquid.

20. Why are aquatic species more comfortable in cold water in comparison to warm water?
At a given pressure the solubility of o xygen in water increases with decrease in temperature. Presence of more
oxygen at lower temperature makes the aquatic species more comfortable in cold water.

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21. Why is the vapour pressure of an aqueous solution of glucose lower than that of water?
In pure liquid water the entire surface of liquid is occupied by the mo lecules of water. When a non-volatile solute,
for examp le glucose is dissolved in water, the fraction of surface covered by the solvent molecules gets reduced
because some positions are occupied by glucose molecules. As a result number of solvent molecules escaping from
the surface also gets reduced, consequently the vapour pressure of aqueous solution of glucose is reduced.

22. How does sprinkling of salt help in clearing the snow covered roads in hilly areas?
When salt is spread over snow covered roads, snow starts melt ing fro m the surface because of the depression in
freezing point of water and it helps in clearing the roads.

23. What is “semi permeable membrane”?

Continuous sheets or films (natural or synthetic) which contain a network of submicroscopic holes or pores through
which s mall solvent mo lecules like water can pass; but the passage of bigger mo lecules of solute is hindered, are
known as semi permeable membrane.

24. Give an example of a material used for making semipermeable membrane for carrying out reverse osmosis.
Cellulose acetate.

25. Give examples of biological and industrial importance of osmosis.

(i) Movement of water from soil into plant roots and subsequently into upper portion of the plant is partly due to
osmosis. (ii) Preservation of meat against bacterial action by adding salt. (iii) Preservation of fruits against bacterial
action by adding sugar. Bacterium in canned fruit loses water through the process of osmosis, shrivels and dies.
(iv) Reverse osmosis is used for desalination of water.

1. (i) 2. (i v) 3. (iii) 4. (ii), 5. (iii) 6. (ii), 7. (i) 8. (ii) 9. (ii) 10. (i) 11. (iii) 12. (i v) 13. (i) 14. (i) 15. (ii) 16. (ii) 17. (i)
18. (ii) 19. (iii) 20. (ii) 21. (i) 22. (i v) 23. (iii) 24. (i), (ii) 25. (iii), (i v) 26. (i), (ii) 27. (ii), (iii) 28. (ii), (iii) 29. (ii),
(iii) 30. (i), (ii)

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SUMMARY

Solutions are ho mogeneous mixtures of two or more co mponents. The co mponent that is present in the largest
quantity is known as solvent. And the components present in the solution other than solvent are called solutes.
Concentration of Solutions
(i) Mass percentage (w/w):
Mass percentage of a component = x 100

(ii) Volume percentage (V/V):

Volume percentage of a component = x 100
(iii) Parts per million (ppm):
Parts per million = x 106
(iv) Mole fraction ( x):
Mole fraction of a component, x =
(v) Molarity (M):
Molarity (M) is defined as number of moles of solute dissolved in one litre of solution.
Molarity = or
=

(vi) Molality (m):

Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent.
Molality (m) = or
=
(vii) Normality (N):
Normality (N) is defined as the number of gram equivalent of solute dissolved in one litre of solution.
Normality (N) =
(viii) Formality (F): It is the number of formula weights of solute present per litre of the solution.
Formality = moles of substance added to solution / volume of solution (in L)

Solubility
Solubility of a substance is its maximu m amount that can be dissolved in a specified amount of solvent at a specified
temperature.
Henry‟s law. The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional
to the partial pressure of the gas present above the surface of liquid or solution.
OR “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the
solution” and is expressed as:
p = KH x
Here KH is the Henry‟s law constant

Raoult’s law
It states that „the partial vapour pressure of an volatile component in a solution is directly proportional to its mole
fraction‟.
OR „partial vapour pressure of an volatile component in a solution is equal to the product of vapour pressure of the
component in pure state and its mole fraction.
ÓR, „the relative lowering of vapour pressure is equal to the mole fraction of non- volatile component present in the
solution‟.
Ideal Solutions:
The solutions which obey Raoult‟s law over the entire rang e of concentration are known as ideal solutions. Here
ΔH mix = 0, ΔV mix = 0

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In the case of ideal solution the force of attraction exist between A & A and B & B is same as that exist between A
& B. So the escaping tendency of A and B from solution found to be independent of each other.
Eg : solution of n-hexane and n-heptane,
bromoethane and chloroethane,
benzene and toluene.
Non-ideal Solutions:
When a solution does not obey Raoult‟s law over the entire range of concentration, then it is called non-ideal
solution.
In case of non-ideal solution with positive deviation from Raoult‟s law, the force of attraction exist
between A & B found to be weaker than those between A & A or B & B. So the escaping tendency of A and B from
solution increases. This will increase the vapour pressure and result in positive deviation.
Here , ΔH mix = +ve , ΔV mix = +ve
Eg: Mixtures of ethanol and acetone,
carbon disulphide and acetone.

In case of non-ideal solution with negative deviation from Raoult‟s law, the force of attraction exist between A & B
found to be greater than those between A & A or B & B. So the escaping tendency of A and B from solution
decreases. This will decrease the vapour pressure and result in negative deviation.
Here , ΔH mix = -ve , ΔV mix = -ve
Eg: mixture of phenol and aniline, a mixture of chloroform and acetone.
Colligative Properties
Properties of solution wh ich depend on the number of solute particles and are independent of their nature are called
colligative properties. They are: (1) relat ive lowering of vapour pressure of the solvent (2) depression of freezing
point of the solvent (3) elevation of boiling point of the solvent and (4) osmotic pressure of the solution.
1. Relative Lowering of Vapour Pressure:
.

is called relative lowering of vapour pressure.

2. Elevation in Boiling Point (ΔT b )
Elevation in boiling point is directly proportional to the molality of the solution.
i.e. △Tb α m or .△Tb=Kb m.
Where Kb is the proportionality constant.
If W2 g of solute of molecular weight M2 , dissolved in W1 g of solvent; then the molality of the solution;
m= W2 1000/W1 M2 .

Therefore .△Tb= Kb

3. Depression in freezing point (ΔTf ) :

Depression in freezing point is directly proportional to the molality of the solution.
i.e. △Tf α m or .△Tf=Kf m.
Where Kf is the proportionality constant.
If W2 g of solute of molecular weight M2 , dissolved in W1 g of solvent; then the molality of the solution;
m= W2 1000/W1 M2 .

Therefore .△Tf= Kf
4. Osmotic pressure (π) :
Osmosis is the spontaneous flow of solvent from solvent to solution or from dilute solution to concentrated solution,
when the two liquids are separated by a semipermeable membrane. The passage of solute molecules from
concentrated solution to dilute solution or from solution to solvent is called diffusion.

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As a result of osmosis, a hydrostatic pressure will be developed on the solution. The maximu m hydrostatic pressure
developed on a solution when it is separated from its solvent by a semipermeable membrane is called osmotic
pressure.
It can also be defined as the minimum pressure to be applied on the solution so that it is just su fficient to prevent the
entry of solvent to solution when the two liquids are separated by a semipermeable membrane.
Π=CRT
Where π- osmot ic pressure, v – volu me of solution containing „n‟moles of a solute, R- gas constant, T- kelv in
temperature c-concentration (molarity) of solution.
Since n =
πv = RT
Abnormal Molecular Masses:
Colligative properties, which depend only on the number of particles of solute, are measured to determine mo lecular
masses of solute. Any factor that affects the nu mber of part icles in solution will therefore give abnormal v alue for a
colligative property. Hence the molecular mass calculated also will be abnormal. The two factors that affects are;
A. Association:
Some solutes in solution undergo association. For example in ben zene, benzo ic acid undergoes dimerization through
hydrogen bonding. Therefore the number o f particles is reduced to half. The observed colligative property will be
half the expected value and molecular mass will be twice the theoretical value.
B. Dissociation :
Electrolytes like acids, bases, salts etc, undergoes dissociation in solu tion. For example, NaCl dissociate into Na+
and Cl- ions. This increases the number of particles in solution. Consequently the measured colligative property will
be higher and molecular mass will be less.
van’t Hoff Factor (i):
van‟t Hoff introduced a factor i, known as the van‟t Hoff factor, to account for the extent of dissociation or
association of solute in solution. It is the ratio of observed value of colligative property to the calculated value of
colligative property.
i = observed value of colligative property / calculated value of colligative property
or i = normal molecular mass / observed molecular mass
or i = number of particles after association or dissociation / number of particles initially
van‟t Hoff factor (i) > 1 for solutes undergoing dissociation and it is < 1 for solutes undergoing association.
For electrolytes the degree of dissociation α is related to the van t Holf factor by the expression,
α = where „r‟is the number of ions produced from 1 mole of the electro lyte. Eg : for NaCl ; r = 2 and for
Na2 CO3 , „r‟is 3.

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