UNIT- 2: SOLUTION
Points to Remember
1. Solution, Solute and Solvent. A homogeneous mixture of two or more chemically non-reacting
substances is known as solution. The substance present in largest amount is called solvent and the
other(s), called solute.
2. Miscible and Immiscible Liquids: Miscible liquids are those, which are completely soluble in each
other in all proportions while immiscible liquids are not soluble in each other.
3. Concentration in per cent.
% by mass =
Mass of solute
×100
Mass of solution
% by mass-volume =
% by volume 
Mass of solute g
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑙
Volume of solute
 100
Volume of solution
× 100
4. Mole fraction.
Sum of mole-fractions of all components is unity i.e.
5. Solubility. Maximum amount of a solute in a saturated solution per 100 g of a solvent. Like
dissolves like. Ionic and polar substances are generally soluble in polar solvents like water and are
insoluble in non-polar solvents like benzene, chloroform, carbon disulphide etc. Similarly, non-polar
solutes like iodine, sulphur, phosphorus and organic substances are soluble in non-polar solvents like
benzene, chloroform etc. and are insoluble in polar solvents like water.
6. Molarity (M). Number of moles of solute per litre of solution.
M=
w ×1000
m× V 𝑚𝑙
g-mol L-1
7. Molality (m). Number of moles of solute per kg of solvent.
8. Equivalent weight
9. Normality. Number of gram-equivalents of solute per litre of solution.
10. Parts per million (ppm). This unit is convenient unit to express the concentration of solute when
it is present in very small amounts.
11. Relation between strength, molarity and normality of a solution.
12. Molarity of a mixture solution (M) is given by:
13. Henry’s law: It states that the solubility of a gas in a liquid is directly proportional to the
pressure of the gas at a given temperature. If X is the mole fraction of gas in solution, then:
where p is partial pressure of gas and
is Henry‟s constant.
14. Vapour pressure. In case of pure liquid, it is the pressure exerted by vapour of liquid over surface
of liquid when vapour and liquid are in equilibrium at a given temperature. It increases with
temperature. In case of a solution of two volatile liquids, each liquid vapour exerts a partial pressure,
and total pressure is equal to sum of partial pressures. Partial pressure of each component is given by
Raoult‟s law.
15. Raoult’s law. For non-volatile solute, relative lowering of vapour pressure,
16. Colligative properties: Those properties of solution, which depend upon the concentration of
solute particles but are independent of nature of solute. These are: relative lowering of vapour
pressure, elevation in boiling point, depression in freezing point and osmotic pressure of solution.
17. Boiling point: The temperature at which vapour pressure of liquid or solution becomes equal to
atmospheric pressure.
18. Molal elevation constant (Kb). The elevation in boiling point of the solution when its molality is
unity. It is also called molal ebullioscopic constant and is represented by Kb. It‟s unit are km–1 or kg
mol–1.
19. Elevation in Boiling point
solvent.
Boiling point of a solution is always higher than that of a pure
20. Freezing point . The temperature at which solid, liquid and vapours are in equilibrium i.e. they
have same vapour pressure.
21. Molal Depression Constant (Kf): It is equal to depression in freezing point of a solution whose
molality is unity i.e. 1 mole of solute is dissolved in one kg of solvent. It is represented by Kf.
.
22. Semi permeable membrane. A membrane which allows a selective passage of solvent molecules
is called a semi-permeable membrane. Cellophane, parchment paper and the walls of living cells are
examples of semi-permeable membrane.
. Osmosis. The passage of solvent molecules from pure solvent (or solution of low concentration) to
the solution of higher concentration through the semi-permeable membrane.
24. Osmotic pressure ( ). The excess pressure that must be applied to the solution side to just
prevent the passage of solvent into the solution through semi-permeable membrane. It is a colligative
property of solution and is given by relation:
where M is molarity of solution.
25. Isotonic solutions. The solutions which have same osmotic pressure.
26. van‟t Hoff factor.
In case of abnormal solutions (solutions showing association or dissociation of solute particles), the
colligative properties are expressed by following relations:
The value of i is equal to 1 in case of normal solution, i > 1 in case of dissociation, and
i < 1 in case of association.
27. The units of
are degrees/molality i.e. K/m or K kg mol–1.
28. Molality, mole-fraction, mass fraction etc. are preferred over molarity, normality, etc. because the
former units involve only masses which do not change with temperature while the latter involve
volumes which change with temperature.
29. A 0.91% NaCl solution is isotonic with red blood corpuscles (RBC). NaCl solution with
concentration greater than 0.91% is called ‘hypertonic’ and less than this is called ‘hypotonic’.
30. „Reverse Osmosis‟ is the process of flow of solvent from solution to solvent side when a pressure
higher than osmotic pressure is applied to solution separated from solvent by a semi permeable
membrane.
31. For comparing the values of colligative properties of different solutions, express all given
concentrations into mol L–1 of ions (in case of electrolytes). Greater the molar concentration of
particles, greater is the value of colligative property.
Very Short Answer Type Questions (1mark)
1.What are solutions?
Ans. Homogeneous mixture containing two or more than two components.
2.What are binary solns?
Ans. Binary solution : Solutions containing only two components.
3. Define mole fraction
Ans.. Ratio of number of moles of a component to the total number of moles of solutions.
4 . Define molarity
Ans. Number of moles of solute dissolved in one litre of solution.
5. Define molality
Ans.Number of moles of the solute dissolved in 1 kg of the solvent.
6. State Henry’s law
Ans.The partial pressure of the gas in vapour phase is directly proportional to the molefraction of the
gas in the solution.
P = KH x
7. Mention one application of Henry’s law
Ans.To increase the solubility of CO2 in soft drinks , the bottle is sealed under high pressure.
8. State Raoutlt’s Law
Ans.For a solution of volatile liquids, the partial vapour pressure of each component in the solution is
directly proportional to its mole fraction.
P1 = p01 x1
9. What are Ideal solutions?
Ans.Solutions which obey Raoult‟s law . solute –solvent interaction is nearly equal to solute –solute or
solvent – solvent interaction. Δmix H = 0 and Δmix V = 0
Eg: n- hexane and n-heptane
10. What are non-ideal solutions?
Ans. Solutions which do not obey Raoult‟s law . Inter molecular force of attraction between solute –
solvent is Weaker or stronger than those between solute –solute or solvent – solvent interaction.
Δmix H ≠ 0 Δmix V ≠ 0
11. Define colligative property
Ans.Properties which depend on the number of solute particle irrespective of their nature relative to
the total number of particle present in the solution.
12. Define Ebullioscopic constant
Ans. Elevation of boiling point of one molal solution.
13. Define Cryoscopic constant
Ans. Depression of freezing point of one molal solution.
14. Define Osmosis
Ans. Flow of solvent molecule through semi permeable membrane from pure solvent to the solution.
This process of flow of the solvent is called osmosis.
15. Define Osmotic Pressure
Ans. Excess pressure that must be applied to a solution to prevent osmosis.
16. What is reverse osmosis
Ans. Pure solvent flows out of the solution through the semi permeable membrane when we apply
pressure larger than the osmotic pressure. Application of reverse osmosis is desalination of seawater.
17. What are isotonic solution
Ans.Two solutions having same osmotic pressure at a given temperature.
18. Define van’t Hoff factor
Ans.
i = Normal molar mass / abnormal molar mass
i= observed colligative property / calculated colligative property.
Short Answer Type Questions (2mark)
1.18 g of glucose, C6H12O6(Molar Mass = 180 g mol-1) is dissolved in 1 kg of water in a sauce pan.
Atwhat temperature will this solution boil? (Kbfor water = 0.52 K kg mol–1, boiling point of pure
water = 373.15 K)
Ans.
2. Explain the following:
(i)Henry’s law about dissolution of a gas in a liquid.
(ii)Boiling point elevation constant for a solvent
Ans. (i)Henry‟s lawstates that the solubility of a gas in a liquid is directly proportional to the pressure
of the gas.
(ii) The boiling point elevation constant may be defined as the elevation in boiling point when
one mole of a non-volatile solute is dissolved in one thousand grams of solvent
3. A solution of glucose (C6H12O6) in water is labeled as 10% by weight. What would be the
Molality of the solution? (Molar mass of glucose = 180 g mol–1)
Ans.
4. Differentiate between molarity and molality for a solution. How does a change in temperature
influence their values?
Ans.
5. Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a
Colligative property? Explain.
Ans.
6. Explain whya) Aquatic animals are more comfortable in cold water than in warm water.
b) Deep sea divers are advised not to come to surface immediately from deep waters.
(Ans) (a) This is because “Kh” (Henry constant) values for both N2 and O2 increase with increase in
temperature indicating that the solubility of gases increases with decrease in temperature.
(b) Deep sea divers are advised not to come to surface immediately from deep waters because
sudden change in outside pressure can be fatal for divers because N2 will bubble out of blood vessels
causing severe pain and can be dangerous.
7. Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous
solution. What should be the molarity of such a sample of the acid if the density of solution is
1.504 g mL–1?
Sol.
Percentage mass (%) and molarity (M) of the solution are related as
Molarity (M) =
% mass  density  10
Molar mass of solute (HNO 3 )
68  1.504( g mL1 )10
=
= 16.23 M
68 (g mol -1 )
8. A sample of drinking water was found to be severely contaminated with chloroform, CHCl 3,
supposed to be carcinogen. The level of contamination was 15 ppm (by mass)
(i)
Express this in percent by mass.
(ii)
Determine the molality of chloroform in the water sample.
Sol.
Mass of solution (wA + wB) = 106 g
Mass of solute (wB)
= 15 g
% mass =
w B  100 15  100
= 1.5 × 10–3%

6
w A  wB
10
Mass of solvent (wA) = 106 – 15 = 999985 g
Molality (m) =
w B  10 3
15(g)  10 3 (g)

MM B  w A 119.5 (g mol -1 )  999985 g
= 1.25 × 10–4 m
9. Non-ideal solutions exhibit either positive or negative deviations from Raoult's law. What
are these deviations and why are they caused? Explain with one example for each type.
Ans.
10.State Raoult’s law for a solution containing volatile components. How does Raoult’s
law become a special case of Henry’s law?
Ans.
Short Answer Type Questions (3 Mark)
1. 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution
freezes at – 0.34°C. What is the molar mass of the material?
(Kf for water = 1.86 K kg mol-1)
Ans.
WB= 15.0g;WA= 450g; ΔKf= 1.86 K kg molL-1
ΔTf = 0°C– (– 0.34°C) = 0.34°C or 0.34 K
2.Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10–2 g of K2SO4 in 2
L of water at 25 oC, assuming that it is completely dissociated.
(R= 0.082 L atm K–1 mol–1, Molar mass of K2SO4 = 174 g mol–1).
Ans.
3) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of
benzene by 0.40 K. Find the molar mass of the solute. (Kf for benzene = 5.12 kg mol–1 )
ans.
4. Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.00 g of
water. (Kb for water = 0.512 kg mol–1),(Molar mass of NaCl = 58.44 g)
ans.
5.A solution prepared by dissolving 1.25 g of oil of wintergreen (methyl salicylate) in 99.0 g of
benzene has a boiling point of 80.31°C . Determine the molar mass of this compound.
(B.P.of pure Benzene =80.10°C and Kb for benzene = 2.53°C kg mol–1)
Ans.
6. What mass of NaCl (molar mass= 58.5 g mol–1) must be dissolved in 65 g of water to lower the
freezing point by 7.5°C ? The freezing point depression constant, Kf , for water is 1.86 K kg mol–
1
.Assume van’t Hoff factor for NaCl is 1.87.
Ans.
7. What concentration of nitrogen should be present in a glass of water at room
Temperature ? Assume a temperature of 25° C, total pressure of 1 atmosphere and mole
fraction of nitrogen in air of 0.78. [KH for nitrogen = 8.42 × 10–7 M/mm Hg]
Ans.
8. Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr2 in 200 g of
water. (Molar mass of MgBr2= 184 g/mol)(Kf for water = 1.86 K kg /mol )
Ans.
9. Vapour pressure of water at 298 K is 17.535 mm Hg. Calculate vapour pressure of water
at 293 K when 25 g of glucose is dissolved in 450 g of water.
Ans.
p 0A = 17.535 mm Hg. Let vapour pressure of solution be ps
p 0A  p s
p 0A
or
 x B or
1-
ps
p 0A
= xB or ps = p0A (1  xB )
ps = p 0A × xA = 17.535 ×
450 / 18
25 / 180  450 / 18
= 17.535 × 0.9945 = 17.438 mm Hg

17.44 mm Hg.
10.(a) What happens to blood cells when they are placed in water containing less than 0.99%
(mass/volume) salt?
(b) What care should be taken while preparing intravenous injection?
(c) Why osmotic pressure is better technique for determination of molar mass of
biomolecules?
Ans.
(a) The blood cells collapse due to loss of water by osmosis when placed in water containing less
than 0.9% (mass/volume) salt.
(b) They should be prepared in aquatic medium and salt concentrations should be same as blood
plasma levels.
(c) Osmotic pressure is better technique for the determination of molar mass of biomolecules because
they are generally not stable at higher temperatures and have poor solubility. Therefore, other
techniques like elevation in boiling point and depression in freezing point cannot be used to determine
molecular masses of biomecules.
HOTS QUESTIONS
Q.1)Two elements A and B, form compounds having molecular formula AB 2 and AB4. When
dissolved in 20 g of C6 H6, 1 g of AB2 lowers the freezing point by 2.3 K, whereas 1.0 g of AB4
lowers it by 1.3 K. The molal depression constant for benzenes is 5.1 K kg mol –1. Calculate
atomic mass of A and B.
Sol.
We know,
MB =
1000  K f  w B
w A  Tf
1000  5.1 1
= 110.87 g mol–1
20  2.3
1000  5.1 1
MAB4 
= 196.15 g mol–1
20  1.3

MAB2 
If a and b are respectively atomic masses of A and B then,
MAB2 = a + 2b
or
a + 2b = 110.87
MAB4 = a + 4b
or
a + 4b = 196.15
Solving the simultaneous equations we have
a = 25.58 u; b = 42.64 u.
Q.2) If density of some lake water is 1.25 g mL–1 and it contains 92.0 g of Na+ ions per kg of
water. Calculate the molality and molarity of Na+ ions in lake.
Sol.
Mol . of Na  ions
Mass of w ater (kg)
Molality =
=
92(g) / 23(g mol -1 )
=4m
1(kg)
Molarity =
=
=
Mol . of Na  ions
Vol. of w ater (L)
92(g) / 23 (g mol -1)
1000 (g) / 1.25  10 3 (gL-1 )
92(g) 1.25  10 3 (g L-1)
23 (g mol -1 )  1000 (g)
= 5M
Q.3) 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point
observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of the acid.
Kf for H2O = 1.86 K kg mol–1 and density of solution is 1.124 g mL–1.
Sol.
Mass of solution = 500 + 19.5 = 519.5 g
Volume of solution =
519 .5 (g)
= 462.18 mL
1.124 (g mL -1 )
19.5  1000
Molarity of solution =
= 0.541 mo L–1
78  462 .18
Tf = i × Kf × m
or
Also
1 78  500
= 1.0753
1.86  19.5  1000
i  1 1.0753  1

diss. =
= 0.0753
m 1
2 1
diss. = K a / c
or
Ka = 2diss. × c
i=
= (0.0753)2 0.541 = 3.07 × 10–3
.4) The air is a mixture of a number of gases. The major components are oxygen and nitrogen
with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium
with air at a pressure of 10 atm. At 298 K, if the Henry’s law constant for oxygen and nitrogen
are 3.30 x 107 mm and 6.51 x 107 mm respectively, calculate the composition of these gases in
water.
Sol.
Consider 1 mol of air and its volume at 10 atm be V
V  20
V  79
= 0.2 V;
Volume of N2 =
= 0.79 V
100
100
10  0.2V
Partial pressure of O2 (pO2 ) =
= 2 atm.
V
10.  0.79 V
Partial pressure of N2 (pN2 ) =
= 7.9 atm .
V
2  760 (mm)
Solubility of O2 x O2 = p O2 / KH 
= 4.606 × 10–6 .
7
3.30  10 (mm)
2  760 (mm)
Solubility of N2 x N2 = pN2 / KH 
= 9.22 × 10–5
7
6.51 10 (mm)
Volume of O2 =


 
Q.5) How many mL of a 0.1 M HCl are required to react completely with 1 g mixture of
Na2CO3 and NaHCO3 containing equimolar amounts of two?
Sol. Let Na2CO3 be x g

x
mol
106
Mol of Na2CO3 (n1) =
NaHCO3 = (1- x) g

Mol of NaHCO3 (n2) =
Na 2CO 3

2HCl  NaCl  CO 2  H2O
2 mol
1mole
NaHCO 3
(1  x )
mol
84

1mole
HCl  NaCl  CO 2  H2O
1mol
HCl required for
x
2x
mol of Na2CO3 =
mol
106
106
HCl required for
(1  x )
(1  x )
mol of NaHCO3 =
mol
84
84
Since given mixture contains equimolar amount

x
(1  x )

or
106
84
x = 0.56
Total HCl required
 2x
1 x 
= 


106 84 
= 0.01578 mol
If V is the volume of HCl required, then
V(L) × 0.1 = 0.01578
or
VL = 0.1578 L
or
2  0.56 1  0.56

106
84
157.8 mL
VALUE BASED QUESTIONS
1. Bharath went to his grandfather’s house in winter this year. As usual he went for fishing.
His grandmother told him there will be no fishes in the lake. He noticed that it was more
difficult to find fishes in winter. The fishes were deep inside the river. Whereas in summer
they were on the surface and hence he was able to catch fishes.
a) Why are fishes on the surface in water than in the depth in summer?
b) What value can be derived from this?
Ans:
a) According to Henry‟s law at low temperature gases are more soluble and hence as
more oxygen gets dissolved in water fishes survive better even in depth of the river.
In summer as the oxygen is less in water the fishes come to the surface.
b) The value that I derive from this is wisdom is superior to knowledge
2. Scuba divers when come towards the surface, the pressure gradually decreases resulting
in the release of dissolved gases leading to formation of bubbles of nitrogen gas in the
blood which blocks the capillaries and thus harmful effects are created. Toavoid bends
and toxic effects of high concentration of nitrogen gas, the air is diluted with helium.
After reading the above passage , answer the following questions.
(i)
Which law is associated with this?
(ii)
Why is the condition of bends overcome by the use of Helium
(iii)
Mention the value associated with providing divers air diluted with helium.
Ans: (i) Henry‟s Law
(ii) Solubility inversely proportional to KH.
(iii )Critical thinking and self awareness