1

Chapter 11
Solutions
Solutions is a homogeneous mixture of two or more substances. It consists of two parts
namely solute and solvent. The substance dissolved is known as solute and the substance in
which dissolution takes place is known as solvent.
Eg : solution of Nacl in water here Nacl is the solute & water is the solvent.
Solute + solvent → solution

Classification of Solution
Based on the physical appearance solution is classified into 3 types.
Solid solution - in which both solute and solvent are solids. Eg : Alloys
Liquid Solution - In which both solute solvent are in liquid state. Eg : Alcohol and water.
Gaseous Solution - In which both solute & solvent are in gaseous state. Eg : Air

Methods of expressing the concentration of a solution

1. Molarity (M) - Molarity is defined as the number of moles of solute present in 1 litre of
solution. It is given by the equation,
𝑊𝑡 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔 𝑊𝑡 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒 ×1000
𝑀= 𝑂𝑅 𝐻 =
𝐺.𝐻 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝑣𝑜𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡 𝐺.𝐻 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝑣𝑜𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝑙

Q. Calculate the molarity of a solution containing 10 gm of KOH in 3 litre water ?

𝑊𝑡 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔 10
Ans : 𝑀 = = = 0.0595 𝑀
𝐺.𝐻 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝑣𝑜𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡 56 ×3

2. Molarity - It is defined as the number of moles of solute present in 1000 gm or 1 kg of

solvent
𝑊𝑡 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔 𝑊𝑡 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔 ×1000
Ans : 𝑚 = 𝑂𝑅 𝑚 =
𝐺.𝐻 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝑤𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑘𝑔 𝐺.𝐻 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝑤𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛𝑔

3. Mole fraction - It is the ratio of number of moles of solute or solvent to the total number of
moles of solution. denoted by x
𝑛𝐴
Mole fraction of solvent 𝑥𝐴 =
𝑛𝐴 +𝑛𝐵
𝑛𝐵
Mole fraction of solvent 𝑥𝐵 =
𝑛𝐴 +𝑛𝐵
𝑛𝐴 𝑛𝐵
For a solution 𝑥𝐴 + 𝑥𝐵 = +
𝑛𝐴 +𝑛𝐵 𝑛𝐴 +𝑛𝐵
𝑛𝐴 +𝑛𝐵
𝑥𝐴 + 𝑥𝐵 = =1
𝑛𝐴 +𝑛𝐵

𝑥𝐴 = 1 − 𝑥𝐵
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𝑥𝐵 = 1 − 𝑥𝐴
4. Mass percentage - It is defined as the mass of solute present in 100 GM's of a solution. It
is given by the equation.
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
Mass % = × 100
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Eg : A solution contain 10 gm of Nacl in 90 gms of water here the total weight of the solution
is 100 gm (90 + 10)
mass of solute = 10 gm
10
mass % = × 100 = 10 %
100

5. Volume percentage - It is defined as the volume of solute present in 100 ml of the solution.
It is given by the equation.
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
Volume % = × 100
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

6. Mass by volume percentage

It is defined as the mass of solute dissolved in 100 ml of a solution. It is given by the
equation.
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
Mass by volume % = × 100
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

7. Parts per million (ppm) - It is defined as the number of parts of a component (solute)
present in 1 million (106 ) of solution.
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 × 106
ppm =
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Solubility of a gas in a solvent (Henry's law) - Henry's law states that the solubility of a gas
in a liquid is directly proportional to the pressure of the gas at constant temperature or the law
states that the mole fraction of a gas is directly proportional to the pressure of the gas over the
solution at constant temperature.

𝑆 ∞ 𝑃 𝑜𝑟 𝑥 ∞ 𝑃
𝑥
𝑥 = 𝐾𝐻 𝑃 𝑜𝑟 𝐾𝐻 = ,
𝑃

𝐾𝐻 𝑖𝑠 𝐻𝑒𝑛𝑟𝑦 ′ 𝑠 𝐿𝑎𝑤 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒, 𝑃 𝑖𝑠 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
The graphical representation of Henry's law consists of the plot of pressure of the gas Vs mole
fraction as shown below
 3

Application of Henry's law - Following are the application of Henry's law.

1. To increase the solubility of 𝐶𝑂2 in soft drinks & soda water, the bottle is sealed under high
pressure.
2. Scuba divers consume excess of gas under water due to high pressure. Then the divers come
towards the surface the pressure gradually decreases. This releases the dissolved gases and
makes the formation of bubbles of 𝑁2 in the blood. It create a disease known as bends. It is
dangerous to life. To avoid bends, the cylinder used by divers are filled with air and helium
(11.7 % He)
3. At high attitude the pressure is very low. This leads to low concentration of oxygen in the
blood and tissues of people living at high attitude or mountaineers. They unable to think clearly
or they become weak. This symptom of the condition is known as anoxia.
Effect of temperature on solubility - Solubility of a gas and temperature are inversely
proportional to each other. ie. Solubility decreases with increase of temperature and vice versa.
Raoult's law of partial pressure -
The law states that partial pressure of any volatile component in a solution is equal
to the product of mole fraction of that substance and the vapour pressure of the substance in
the pure state.
Consider a solution containing volatile solvent A & solute B. If 𝑃𝐴 & 𝑃𝐵 are partial
pressures of A & B respectively. According to Raoult's law
𝑃𝐴 = 𝑥𝐴 𝑃0𝐴 𝑎𝑛𝑑 𝑃𝐵 = 𝑥𝐵 𝑃0 𝐵
𝑃𝐵 = 𝑥𝐵 𝑃0 𝐵 Where
𝑥𝐴 & 𝑥𝐵 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛, 𝑃0𝐴 & 𝑃0 𝐵 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑣𝑎𝑝𝑜𝑢𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑠 of A & B in the pure
state. The total pressure
𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃𝐴 + 𝑃𝐵 = 𝑥𝐴 𝑃0𝐴 + 𝑥𝐵 𝑃0 𝐵
Consider a non volatile solute B in a solution of solvent A in such case the total pressure
𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃𝐴 (𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝐴 𝑜𝑛𝑙𝑦)
Substitute the value of 𝑥𝐴 in the above equation.
𝑃𝑡𝑜𝑡𝑎𝑙 = (1 − 𝑥𝐴 )𝑃0𝐴 𝑜𝑟 𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃0𝐴 − 𝑥𝐵 𝑃0𝐴
𝑥𝐴 + 𝑥𝐵 = 1 or 𝑥𝐵 𝑃0𝐴 = 𝑃0𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙

𝑃0 𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙
𝑥𝐴 = 1 − 𝑥𝐵 ∴ 𝑥𝐵 =
𝑃0 𝐴
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𝑃0 𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙
Where is known as Relative lowering of vapour pressure hence
𝑃0 𝐴
Raoult's law states that Relative lowering of vapour pressure of a solution containing a non -
volatile solute is equal to its mole fraction.
Graphical representation of Raoult's law

Here the first graph represent the variation of vapour pressure of the substance A by the
addition of solute B. Second graph represent the variation of vapour pressure of B by the
addition of solvent A. Graph number III represent the total pressure of solution.
Ideal & non - ideal Solution -
A solution is said to be ideal if
a) It should obey Raoult's law at all conditions of temperature & pressure.
𝑃𝐴 = 𝑥𝐴 𝑃0𝐴 𝑎𝑛𝑑 𝑃𝐵 = 𝑥𝐵 𝑃0 𝐵
b) The change in volume of mixing of the two solutions is equal to zero. ∆𝑉𝑚𝑎𝑥 = 0
c) The change in enthalpy of mixing two solutions is equal to zero. ∆𝐻𝑚𝑎𝑥 = 0
Following graph represent ideal solution.

Eg : Benzene & toluene, ethyl iodide & ethyl bromide.

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Non - ideal solution - Following are the conditions of non - ideal solution.
1) It does not obey Raoult's law.
𝑃𝐴 ≠ 𝑥𝐴 𝑃0𝐴 𝑃𝐵 ≠ 𝑥𝐵 𝑃0 𝐵
2) The change in volume of mixing of two solution is ∆𝑉𝑚𝑖𝑥 = 0
3) The change in enthalpy of mixing of two solution is ∆𝐻𝑚𝑖𝑥 = 0
Deviation of solution from Ideal behaviour (Non - ideal solution)
Non-ideal solutions are classified into types
a) Solution showing +ve deviation.
b) Solution showing -ve deviation.
a) Solution showing +ve deviation - In this type of solution following are the conditions
a) The experimental vapour pressure is greater than the product of mole fraction and
vapour pressure of the substance in the pure state.
𝑃𝐴 > 𝑥𝐴 𝑃0𝐴 𝑃𝐵 > 𝑥𝐵 𝑃0 𝐵
b) The change in volume of mixing of two solution is greater than zero.

∆𝑉𝑚𝑖𝑥 > 0

c) The change in enthalpy mixing of two solution greater than 0

∆𝐻𝑚𝑖𝑥 > 0
Eg : A mixture of ethanol and normal hexane.
In ethanol the molecules are held together by strong inter molecular hydrogen bonding
as shown below

When n-hexane is added into ethanol, the n-hexane molecules occupy in between molecules
of ethanol, thereby breaking hydrogen bonds, this result more evaporation vapour pressure
increases or expansion of liquid take place.
Eg : Acetone & CS2, Acetone & Alcohol, Alcohol & water.
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Graphical representation of solution showing +ve deviation is as shown below.

b) Solution showing -ve deviation -

Following are the conditions of this type of Solution.
1. 𝑃𝐴 < 𝑥𝐴 𝑃0𝐴 2. 𝑃𝐵 < 𝑥𝐵 𝑃0 𝐵 3. ∆𝑉𝑚𝑖𝑥 < 0 4. ∆𝑉𝑚𝑖𝑥 < 0
Eg : A mixture of Acetone & chloroform
When Chloroform is added to Acetone, the molecules of chloroform & Acetone are held
together by intermolecular hydrogen bonding, this result decrease in vaporization & contract
the volume of liquid. The vapour pressure of the solution decreases. The formation of hydrogen
bonding can be represented as shown below.

Graphical representation of this type of solution is as shown below.

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Colligative properties - The properties of solution depends only on the number of moles of
solute & not depends on the nature of solute are called colligative properties.
Following are the colligative properties of solution
1. Relative lowering of vapour pressure.
2. Elevation of Boiling point.
3. Depression of freezing point.
4. Osmosis & osmotic pressure.
The above properties are used for the determination of molecular weight of an unknown
solute.
1. Determination of molecular weight of an unknown solute from Relative lowering of
vapour pressure
According to Roault's law the Relative lowering of vapour pressure is equal to the mole
fraction of a non-volatile solute.
𝑃0 𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙 𝑛𝐵
𝑥𝐵 = → (1) But 𝑥𝐵 =
𝑃0 𝐴 𝑛𝐴 + 𝑛𝐵
𝑊𝐴 𝑊𝐵
𝑛𝐴 = 𝑛𝐵 =
𝑀𝐴 𝑀𝐵
𝑊𝐵
𝑀𝐵
∴ 𝑥𝐵 = 𝑊𝐴 𝑊𝐵 → (2)
+
𝑀𝐴 𝑀𝐵

𝑊𝐵
But in the case of a dilute solution the weight of solute (𝑊𝐵 ) is very small term in the
𝑀𝐵
denominator of eq(2) can be neglected.
𝑊𝐵 𝑀𝐴
𝑥𝐵 = × → (3)
𝑀𝐵 𝑊𝐴

Substitute the value of 𝑥𝐵 in eqn

𝑃0 𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙 𝑊𝐵 𝑀𝐴
= × OR
𝑃0 𝐴 𝑀𝐵 𝑊𝐴

(𝑃0 𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙 ) 𝑀𝐵 . 𝑊𝐴 = 𝑊𝐵 . 𝑀𝐴 . 𝑃0𝐴

(𝑊𝐵 .𝑀𝐴 .𝑃0 𝐴 )
𝑀𝐵 =
(𝑃0 𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙 ) .𝑊𝐴
 8

Here 𝑃 0𝐴 is the vapour pressure of the solvent in the pure state 𝑊𝐵 is the weight of
solute. 𝑀𝐵 is the molecular weight of solute. 𝑃0𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙 is the lowering of vapour
pressure. 𝑊𝐴 is the weight of solvent.
Pbm :- At 298k the vapour pressure of water is 23.75mm of mercury.
Ques. Calculate the vapour pressure at the same temperature of 5% aqueous solution of urea.
(molecular weight of urea is 60) ?
From relative lowering of vapour pressure we can write

(𝑊𝐵 .𝑀𝐴 .𝑃0 𝐴 ) 23.75 ×5 ×18

Ans: 𝑃0𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙 = = = 0.3
𝑀𝐵 .𝑊𝐴 60×100

𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃0𝐴 − 0.356 = 23.75 − 0.356 = 23.394

Pbm :- 2.6g of a non-volatile solute was dissolved in 50g of Acetone. The vapour pressure of
the solution is 182.5mm of mercury. Calculate the molecular weight of the solute ? Given that
molecular weight of Acetone is 58 & vapour pressure is 185 mm ?

(𝑊𝐵 .𝑀𝐴 .𝑃0 𝐴 ) 185×2.6×55

Ans: 𝑀𝐵 = = = 223.15
(𝑃0 𝐴 − 𝑃𝑡𝑜𝑡𝑎𝑙 ) .𝑊𝐴 (185−182.5)×50

2. Determination of molecular weight of a solute from elevation of Boiling point

Boiling point of a liquid is the temperature at which vapour pressure becomes equal 1 atm.
Let 𝑇1 be the boiling point of pure solvent & 𝑇2 is the boiling point of solution (𝑇1 > 𝑇2 )
Elevation of Boiling point 𝑇2 − 𝑇1 = ∆ 𝑇𝑏
The graphical representation of the variation of vapour pressure with respect to temperature of
the pure solvent & solution are as shown below

From the above graph elevation of Boiling point ∆ 𝑇𝑏 = 𝑇2 − 𝑇1

𝑊𝐵 .1000
∆ 𝑇𝑏 ∞ 𝑚 𝑜𝑟 ∆ 𝑇𝑏 = 𝑘𝑏 𝑚 → 1 But 𝑚=
𝑀𝐵 .𝑊𝐴
𝑘𝑏 .𝑊𝐵 1000 𝑘𝑏 .𝑊𝐵 1000
∆ 𝑇𝑏 = ∴ 𝑀𝐵 =
𝑀𝐵 .𝑊𝐴 ∆ 𝑇𝑏 .𝑊𝐴
 9

𝑊𝐴 = 𝑊𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑊𝐵 = 𝑊𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑘𝑏 is called molal elevation constant or ebullioscopic constant
Pbm:-1.gm of a substance dissolved in 100gm of water boiled at 100.307° C. Molal elevation
constant of water is 0.513. Calculate molecular weight of solute?
𝑘𝑏 .𝑊𝐵 1000 0.513×1×1000
Ans: 𝑀𝐵 = = = 16.7
∆ 𝑇𝑏 .𝑊𝐴 0.307×100

∆ 𝑇𝑏 = 𝑇2 − 𝑇1 = 100.307 − 100 = 0.307

Pbm:-A solution of an organic solute in benzene boils at 0.063° C higher than that of benzene
kb=0.53. Calculate the molality of the solution ?
∆𝑇𝑏 0.063
Ans: ∆ 𝑇𝑏 = 𝑘𝑏 𝑚 : 𝑚=
𝑘𝑏
= 0.53
= 0.118
Prblm: Boiling point of benzene is 353.23k when 1.8g of solute dissolved in 90g of benzene,
the boiling point is raised to 354.11k. Find the 𝑀𝐵 ? kb = 2.53?
𝑘𝑏 .𝑊𝐵 1000 2.53×1.8×1000
Ans: 𝑀𝐵 = = = 0.575
∆ 𝑇𝑏 .𝑊𝐴 0.88×900

Molal elevation Constant or Ebullioscopic Constant (𝑘𝑏 )

It is defined as elevation of boiling point of a solution containing 19 mole of solute in 1000g

of solvent. Or This is the elevation of boiling point of 1 molal solution. The unit of molal
elevation constant is k/m or c/m.
Application of elevation of boiling point-Pressure cooker reduces cooking time. At higher
pressure over the liquid (due to the weight of pressure cooker lid) boils at higher temperature.
Cooking occurs faster.
3.Derivation of molecular weight of Solute from depression of F.P method
Let 𝑇1 be the Freezing point of Pure Solvent & 𝑇1 is the Freezing point of solution.

Therefore, Depression of F.P =∆ 𝑇𝑓 = 𝑇1 − 𝑇2

The graphical representation for the variation of vapour pressure of solvent & solution with
respect to temperature is as shown below.
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From the graph we can write

∆ 𝑇𝑓 = 𝑇1 − 𝑇2 .
Depression of Freezing Point depends of molality of solution.
𝑊𝐵 .1000
∆ 𝑇𝑓 ∞ 𝑚 𝑜𝑟 ∆ 𝑇𝑓 = 𝑘𝑓 𝑚 → 1 𝑚=
𝑀𝐵 .𝑊𝐴
Substitute the value of m in eq (1)
𝑘𝑓 .𝑊𝐵 1000 𝑘𝑓 .𝑊𝐵 1000
∆ 𝑇𝑓 = ∴ 𝑀𝐵 =
𝑀𝐵 .𝑊𝐵 ∆ 𝑇𝑓 .𝑊𝐵

Where 𝑀𝐵 is molecular weight of solute k is molal depression constant or cryoscopic constant

𝑊𝐵 weight of solvent.
Molal depression constant or Cryoscopic Constant- It defined as depression freezing point of a
solution containing 1g mole of solute in is 1000g of the solution.
Or
The depression of freezing point of 1 molal solution is known as molal depression
constant. The unit of molal depression constant is k/m or "c/m.
Application of depression of Freezing Point
1.Antifreeze solution- Water is used in radiators of vehicles, if vehicle is used in places where
the temperature is less than zero, then the water should freeze in the radiators. Addition of
antifreeze such as Ethylene glycol to radiators prevent the freezing of water due to the
depression of freezing point. Ethylene glycol is used as a coolant.
2.Clearing of ice from roads in hills-The purpose of spreading NaCl or CaCl2 on road ways in
the winter season is to depress the freezing point of water & reduce the temperature at which
ice is expected to be formed. NaCl can melt ice at a temperature below -21°.
Pbm- A solution of 0.911g of an organic solute in 50g of benzene produced a freezing point
depression of 0.603k. Calculate the molecular weight of solute. kr of benzene is 5.12?
 11

𝑘𝑓 .𝑊𝐵 1000 5.12×0.911×1000

Ans : 𝑀𝐵 = = = 154.7
∆ 𝑇𝑓 .𝑊𝐵 0.602×50

Pbm- a 0.25 molal solution of certain solute in nitrobenzene causes a freezing point
depression of 2 degree. Calculate the value of kf of nitro benzene?

Ans : ∆ 𝑇𝑓 = 𝑘𝑓 𝑚 𝑘𝑓 = ∆ 𝑇𝑓 /𝑚 = 2/0.25 =8
Pbm : Find the molality of a water solution which freezes at 263.15k 𝑘𝑓 of water is 1.86?

Ans: ∆ 𝑇𝑓 = 𝑘𝑓 𝑚 𝑚 = ∆ 𝑇𝑓 /𝑘𝑓
Here ∆ 𝑇𝑓 = 𝑇1 − 𝑇2
∆ 𝑇𝑓 = 273 − 263.15 = 9.85
9.85
So, 𝑚= = 5.2
1.86
4.Osmosis & Osmotic pressure-When a solvent and a solution are separated by a semi
permeable membrane, then the solvent molecule pass to the solution side through pores of the
membrane. This phenomenon of passage of solvent molecule to the solution side through the
pores of membrane is known as osmosis.
The process of osmosis can be stopped if some extra pressure is Supplied on the surface of
solution. This hydrostatic pressure that just stop the flow of solvent to the solution side through
the pores of the membrane is known as osmotic pressure and denoted by 𝜋.

Derivation of molecular weight of solute from osmotic pressure n

Osmotic pressure is a colligative
𝑛
𝜋 𝛼 𝑐 𝑎𝑛𝑑 𝜋 𝛼 𝑇 ∴ 𝜋 𝛼𝐶𝑇 𝑏𝑢𝑡 𝑐 = 𝑣 , 𝜋 = 𝑅𝐶𝑇

𝑛 𝑛𝑅𝑇 𝑊 𝑅𝑇
𝜋 = 𝑣 𝑇𝑅 𝑜𝑟 𝜋𝑣 = 𝑛𝑅𝑇 𝜋= = 𝑀𝐵 𝑣
𝑣 𝐵

𝑊𝐵 𝑅𝑇
𝑀𝐵 = // 𝑊𝐵 = 𝑤𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒, 𝑣 = 𝑣𝑜𝑙 𝑖𝑛 𝑙𝑖𝑡,
𝑣𝜋
 12

𝑇 = 𝐴𝑏𝑠. 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒, 𝜋 = 𝑜𝑠𝑚. 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

Isotonic, hypertonic and hypotonic sólution_ -If two solutions having same are called
Isotonic Solution.

If 𝑛1 & 𝑛2 are osmotic pressure of two solutions, if they are isotonic then 𝑛1 = 𝑛2 .

For isotonic solution the number of moles of solutes are also same. ie𝑛1 = 𝑛2 ; or W1/M1 =
W2/M2
Osmotic pressure of a solution is greater than osmotic pressure of other solution then it is said
to be hypertonic If 𝑛1 > 𝑛2 then the solution having osmotic pressure n, is hypertonic.
Osmotic pressure of a solution less than osmotic pressure of other solution then it is said to
be hypotonic. If 𝑛1 > 𝑛2 then the solution having osmotic pressure 𝑛2 is hypotonic.
Reverse Osmosis (R/0)-The process of osmosis can be reversed by applying pressure larger
than the osmotic pressure of the solution. The pure solvent goes out of the solution through the
semi permeable membrane. This phenomenon is called reverse osmosis. It is used for
desalination of sea water.

Application of Osmosis-Following are the important application of osmosis.

1. A person suffering from high blood pressure is advised to take minimum quantity of NaCl
or common salt. Since the body fluid contain Na+ & Cl- ions. The excessive intake of NaCl
causes increase in the concentration of these ions which result is increase in osmotic pressure.
The high osmotic pressure may cause bursting of blood cells.Thus patients are advised to take
minimum quantity of NaCI
2. A row of mango placed in a solution of NaCl loses water due to Osmosis and then to shrink.
3. After removing the outer shell of two eggs, one is placed in distilled water and the other is
placed in NaCl solution. It is observed that the egg in water will swell while the other will
shrink because of osmosis.
4.An excessive use of chemical fertilizers damages the plants since the osmotic pressure of the
fertilizer solution become higher than that of the cell sap. This causes the out flow of water
from plant cell into the soil leading to the drying up of the plants.
 13

5.When dried fruits and vegetables are placed in water they slowly swell and return to original
form.
6. 0.16 molar sodium chloride solution (91 g or 0.91%) is isotonic with blood cells. Which
neither shrink nor swell when placed in 91% NaCI solution. When blood cells are placed in
distilled water, water flows into the cell and they swell.
III. when blood cells are placed in 5% NaCI solution, water comes Out of the cell and they
shrink.
7. 0.91% solution of NaCI is isotonic with human blood. During intravenous injection the
medicines mixed with saline water before the injection.
8.Plant absorb water from the soil through their roots due to osmosis
1. An aqueous solution containing 1 g of a protein per litre executed an osmotic pressure of
3.1mm of mercury at 25 degree C. Calculate the molecular weight of the protein.

𝑊𝐵 𝑅𝑇 1×0.0821×298
𝑀𝐵 = = = 6116.45
𝑣𝜋 0.004×1
2.Calculate osmotic pressure of 5% solution of glucose at 270C given that molecular weight of
glucose is 180?

𝑊𝐵 𝑅𝑇 𝑊𝐵 𝑅𝑇 5×0.0821×298
𝑀𝐵 = 𝑜𝑟 𝜋 = =
𝑣𝜋 𝑣𝑀𝐵 180×0.1
3. 1.78% of solution of urea is isotonic with 10% solution of cane sugar molecular weight of
sugar is 342). Calculate molecular weight of urea?
Since the solutions are isotonic
𝑊1 𝑊2
𝑛1 = 𝑛2 𝑜𝑟 𝜋1 = 𝜋2 or = or 𝑊1 𝑀2 = 𝑊2 𝑀1
𝑀1 𝑀2

𝑊1 𝑀2
𝑀1 = = 1.78 ∗ 342 = 60.8
𝑊2
Abnormal Molecular Weight of Solute: Molecular weight of a solute can be experimentally
determined by any one of the colligative properties. In certain solutes the normal molecular
weight and observed molecular weight are different. This is known as abnormal molecular
weight of solute.
Because of the following reasons:

1. Molecular Association Here observed molecular weight is greater than normal

molecular weight because two or more molecules of the solute associate to form a dimer
by inter molecular hydrogen bonding Here the effective number of particles decreases.
 14

Eg: Normal molecular weight of 𝐶𝐻3 𝐶𝑂𝑂𝐻 (Acetic acid or Ethanoic acid) is 60 and
the observed molecular weight in benzene is 120. This is due to the formation of a dimer
by molecular association or due to intermolecular hydrogen bonding as shown below.

Here observed molecular weight 2x normal molecular wt 2x60 = 120 Similarly normal
molecular weight of benzoic acid is 122 and in benzene, the observed molecular weight
of benzoic acid is 244. This is also due to the formation of a dimer by molecular
association or due to intermolecular hydrogen bonding as shown below.

Here observed molecular weight = 2 x normal weight = 2 x 122 244 Or in general in

the molecular association.
Observed molecular weight = n x normal molecular weight.
In molecular dissociation the observed molecular weight is less than normal molecular
weight
Eg: Normal molecular weight of NaCl is 58.5 but Observed molecular weight of Nacl
in water is 29.25. Here NaCl decompose to 𝑁𝑎+ and 𝐶𝑙 −

𝑁𝑎𝐶𝑙 → 𝑁𝑎+ + 𝐶𝑙 −

Observed molecular weight= normal molecular weight/2 = 58.5/2 = 29.25

Eg for 𝐾𝐶𝑙
Normal molecular weight of 𝐾𝐶𝑙is 74.5 but Observed molecular wt in water = 37.25
here KCI decompose to 𝐾 + and 𝐶𝑙 −
𝐾𝐶𝑙 → 𝐾 + + 𝐶𝑙 −
Observed molecular weight = 74.5/2=37.25

In general in molecular dissociation

Observed molecular weight = normal molecular weight/n where n is the number of
ions formed from one molecule of the solute

Van't Hoff's factor (i)

It is defined as the ratio of normal molecular weight of a solute to the observed
molecular weight of solute.
Normal molecular weight Obsrved colligative property
𝑖= or 𝑖 =
Observed molecular weight Normal colligative property
 15

60
Eg: for acetic acid 𝑖 = 120 = 0.5
122
For benzoic acid 𝑖 = 244 = 0.5
It has been found that in molecular association I < 1

58.5 74.5
But for NaCI 𝑖 = 29.25 = 2 , For 𝐾𝐶𝑙 𝑖 = 37.25 = 2

It has been found that in molecular dissociation i> 1.

For non ionic solute such as urea (𝑁𝐻2 - 𝑁𝐻2 ) i = 60/60 = 1

Here no molecular association or dissociation takes place.

180
Similarly for glucose (𝐶6 𝐻12 𝑂6 )𝑖 = 180 = 1
It has been found that if solute has no association or dissociation i = 1

Relation between colligative properties and Van't Hoff factor

They are related by the equation

(𝑃0 𝐴 − 𝑃𝐵 )
1) Relative lowering vapour pressure = 𝑖𝑥𝐵
𝑃0 𝐴

2) Elevation of boiling Point ∆ 𝑇𝑏 = 𝑖𝑘𝑏 𝑚

3) Depression of freezing point ∆ 𝑇𝑓 = 𝑖𝑘𝑓 𝑚
4) Osmotic Pressure = 𝑖𝐶𝑅𝑇
𝑖−1
Degree of Dissociation - It is given by the equation 𝛼 = and van't Hoffs factor
𝑛−1

i=1+(n-1) where n is the number of particles formed from one molecule of solute
eg: for NaCI n = 2 and for 𝐵𝑎𝐶𝑙2 n=3
 16

Chapter 2-Solutions (Questions)

1) ___Unit has been used to express the concentration of very diluted solution?

Ans:-parts per million (ppm)

2) Name the concentration term which is a temperature dependent.

Ans:- Molarity

3) Calculated the molarity of a solution containing 5.5g 𝐻2 𝑆𝑂4 in 300ml of

solution?
Ans: M= ( 𝑊𝐵 x 1000)/ (𝑀𝐵 ∗ V in ml) = (5.5 x 1000)/ (98 x 300) = 0.187

4) Anoxia is due to ______?

Ans: - Low concentration of oxygen in our blood.

5) What are azeotropic mixtures?

Ans:- Mixture of two liquid which boils at a constant temperature is called

azeotropic mixture or constant boiling mixture. It is then classified in to two types

a. Maximum boiling azeotropes: -These are formed by non-ideal solutions

showing8 negative deviation. Eg- 𝐻𝑁𝑂3 (68%) + 𝐻2 𝑂 (32%).BP of these liquid
mixture is more than either of two pure components. Here BP is 393.5K.

b. Minimum boiling azeotropes: - These are formed by non-ideal solutions

showing positive deviation. Eg- Ethanol (95.6%) + 𝐻2 𝑂 (4.4%). BP of these
liquid mixture is less than either of two pure components. Here BP is 351.15K.

6) Find the boiling point of 2M solution of a non-volatile solute in chloroform. (BP

of chloroform is 61.2°C and 𝑘𝑏 =3.63 𝑘𝑚 −1
Ans:-BP of solution is 𝑇2 , ie ∆ 𝑇𝑏 = 𝑇2 − 𝑇1 𝑇2 = ∆ 𝑇𝑏 + 𝑇1

But ∆ 𝑇𝑏 = 𝑘𝑏 𝑚 = 3.63x2 = 7.26 So, 𝑇2 = 7.26+ 61.2 = 68.46°C

7) Calculate the freezing point lowering of 5% solution of glucose (Mol wt.=180) i

water (𝑘𝑓 =1.86 K/m)
𝑘𝑓 .𝑊𝐵 1000
Ans: ∆ 𝑇𝑓 = = (1.86 x 5 x 1000)/ (180 x 95) = 0.544 K
𝑀𝐵 .𝑊𝐵
 17

8) The chemical substance used for the preparation of semi permeable membrane
is_?Ans:-Copper Ferro Cyanide

9) Write Van't Hoff's solution equation. Ans: 𝜋 = 𝐶𝑅𝑇,

𝑛
𝜋 = ( 𝐵 ) . 𝑅𝑇 𝑜𝑟 𝜋𝑣 = 𝑛𝐵 𝑅𝑇
𝑣

10) Molecular mass of polymers or macromolecules can be determined by __

method. ?
Ans: Osmotic pressure method.

11) Human blood is isotonic with ____?

Ans: 0.91% NaCl solution

12) 200𝑐𝑚3 of an aqueous solution of protein contains 1.26g of protein. The osmotic
pressure of solution is 30OK is found to be 2.57x 10−3 bar. Calculate the Mol.wt
of protein.

𝑊𝐵 𝑅𝑇
Ans:- 𝑀𝐵 = here, R= 0.0821 lit atm, T=300K, v= 20Ocm =200ml
𝑣𝜋
= 0.2 lit
𝑀𝐵 = −3
(1.26x0.0821x300)/ (2.57x10 x0.2) = 61.02g/mol

13) 1.8% solution of glucose is isotonic with solution of raffinose (a type of sugar).
Calculate the percentage of raffinose. (Mol.wt of glucose = 0180 and raffinose
596)
𝑊 𝑊2
Ans:- For isotonic solution 𝜋1 = 𝜋2 or 𝑀1 = 𝑀2
or 𝑊1 𝑀2 = 𝑊2 𝑀1
1
1.8∗596
𝑊1 = = 5.96%
18
14) What is the relationship between colligative property and Van't Hoff's factor?
(𝑃0 𝐴 − 𝑃𝐵 )
Ans:- a) Relative lowering of Vapour pressure, = 𝑖𝑥𝐵
𝑃0 𝐴
b) Elevation of boiling point∆ 𝑇𝑏 = 𝑖𝑘𝑏 𝑚
c) Depression of freezing point ∆ 𝑇𝑓 = 𝑖𝑘𝑓 𝑚
d) Osmotic pressure = 𝑖𝐶𝑅𝑇
15) Concentrated Nitric acid is 68% by mass in aqueous solution. Find the molarity
of acid if density is 1.504g/m𝑙 −1.

Ans: -Mass of 1 lit of solution = dxV= 1.504x1000 1504g

Mass of 68% 𝐻𝑁𝑂3 = (1504x68)/ 100 1022.72
 18

Molarity = (Mass of solute in g/lit)/ (Molecular mass of solute) 1022.72/ 63 =

16.23M

16) Arrange the following solution in the increasing order of BP

1𝑀 𝐶𝑎𝐶𝑙2 , 1𝑀 𝑁𝑎𝐶𝑙, 1𝑀 𝐶𝐻3 − 𝐶𝑂𝑂𝐻, 1𝑀 𝑢𝑟𝑒𝑎. Explain the reason.

Ans: 1𝑀 𝑢𝑟𝑒𝑎 < 1𝑀 𝐶𝐻3 − 𝐶𝑂𝑂𝐻 < 1𝑀 𝑁𝑎𝐶𝑙 < 1𝑀𝐶𝑎𝐶𝑙2 ,.because
∆ 𝑇𝑏 ∝ 𝑉𝑎𝑛𝑡 ′ 𝐻𝑜𝑓𝑓 ′ 𝑠 𝑓𝑎𝑐𝑡𝑜𝑟

Solute Solvent Normal Mol.wt Observed

Mol.wt
NaCl Water A A/2
Benzoic Acid Benzene B -
Urea Water C -
Acetic Acid Benzene D 2D
𝐶𝑎𝐶𝑙2 Water E -
Glucose Water F -
𝐴𝑙2 (𝑆𝑂4 )3 Water G G/5

Ans: 2B, C, E/3, F

17) Calculate the wt. of 𝐶𝑎𝐶𝑙2 (i=2.47) dissolved in 2.5lit of water such that its
osmotic pressure is 0.75atm at 27°C.

𝑖(𝑛𝐵 𝑅𝑇) 𝑖(𝑊𝐵 .𝑅𝑇)

Ans:- 𝜋 = 𝑖. 𝑐𝑅𝑇 𝑜𝑟 𝜋 = ,𝜋 =
𝑉 𝑀𝐵 .𝑉
𝑀𝐵 .𝑉 0.75×𝟏𝟏𝟏×𝟐.5
𝑊𝐵 = = = 3.42 𝑔
𝑖.𝑅𝑇 2.47×0.0821×300

18) 18g of glucose dissolved in 1 Kg of water. At what temperature will water boils
at 1.013 bar? 𝑘𝐵 =0.52

Ans: BP of solution 𝑇2 = ∆ 𝑇𝑏 + 𝑇1 , BP of water = 373K, ∆ 𝑇𝑏 = 𝑘𝑏 𝑚

Number of moles of glucose =18/180 = 0.1, Wt. of water = 1 Kg
Molarity of solution = 0.1/1 =0.1
∆ 𝑇𝑏 =0.5x0.1= 0.052, T2 = 373+0.052 373.052K
 19

Solutions
Questions and answers
1. Name the concentration unity which is used for expressing concentration in industrial
chemical reaction
Ans :- mass percentage
2. The unit of concentration term widely used in medicine or pharmacy is .....
Ans :- mass by volume percentage
3. Concentration of pollutants in water or atmosphere is expressed in terms of .....
Ans :- parts per million (ppm)
4. Which concentration unit is temperature dependent
Ans :- Molarity
[Because volume depends on temperature]
5. Calculate the mole fraction benzene in a solution containing 30% by of carbon
tetrachloride.
Ans :- Let total mass of solution = 100g
Mass of benzene = 30g
Mass of 𝐶𝐶𝑙4 = 100 - 30 = 70g
Number of moles of benzene
𝑛𝐶6 𝐻6 = (𝑚𝑎𝑠𝑠 𝑜𝑓 𝑏𝑒𝑛𝑧𝑒𝑛𝑒)/ 𝑚𝑜𝑙 𝑤𝑡
= 30/78 = 0.385 mol
Number of moles of 𝐶𝐶𝑙4
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝐶𝑙 70
𝑛𝐶𝐶𝑙4 = 𝑚𝑜𝑙 𝑤𝑡 𝑜𝑓 𝐶𝐶𝑙4 = 154 = 0.455 𝑚𝑜𝑙
4

Mole fraction of benzene

𝑛𝐶6 𝐻6 0.385 0.385
= = = = 0.458
𝑛𝐶6 𝐻6 + 𝑛𝐶𝐶𝑙4 0.385+0.455 0.84

6. Calculate the Molarity of 30ml 0.5m 𝐻2 𝑆𝑂4 . diluted to 500ml.

Ans :- 𝑀1 𝑉1 = 𝑀2 𝑉2
𝑀1 𝑉1
𝑀2 = = 0.5 × 30 / 500
𝑉2

= 0.03 m
 20

7. Calculate the molality of 2.5g of ethanoic acid (𝐶𝐻3 − 𝐶𝑂𝑂𝐻) in 74g of Benzene.
Ans:- Number of moles of ethanoic acid = W/M = 2.5/60 = 0.0417
Wt of solvent benzene = 75g
𝑛𝐵 0.0417×1000
Molality = × 1000 = = 0.556 𝑚
𝑊𝐴 75

8. Concentrated nitric acid used for laboratory works is 68 % by mass in aqueous solution.
Calculate the molarity of the acid 18 its density is 1.5 g/mol.
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒
Ans :- Molarity = 𝑀𝑜𝑙 𝑤𝑡 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑒 ×𝑉

Mass of 1 lit solution = d × v

= 1.5 × 1000 = 1500g
Mass of 𝐻𝑁𝑂3 in 100g = 68g
68×1500
Mass of. 𝐻𝑁𝑂3 in 1500g = = 1020𝑔
100
1020
Molarity = = 16.19M
63×1

9. Elevation of boiling point of a solution is ∆ 𝑇𝑏 = 𝑘𝑏 𝑚 Derive the equation for

molecular weight of a non volatile solute from the above equation

Ans :- ∆ 𝑇𝑏 = 𝑘𝑏 𝑚 ----- (1)

𝑛 𝑊𝐵 ×1000
But m = 𝑊𝐵 × 1000 = ---- (2)
𝐴 𝑀𝐵 .𝑊𝐴

Substitute the value of m in eq(1)

𝑘𝑏 .𝑊𝐵 ×1000
∆ Tb = 𝑀𝐵 .𝑊𝐴

𝑘𝑏 .𝑊𝐵 ×1000
𝑀𝐵 = ∆ Tb.𝑊𝐴

Here 𝑀𝐵 = Mol.wt of solute

𝑘𝑏 = Knows as molal elevation constant
∆ Tb = Elevation of BP
𝑊𝐴 = weight of solvent in gram.
10. Seawater freezes at a lower temperature than distilled water. Why ?
Because sea water contains solutes (Nacl) than in distilled water.
11. Solutions having same osmotic pressure are called .....
Ans :- isotonic solutions
 21

12. A 5% solution of cane sugar in water has F.P of 271k. Calculate the F.P of 5% glucose in
water. If F.P of pure water is 273.15k.
Ans :- Molecular mass of sugar = 342
Molecular mass of glucose = 180
F.P of cane sugar solution = 271k
F.P pure water = 273.15k

Depression in F.P , ∆ 𝑇𝑓 = 𝑇1 − 𝑇2
∆ 𝑇𝑓 = 273.15 − 271 = 2.1
𝑘𝑓.𝑊𝐵 ×1000
, ∆ 𝑇𝑓 = OR
𝑀𝐵 .𝑊𝐴

,∆ 𝑇𝑓 .𝑀𝐵 .𝑊𝐴 2.15 ×342 × 95

𝑘𝑓 = =
𝑊𝐵 ×1000 5 ×𝟏𝟎𝟎𝟎

= 13.96
For glucose solution
𝑘𝑓.𝑊𝐵 ×1000 13.96 ×5 ×1000
∆ 𝑇𝑓 = 𝑀𝐵 .𝑊𝐴
=
180×95
= 4.08

F.P of glucose solution = 273.15 - 4.08 = 269.07 K

13. When dried fruits and vegetables are placed in water, They slowly swell. What will be
the effect of temperature on this process ?
Ans :- When dry fruits and vegetables are placed in water, they slowly swell due to
movement of water (solvent) to the dry fruit (solutent) due to Osmosis. When temperature
increases the rate of osmosis increases.
14. Calculate the freezing point of a solution of 54g glucose in 250 g water. Given that
Mol.wt of glucose is 180 and 𝑘𝑓 = 1.86 k mo𝑙 −1

Ans :- Freezing point of solution

∆ 𝑇𝑓 = 𝑇1 − 𝑇2
𝑇2 = 𝑇1 − ∆ 𝑇𝑓
But 𝑇1 = 0° 𝑐
𝑘𝑓.𝑊𝐵 ×1000 1.86 ×54×1000
∆ 𝑇𝑓 = 𝑀𝐵 .𝑊𝐴
= 180×250
= 2.23
 22

Freezing point of solution

𝑇2 = 𝑇1 − ∆ 𝑇𝑓
= 0 - 2.23
= -2. 23°c
15. Calculate osmotic pressure in pascals exerted by a solution prepared by dissolving 1g of
polymer of molecular mass 185000 in 450 ml of water at 37°c
𝑊𝐵 𝑅𝑇
Ans :- Osmotic pressure, 𝜋 = 𝑀𝐵 .𝑉

𝑊𝐵 = 1𝑔, 𝜋 = 0.0821 𝑙𝑖𝑡 𝐴𝑡𝑚

T = 37°c = 37 + 273 = 310 k
V = 450 ml = 450/ 1000 = 0.45 lit
𝑀𝐵 = 185000
1×0.0821×310
𝜋= = 3.08 × 10−4 𝑏𝑎𝑟 = 3.08 × 10−4 × 105 𝑝𝑎 = 30.9 𝑝𝑎
18500×0.450

16. At a 300k,36 g of glucose present in one litre of solution has an Osmotic pressure of 4.98
bar. If the Osmotic pressure of the solution is 1.52 bar at the same temperature. What would
be the concentration?

Ans :- osmotic pressure, 𝜋 = 𝑐𝑅𝑇

R & T are two constant
𝜋1 𝐶1 𝜋1
𝜋 ∝ 𝐶 𝑂𝑅 = 𝑂𝑟 𝐶2 = 𝐶1
𝜋2 𝐶2 𝜋2

36
𝑏𝑢𝑡 𝐶1 = 36 𝑔 = = 0.2 𝑚𝑜𝑙
180
1.42
𝐶2 = 0.02 = 0.06 𝑚𝑜𝑙 𝑙𝑖𝑡
4.98

17. Calculate the Osmotic pressure of a solution prepared by dissolving 25 mg of 𝐾2 𝑆𝑂4 in

2 litre water at 25°c assuming that it is completely dissociated.
𝑊𝐵
Ans :- , 𝜋𝑉 = 𝑖𝑐𝑅𝑇 𝐶= 𝑀𝐵
= 𝑛𝐵

𝑊𝐵 𝑅𝑇
𝜋𝑉 = 𝑖 𝐾2 𝑆𝑂4 → 2𝐾 + + 𝑆𝑂4 2− ; 𝑖 = 3,
𝑀𝐵 .𝑉
3×0.025×0.0821×298
𝜋= 𝑅 = 0.0821; 𝑇 = 298𝑘; 𝑀𝐵 = 174
174×2
 23

= 5.2 × 10−3 𝑎𝑡𝑚 𝑉 = 2 𝑙𝑖𝑡 ; 𝑊𝐵 = 25𝑚𝑔 = 0.025𝑔

18. Calculate the molality of KCl solution in water such that the freezing point be depressed
by 2k. 𝑘𝑓 = 1.86
Ans :- ∆ 𝑇𝑓 = 𝑖𝑘𝑓 𝑚 𝑚 = ∆ 𝑇𝑓 /𝑖𝑘𝑓
Here, ∆ 𝑇𝑓 = 2𝑘 𝑘𝑓 = 1.86
2
𝑚= = 0.537 𝑚𝑜𝑙 𝑘𝑔−1
2×1.86
19. Calculate the mass percentage of benzene and 𝐶𝐶𝑙4 if 22g of benzene is dissolved in
122g of 𝐶𝐶𝑙4
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑏𝑒𝑛𝑧𝑒𝑛𝑒
Ans :- Mass percentage of Benzene = × 100
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠
22
= 22+122 × 100 = 15.27%

Mass % of 𝐶𝐶𝑙4 = 100 − 15.27 = 84.72%

20. Calculate the mass of urea (𝐻2 𝑁 − 𝐶𝑂 − 𝑁𝐻2 ) required in making 2.5 kg of 0.25 molal
aqueous solution.
𝑛𝐵
Ans :- Molality m = 𝑊𝐴

𝑊𝐵 𝑊𝐵
But 𝑛𝐵 = 𝑜𝑟 𝑚 =
𝑀𝐵 𝑀𝐵 .𝑊𝐴

𝑊𝐵 = 𝑚 × 𝑀𝐵 × 𝑊𝐴 = 0.25 × 60 × 2.5
Wt of urea 𝑊𝐵 = 37.5g
21. Boiling point of water at 750 mm Hg is 99.63°. How much sucrose is to be added to
500g of water such that it was at 100°c.
∆ 𝑇𝑏 𝑊𝐴 𝑀𝐵 0.37×342×500
Ans :- 𝑊𝐵 = = = 122𝑔
𝑘𝑏 ×1000 0.52×1000