answer for cbse chenistry

1. 1
Solutions
Solution and Their types
1. Define ‘solution’?
Solution is a homogeneous mixture of two or more components. (Chemically non-reactive substances)
2. What is binary solution?
A solution containing only one solute dissolved in a solvent is called binary solution.
Solvent: The component that is present excess in the solution
Solute: The other components other than solvent are called solute.
3. What are the different types of solutions? Write briefly about each type with an example.
A solution can be solid, liquid or a gas depending upon the physical state of the solvent.
There are three types of solutions.
(i) Gaseous solution:
The solution in which the solvent is a gas is called a gaseous solution. The solute in these solutions may
be liquid, solid, or gas.
For example
Gas in gas oxygen and nitrogen gas is a gaseous solution.
Liquid in gas Chloroform and nitrogen gas
Solid in gas Camphor in nitrogen gas
(ii) Liquid solution:
The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions
may be gas, liquid, or solid.
For example
Gas in liquid oxygen in water
Liquid in liquid Ethanol in water
Solid in liquid glucose in water
(iii) Solid solution:
The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid
or solid.
For example,
Gas in solid hydrogen in nickel
Liquid in solid mercury with sodium
Solid in solid copper in gold
Concentration of solutions
4. Define the term ‘concentration?
Composition of a solution is expressed in terms of concentration. It is the amount of the solute
present in a known quantity of the solvent or solution.
Qualitatively it is called dilute or concentrated solution.
Quantitatively they are expressed differently.
5. Define the following terms:
(i) Volume percentage ii) Mass percentage iii)mass by volume percentage
Mass percentage: (w/w) The mass of the solute (in grams) present in 100 g of the solution.
𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒+ 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
x 100

2. 2
Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22g of
benzene is dissolved in 122 g of carbon tetrachloride.
𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒+ 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
x 100
𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑏𝑒𝑛𝑧𝑒𝑛𝑒 =
22
144
𝑥 100 = 15.28%
𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑏𝑒𝑛𝑧𝑒𝑛𝑒 =
122
144
𝑥 100 = 84.72%
Volume percentage: (v/v) The volume of the solute (in ml) present in 100 ml of the solution.
𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑚𝑙
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
=
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑚𝑙
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑥100
Mass by volume percentage: (w/v) The mass of the solute present (in g) in 100 ml of the solution.
Mass volume percentage =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒+ 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑥 100
2 ml distilled water is mixed with 6g of substance and the final volume become 50ml. What is the
mass by volume percentage?
Mass volume percentage =
6
50
𝑋 100 = 12 % or 12 gms per 100ml
6. Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Parts per million
Mole fraction (x): It is defined as the ratio of the number of moles of a particular component to the
Total number of moles of all the component present in the solution.
𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
Calculate the mole fraction of ethanol and water in a sample of rectified spirit which contains
46% ethanol by mass?
Mass of ethanol = 46 g
Mass of water = 100 – 46 = 54g
𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝐸𝑡ℎ𝑎𝑛𝑜𝑙 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑒𝑡ℎ𝑎𝑛𝑜𝑙
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒𝑡ℎ𝑎𝑛𝑜𝑙+ 𝑤𝑎𝑡𝑒𝑟
𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝐸𝑡ℎ𝑎𝑛𝑜𝑙 =
46
46
⁄
46
46
⁄ + 54
18
⁄
= 0.25
Calculate mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1
.
Moles of KI =
20
166
= 0.12 moles
Moles of solvent in kg =
80
18
= 4.44
Mole fraction of KI =
0.12
4.56
= 0.96
Molarity (M): It is defined as the number of moles of solute present in 1000 ml of the solution.

3. 3
M𝒐𝒍𝒂𝒓𝒊𝒕𝒚 =
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆
𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒎𝒍
𝒙 𝟏𝟎𝟎𝟎
0.25 M solution of NaOH means that 0.25 mol of NaOH is dissolved in one litre of solution.
Calculate the molarity of the following solutions containing 30 g of Co(NO3)2. 6H2O in 4.3 L of
solution
Moles of cobalt nitrate =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑜((𝑁𝑂3)26𝐻2𝑂
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐶𝑜((𝑁𝑂3)26𝐻2𝑂
=
30
291
= 0.103 𝑚𝑜𝑙
Molarity =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒
=
0.103
4300
𝑥 1000 = 0.023 𝑀
It means 0.023 moles present per litre.
Calculate the molarity of the following solutions containing 30 mL of 0.5 M H2SO4 diluted to 500
mL.
Dilution formula
V1M1 = V2M2
X (500) = 30 x 0.5
X = 0.03 M
Calculate molarity of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1
.
Mass of KI = 20g
Molar mass of KI = 166g
Molarity =
𝒅𝒆𝒏𝒔𝒊𝒕𝒚 𝒙 𝟏𝟎𝟎𝟎
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
Molarity =
𝒅𝒆𝒏𝒔𝒊𝒕𝒚 𝒙 𝒑𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒙 𝟏𝟎
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
=
𝟏.𝟐𝟎𝟐 𝐱 𝟐𝟎 𝐱 𝟏𝟎
𝟏𝟔𝟔
= 1.44 M
Molality (m): It is defined as the number of moles of the solute present in 1000 grams of the solvent.
M𝒐𝒍𝒂𝒍𝒊𝒕𝒚 =
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝒊𝒏 𝒈𝒓𝒂𝒎𝒔
𝒙 𝟏𝟎𝟎𝟎
Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous
solutions.
Molar mass urea = 60 g mol-1
Mass of solute = 0.25 x 60 = 15 g of urea
0.25 molal aqueous means 1015g of solution contains 15 g of urea.
Therefore for 2500g of aqueous solution =
15 𝑥 2500
1015
= 36.95 g of urea
Calculate molality of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1
.
Mass of KI = 20g
Molar mass of KI = 166g
Moles of KI =
20
166
= 0.12 moles
Mass of solvent in kg =
80
1000
= 0.08
Molality =
0.12
0.08
= 1.5 m
Parts per Million: It is defined as the mass of the solute present in one million parts by mass of the
solution
ppm =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒈𝒓𝒂𝒎𝒔
𝑿 𝟏𝟎𝟔

4. 4
ppm is usually used to express very dilute solution
One litre of sea water weight 1030g and contains about 6×10-3
g of dissolved 02. Calculate the
concentration of dissolved oxygen in ppm?
Mass of oxygen = 6 x 10-3
g
ppm of oxygen in 1030g of sea water =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑒𝑎 𝑤𝑎𝑡𝑒𝑟
𝑥 106
=
6 𝑥 10−3
1030
𝑥 106
= 5.8 𝑝𝑝𝑚
7. What is the advantage of molality over molarity?
Molarity, normality and volume percentage change with temperature since volume is directly
proportional to temperature. But molality does not change since it has no volume term.
8. What is formality?
It is the number of formula mass of solute present in 1000 ml of the solution.
F𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 =
𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝑙
𝑥 1000
Liquid Solutions
9. What is liquid solution? Mention their types
The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions
may be gas, liquid, or solid. It is very common types of solutions.
For example,
Solid in liquid ex: sodium chloride in water
Gas in liquid ex ; carbondioxide in water
Liquid in liquid ex: ethanol in water.
10. What is saturated solution?
Solution in which no more solute can be dissolved at a given temperature and pressure is called a
saturated solution. At this stage the crystallization and dissolution are in dynamic equilibrium.
11. Define the term – solubility
The maximum amount of a substance that can be dissolved in a specified amount (100g) of solvent
at equilibrium (dissolution and crystallization) is called its solubility.
Solid in Liquid
12. Name the factors affecting solubility of a solid substance in a liquid?
Nature of solute
Nature of solvent
Temperature
13. How nature of solute and solvent affect the solubility?
Like dissolves like
Ionic compounds like NaCl soluble in polar solvents like water
Covalent compounds like Ether soluble in non-polar solvents like benzene
Ionic compound like NH4Cl insoluble in non-polar solvents like benzene
14. What is the effect of temperature on the solubility of a solid in a liquid?
In a nearly saturated solution, if the dissolution process is endothermic (Δ Hsol > 0), the solubility
would increase with rise in temperature
NH4Cl, KNO3, NaCl, NaNO3 and AgNO3 are more soluble at higher temperature since the process of
dissolution is endothermic.
In a nearly saturated solution, if the dissolution process is exothermic (ΔH sol < 0) the solubility would
decrease with temperature.
The solubility of Li2SO4, Ce2(SO4)3 and Na2CO3 decrease with increase in temperature.

5. 5
15. What is the effect of pressure on the solubility of a solid in a liquid?
Since solids and liquids are highly incompressible, pressure does not have any significant effect on
solubility of solids in liquids.
Gas in Liquid
16. Name the factors affecting solubility of a gaseous substance in a liquid?
Nature of solute
Nature of solvent
Temperature
Pressure
17. How the nature of gas and the solvent affects the solubility of gas in a liquid?
Gases like nitrogen, oxygen, helium and hydrogen is less soluble in water since they do not form a
compound with water.
But gases like ammonia, sulphur dioxide, Hydrogen chloride are highly soluble in water since they
form compound ammonium hydroxide, sulpurous acid and hydrochloric acid respectively.
Hence the most soluble gases in solvent are the one which forms a compound with solvent.
18. What is the effect of temperature on the solubility of a gas in a liquid?
The dissolution of a gas in a liquid is an exothermic process
Gas + solvent ⇋ solution + heat
Therefore, According the Lechatlier’s principle gases dissolves more at low temperature.
19. Why aquatic animals are more comfortable in cold water than warm water?
The dissolution of gas in water is exothermic. So, according to lechatlier’s principle when the
temperature is less, solubility of oxygen in the water is more than warm water. So the aquatic animals
are more comfortable in cold water than warm water.
20. What is the effect of pressure on solubility of a gas in a liquid?
The solubility of a gas increases with increase of pressure. The effect of pressure on the solubility of
gas is given by Henry’s law.
21. State Henry’s Law.
The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the
solution at constant temperature.
P 𝛼 S
P = KH S KH is the Henry’s law constant.
22. Mention the significance of Henry’s constant
1. Henry’s constant is a function of the nature of gases. Different gases have different KH values.
2. Greater the value of KH, lower the solubility of the gas at constant pressure at a particular
temperature
3. For a particular gas, The value of KH increase with increase in temperature implying that the
solubility decrease with increase of temperature at constant pressure
23. Mention the applications of Henry’s constant
In the production of carbonated beverages: To increase the solubility of CO2 in soft drinks and soda
water, the bottle is sealed under high pressure.

6. 6
In the deep sea diving: Scuba divers carry oxygen cylinders to breath under deep sea. At high pressure
under water, the solubility of atmospheric gases (oxygen and nitrogen) in blood increases. Oxygen is
consumed for metabolic process. When the divers come towards surface, the pressure gradually
decreases. This releases the dissolved gas (nitrogen but unutilised) and leads to the formation of
bubbles in the blood. This blocks capillaries and creates a medical condition known as bends. To avoid
bends the cylinders used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2%
nitrogen and 32.1% oxygen). Because He is less soluble than nitrogen as well as small in size which
can easily pass through the capillaries.
For people living in higher altitudes or climbers: At high altitudes the partial pressure of oxygen in
the air is less than that at the ground level. This leads to low concentrations of oxygen in the blood and
tissues of people living at high altitudes or climbers. Low oxygen in the blood causes climbers to
become weak and unable to think clearly. This condition is known as anoxia.
24. What are the limitations of Henry’s law?
Henry’s law applicable only
When Pressure of a gas is not too high and temperature is not too low.
The gas should not undergo any chemical change.
The gas should not undergo association or dissociation in the solution.
25. The partial pressure of ethane over a solution containing 6.56 x 10-2
g of ethane is 1 bar. If the
solution contains 5.00 x 10-2
g of ethane, then what shall be the partial pressure of the gas?
Molar mass of ethane = 30 g mol-1
Moles of ethane =
6.56 𝑥 10−3
30
= 2.187 𝑥 10−4
Henry’s law = p = KH X
𝑃1
𝑋1
=
𝑃2
𝑋2
1
6.56 𝑥 10−3
30
=
𝑃2
5 𝑥 10−2
30
P2 = 0.7622 bar
Liquid in Liquid
(Both solvent and solute are volatile)
26. State Dalton’s law of partial pressure
The total pressure of mixture of non-reacting gases is equal to the sum of the partial pressures of
components 1 and 2
Ptotal = P1 + P2
27. State Raoult’s law
The partial vapour pressure of any volatile components over the solution is equal to the vapour pressure
of the pure component multiplied by its mole fraction.
P1 α X1 P2 α X2 (Raoult’s law )
P1 = P1
0
X1 P2 = P2
0
X2
Ptotal = P1 + P2 (Dalton’s law of partial pressure)
Ptotal = P1
0
X1 + P2
0
X2
Ptotal = P1
0
(1- X2) + P2
0
X2
Ptotal = P1
0
- P1
0
X2 + P2
0
X2
Ptotal = P1
0
- ( P2
0
- P1
0
) X2

7. 7
28. Plot a graph between vapour pressure and mole fraction of a solution obeying Raoult’s Law at
constant temperature?
In the plot of vapour pressure (p) versus mole fraction(x) diagram described below
X1 and X2 denote the molefraction of the constituents present in the liquid state
Since, both the constituents are volatile, the vapour above them will contain both of them but richer
in the more volatile component.
i ) Ptotal = P1
0
- ( P2
0
- P1
0
) X2
Since P2
0
, P1
0
are the constant at a particular temperature , the total pressure is the linear
function of the mole fraction X2 (or X1) , thus a straight line is obtained.
ii ) When X1 = 1, the liquid is pure contains only component 1
P = P1
0
When X2 = 1, the liquid is pure contains only component 2
P = P2
0
iii) The composition of the vapour phase in equilibrium with the liquid can be calculated from the
partial pressures of the two components (in the vapour phase). If Y1 and Y2 represent the mole
fractions of the components 1 and 2 respectively in the vapour phase, then
since
Partial pressure = Molefraction X Total pressure
Y1 Ptotal = P1 and Y2 Ptotal = P2
iv) Depending upon the vapour pressures of the pure components, total vapour pressure over the
solution may increase or decrease with increase in the mole fraction of a component.
29. Give the characteristics of ideal solution?
An ideal solution is formed from two liquids only when –
a) They obey Raoult’s Law Pobs = PRaoults ( PA
o
XA + PB
o
XB)
b) ΔHmix = 0
c) ΔVmix = 0
d) The various inter molecular forces are identical.
E.g. solutions of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene etc.

8. 8
30. What are the possible deviations from ideal behaviors?
A solution which does not obey Raoult’s law is called non-ideal solution.
Positive deviation – Pobs > PRaoults ( PA
o
XA + PB
o
XB)
Negative deviation –Pobs < PRaoults ( PA
o
XA + PB
o
XB)
a) Positive deviation:
In the case of positive deviation from Raoult’s law,
 A-B interactions are weaker than A-A and B-B interactions.
 i.e., in this case solute-solvent interactions are weaker than solute-solute and solvent-
solvent
 So more molecules are escaped to vapour phase and hence the vapour pressure of the
solution increases.
 E.g. solutions of ethanol and acetone, acetone and CS2, acetone and CCl4 etc.
b) Negative deviation:
In case of negative deviation from Raoult’s law,
 The A-B interactions are stronger than A-A and B-B interactions.
i.e. solute-solvent interactions are stronger than solute-solute interaction and solvent-
solvent interaction.
 So number of molecules escaped to vapour phase decreases and hence the vapour
pressure of the solution decreases.
 E.g. solution of phenol and aniline, chloroform and acetone etc.
31. Define the term azeotrope?
Azeotropes
They are binary mixtures having the same composition both in liquid phase and vapour phase and boil
at a constant temperature. For such solutions, it is not possible to separate the components by fractional
distillation.
There are two types of azeotropes:
i) Minimum boiling azeotrope
The solutions which show a large positive deviation from Raoult’s law form minimum
boiling azeotrope at a particular composition.
E.g. 95% ethanol solution by volume.
ii) Maximum boiling azeotrope
The solutions which show large negative deviation from Raoult’s law form maximum
boiling azeotrope at a particular composition.
E.g. a mixture of 68% Nitric acid and 32% water by mass forms a maximum boiling
azeotrope at 393.5 K.

9. 9
Solid in Liquid
(Solvent is volatile and Solute is non-volatile)
32. Define the term colligative properties?
The properties of solution which depends upon number of particles present in the solution are called
colligative properties.
For example when sodium chloride is added to the solvent the properties of solvent such boiling point and
melting point changes.
33. Name different colligative properties?
The colligative properties are the properties of solvent when a non-volatile solute is added to
solvent.
a) Relative lowering of vapour pressure.
b) Elevation in boiling point.
c) Depression in freezing point
d) Osmotic pressure
34. Obtain a relationship between relative lowering of vapour pressure and mole fraction of solute?
On adding non-volatile solute to a solvent, it vapour pressure always decreases. This decrease in vapour
pressure of solvent is called lowering of vapour pressure.(∆𝑃 = 𝑃 0
-P)
According to Raoult’s law
P1 = X1P1
0
∆𝑃 = 𝑃1
0
− 𝑃
∆𝑃 = 𝑃1
0
− X1P1
0
= 𝑃1
0
(1 − X1) = X2
∆𝑃
𝑝1
0 =
𝑃1
0− 𝑋1 𝑃1
0
𝑝1
0 =
𝑃1
0(1− X1)
𝑝1
0 = 𝑋2
𝑷𝟎− 𝑷
𝑷𝟎
=
𝒏𝟐
𝒏𝟐+ 𝒏𝟏
𝑷𝟎− 𝑷
𝑷𝟎
=
𝒘𝟐
𝑴𝟐
𝒙
𝑴𝟏
𝒘𝟏
𝐏𝟎− 𝐏
𝐏𝟎 = = Molarity x
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭
𝟏𝟎𝟎𝟎
35. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850
g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

10. 10
36. Why is the boiling point elevated when a non – volatile solute is dissolved in a liquid?
When a non – volatile solute is added the vapour pressure decreases and the solution is heated to a
higher temperature, increasing the boiling point.
37. How is boiling point changed when mass of solvent is doubled?
38. 0.90g of a non – electrolyte was dissolved in 87.90g of benzene. This raised the boiling point of
benzene by 0.250C. If the molecular mass of non – electrolyte is 103.0 g/mol, calculate the molal
elevation constant for benzene?
39. Show graphically the depression in freezing point on adding a non volatile solute?

11. 11
40. Define Cryoscopic constant?
When 1 mole of a solute (that neither dissociates nor associates) is dissolved in 1kg of
solvent, the depression in freezing point is called cryoscopic constant.
41. When 20g of a non – volatile solid is added to 250 ml of water, the freezing point of water
becomes -0.90
C. Calculate molecular mass of the solid if kf of water is 1.860
Ckg/mol.
42. Define osmosis
The spontaneous flow of the solvent molecules from the less concentrated solution to a more
concentrated solution through a semi permeable membrane is called osmosis.
43. What happens when red blood cells are placed in 0.1% NaCl solution?
Water from NaCl solution passes into cells & they swell. Finally they will burst.
44. Define Osmotic pressure
The minimum excess pressure that has to be applied on the solution to prevent the entry of the
solvent into the solution through semipermeable membrane is called the osmotic pressure.
45. How is osmotic pressure of a solution related to its concentration?
46. What are isotonic solutions?
Solutions which have the same osmotic pressure at the same temperature are called isotonic solutions.
47. Calculate the osmotic pressure of 0.25 M solution of urea at 370
C. R = 0.083 L bar/mol/k.

12. 12
48. 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.
49. When the measurement of colligative property does leads to abnormal molecular mass?
When the solute undergoes either association or disassociation abnormal molar mass is obtained.
50. Give various expressions for van’t Hoff factor?
For non –electrolyte
i = 1
For a strong electrolyte
i = n
HCl → H+
+ Cl-
i = 2
NaOH → Na+
+ OH-
i = 2
H2SO4 →2H+
+ SO4
2-
i = 2
Ba(OH)2 →Ba+
+ 2OH-
i = 3
K4[Fe(CN)4] → 4K+
+ [Fe(CN)4] i = 5
For a weak electrolyte
Case 1: For dissociation
i = 1+ (n-1) 𝛼
n – Number of moles of particles or ions
𝛼 - degree of dissociation
Let us consider the dissociation of acetic acid is 10%
𝛼 =
10
100
= 0.1
i = 1+ (n -1) 𝛼
i = 1+ (2-1) 0.1 = 1.1

13. 13
Case 2: For association
Acetic acid is dimerised in benzene upto 10%
𝛼 =
10
100
= 0.1
i = 1 + (
𝟏
𝒏
− 1) 𝛼
= 1 + (
𝟏
𝟐
− 1) 0.1 = 0.95
i =
normal molar mass
abnormal molar mass
i =
total number of particles after dissociation or association
total number of particles before dissociation or association
i =
observed colligative property
theroectical colligative property
51. How are the various colligative properties modified after consideration of van’t Hoff factor?
52. The boiling point elevation of 0.6 g acetic acid in 100g benzene is 0.1265k. What conclusion can
you draw about the state of solute in solution? Molar elevation constant for benzene is 2.53 deg
per molar?
53. When is the value of ‘i’ less then unity?
When the solute under goes association in solution, ‘i’ is less then unity.

14. 14
54. The molecular mass of a solute is 120 g/mol and van’t Hoff factor is 4. What is its abnormal
molecular mass?