1. The document discusses various terms used to express the concentration of solutions such as percentage, molarity, molality, normality, and mole fraction. It provides examples and formulas to calculate these quantities. 2. Several numerical problems are given related to calculating concentration based on the mass or volume of components in a solution. This includes problems determining the percentage or mole fraction composition of mixed solutions. 3. Additional "HOTS" or higher-order thinking skills problems are presented involving multiple steps to determine volumes or densities based on given molarities, molalities, or percentages of solutions.

1. SOLUTIONS
CLASS XII (Chemistry)
Arunesh Gupta
PGT (Chemistry)
KV Barrackpore (AFS)

2. Solution
 Solution: It is a homogeneous mixture of two or more chemically non
reacting substances whose composition is same throughout. The
composition can be varied within a certain limits.
 Binary solution: It is a solution of two non-reacting chemical substances.
 The chemical substance present in large amount (mass, volume or mole)
is called solvent & the substance is smaller amount is called solute.
 Binary solution = Solvent (A) + solute(B).
 Example: ethyl alcohol (B) added in water as solvent (A)

3. BINARY SOLUTION:
• Example: 50 g water & 50 g ethyl alcohol forms a solution where
amount of water (50 ÷ 18 = 2.77 moles ) is more than that of ethyl
alcohol (= 50 ÷ 46 = 1.08 moles) Here nwater > nethyl alcohol
(molar mass of water & ethyl alcohol are 18 & 46 g/mol)
• Here water is a solvent & ethyl alcohol is a solute.
• Aqueous solution is a solution with water as solvent & non aqueous
solution is a solution with solvent other than water like ether, benzene,
alcohol, CCl4, liquid ammonia etc.
E.g. : Solution of iodine in benzene or carbon disulphide.

4. Types of solutions
Types of
solutions
Solute
(B)
Solvent
(A)
Examples
Solid
solutions
1. Solid Solid Alloys,e.g. brass (Cu + Zn), German silver
(Cu+Zn+Ni), bronze (Cu+Sn) etc
2. Liquid Solid Amalgam of Hg with Na, Hydrated salts
3. Gas Solid Dissolved gases in minerals
Liquid
solutions
1. Solid Liquid Sugar in water, Iodine crystals in alcohol
2. Liquid Liquid Alcohol in water
3. Gas Liquid Oxygen dissolved in water, aerated drinks
Gaseous
solutions
1. Solid Gas Iodine vapour in air
2. Liquid Gas Water vapour in air
3. Gas Gas Air (N2 + O2)

5. Concentration of a solution: Methods to express it in different ways:
(i) Percentage:
(a) Mass % or percentage by mass (%w/w) (for solid in liquid or liquid in liquid solution): means mass of
solute in kg (or g) in 100 kg (or g) of solution. It is temperature independent. It has no unit.
Let WB g of solute B is dissolved in WA g of solvent A. Mass of solution = (WA + WB) g
Mass % =
𝐖𝐁
𝐖𝐀+𝐖𝐁
𝐱𝟏𝟎𝟎
(b) Volume % or percentage by volume(%v/v) (for liquid in liquid solution) means volume of solute in L (or
mL) in 100 L (or mL) of solution. It is temperature dependent. It has no unit.
Let VB mL of solute B is dissolved in VA mL of solvent A. Volume of solution = (VA + VB) mL
Volume % =
𝐕𝐁
𝐕𝐀+𝐕𝐁
𝐱𝟏𝟎𝟎
(c) Mass by volume % (% w/v) means mass of solute in kg (or g) in 100 L (or 100 mL ) of solution
(ii) Strength: It is the amount of solute in g in one litre of solution. Its unit is g/L & it depends on
temperature. Strength (g/L) =
Mass of solute in g
Volume of solution in litre

6. Concentration of a solution:
(iii) Molarity (M) : It is the number of moles of solute present in one litre of solution. Its unit is mol/L. It is temperature
dependent. Molarity ( in mol/L) =
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐢𝐧 𝐥𝐢𝐭𝐫𝐞
or, Msoln =
𝐧𝐬𝐨𝐥𝐮𝐭𝐞
𝐕𝐋
Example:
Molarity of NaOH solution is 1 M means 1 L of NaOH solution contains 1 mole = 40 g of NaOH crystals.
[Molarity of a solution is the number of milli moles of solute present in one milli litre of solution.]
(iv) Molality (m) : It is the number of moles of solute present in one kg ( or 1000g) of solvent. Its unit is mol/kg. It is
temperature independent. Molality (mol/kg) =
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
𝐌𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 𝐢𝐧 𝐤𝐠
Example: Molality of ethyl alcohol solution is 1 m means 1 kg or 1000 g of water of 1 m ethyl alcohol solution contains 1 mole
or 46 g of ethyl alcohol.
**(v) Formality (F): It is the number of formula weights of the solute(ionic salt) present per litre of the solution. It is used in
case of ionic solid as solute in solution. We take the formula mass of an ionic solid crystal.
Example: Formality of NaCl solution is 1 F means 1L of NaCl solution contains 1 mole = 58.5 g of NaCl crystals.

7. **(vi) Normality (N): It is the number of gran equivalents of solute present in one litre of solution.
Normality (g eq/L) =
𝐍𝐨.𝐨𝐟 𝐠 𝐞𝐪.𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐢𝐧 𝐥𝐢𝐭𝐫𝐞
Equivalent weight (or mass) of an acid =
𝐌𝐨𝐥. 𝐌𝐚𝐬𝐬 𝐨𝐟 𝐚𝐜𝐢𝐝 𝐚𝐬 𝐬𝐨𝐥𝐮𝐭𝐞 𝐢𝐧 𝐠
𝐁𝐚𝐬𝐢𝐜𝐢𝐭𝐲 𝐨𝐟 𝐚𝐜𝐢𝐝
(Basicity of an acid means no. of replaceable H+ in solution).
For HNO3 , basicity = 1, H2SO4 basicity = 2 & H3PO4, basicity = 3
Equivalent weight (or mass) of a base =
𝐌𝐨𝐥. 𝐌𝐚𝐬𝐬 𝐨𝐟 𝐛𝐚𝐬𝐞 𝐚𝐬 𝐬𝐨𝐥𝐮𝐭𝐞 𝐢𝐧 𝐠
𝐀𝐜𝐢𝐝𝐢𝐭𝐲 𝐨𝐟 𝐛𝐚𝐬𝐞
(Acidity of a base means no. of replaceable OH- in solution.
For NaOH, acidity = 1, Ca(OH)2, Acidity = 2 & Al(OH)3, Acidity = 3
Equivalent weight (or mass) of a salt =
𝐌𝐨𝐥. 𝐌𝐚𝐬𝐬 𝐨𝐟 𝐬𝐚𝐥𝐭 𝐚𝐬 𝐬𝐨𝐥𝐮𝐭𝐞 𝐢𝐧 𝐠
(𝐍𝐨. 𝐨𝐟 𝐜𝐚𝐭𝐢𝐨𝐧) 𝐱 (𝐢𝐭𝐬 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐜𝐡𝐚𝐫𝐠𝐞)
For Al2(SO4)3 its eq. weight =
𝟐𝐱 𝟐𝟕+𝟑(𝟑𝟐+𝟒𝐱𝟏𝟔)
𝟐 𝐱 𝟑
=
𝟓𝟒+𝟑𝐱𝟗𝟔
𝟔
= 342/6 = 57 g / eq.

8. Concentration of a solution:
(vii) Mole fraction (χ) of a component of a solution is the number of the moles of that component divided
by the total no. of moles of all components present in the solution.
The sum of mole fraction of all comonents in a solution is equal to one. It has no unit.
For a solution of components A, B, C, D…. χA + χB + χC + χD+………. = 1
For a binary solution of two components as Solvent (A) & Solute(B) ,
Mole fraction of solvent (χA) =
𝐧𝐀
𝐧𝐀+𝐧𝐁
& Mole fraction of solvent (χB) =
𝐧𝐁
𝐧𝐀+𝐧𝐁
So, χA + χB = 1
Remember: Mole percentage = Mole fraction x 100.
(viii) Parts per million (ppm) : Ppm of a solute by mass in the solution is the mass of solute present in
million parts by mass of solution. It can be ppm by mass or ppm by volume. It has no unit.
For a solute (B) ppm by mass (ppm)B =
𝐌𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 𝐁 𝐢𝐧 𝐠
𝐌𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐢𝐧 𝐠
x 106
Similarly,
ppm by volume is the volume of solute in mL (or L) in 106 mL (or L) solution at a given temperature.

9. Concentration of Solution:
Remember:
(a) No. of moles of solute in solution = Molarity x Volume of solution in L.
(b) No. of milli moles of solute in solution = Molarity x Volume of solution in mL.
(c) Normality = Molarity x n (where n= basicity of acid or acidity of a base or total charge of all cations in a salt).
(1) Relationship between molality (m) of a solution & mole fraction of solute (χB) in solution:
χB =
m MA
1+mMA
MA is the molar mass of solvent.
(2) Relationship between molality (m) & molarity (M) of a solution (density=d kg/L) & solute of molar mass = MB
kg/mol.
d is in g/mL & mol mass of solute is in g/mol OR
m =
𝑀
𝑑−𝑀𝑀𝐵
m =
𝑀
1000𝑑−𝑀𝑀𝐵
x 1000
(3) For a reaction : a A + b B  c C + d D
( a moles of A reacts with b moles of B to form c moles of C & d moles of D) Let VA mL of A of molarity MA reacts with VB of B of molarity
MB. Therefore,
𝐕𝐀𝐌𝐀
𝐕𝐁𝐌𝐁
=
𝐚
𝐛
& let VA mL of A of normality NA reacts with VB of B of normality, so VA.NA = VB.NB
Remember: (i) On dilution, (adding water in a solution, volume of solution increases, molarity & normality of solution decreases, but no. of
moles of solute remains same.
(ii) In general one molar (1 M) solution is more concentrated than one molal (1 m) solution in aqueous solution (density of water = 1g/mL).
But for a non-aqueous solution, 1 M = 1m or 1M > 1m or 1M < 1m, depends on the density of the solution.

10. Example: What are the mole fractions of the components of the solution formed when 92 g glycerol (92 g/mol) is
mixed with 90 g water (18 g/mol)?
Components Step 1  Step 2  Step 3
Glycerol (A) Convert mass into moles
Mole =
mass in g
molar mass
= 92/92 = 1
mol
Total moles = 1+5 = 6
moles
(χA) =
𝐧𝐀
𝐧𝐀+𝐧𝐁
=
𝟏
𝟔
(Ans)
Water (B) Mole = 90 /18 = 5 mol (χB) =
𝐧𝐁
𝐧𝐀+𝐧𝐁
=
𝟓
𝟔
(Ans)
NUMERICAL PROBLEMS to Practices:
(1) Calculate the % composition in terms of mass of a solution obtained by mixing 300g of a 25% & 400 g of a 40%
solution by mass. [Ans: 33.57%, 66.43%].
(2) What volume of 95% by mass of sulphuric acid (density=1.85 g/mL ) & what mass of water must be taken to
prepare 100 mL of 15% by mass of sulphuric acid solution (density=1.10 g/mL) [Ans: 9.4mL, 92.6 g].
(3) Calculate the mole fraction of ethanol & water in a sample of rectified spirit which contains 95% ethanol by mass.
(At. Wt. C=12, H =1, O=16) [ Ans. ethanol = 0.88, water = 0.12]
(4) What volume of 10% (w/v) solution of Na2CO3 will be required to neutralise 100 mL of HCl solution containing 3.65
g of HCl. (At. wt. Na=23 , Cl = 35.5, H = 1) [Ans. 53.19 mL]
(5) Calculate the mole fraction of water in a mixture of 12 g water, 108 g acetic acid & 92 g of ethanol. [Ans.0.15]

11. NUMERICAL PROBLEMS to Practices:
(6) The molality of a solution of ethanol in water is 1.55 m. How many grams of ethanol are dissolved in 2 kg of water?
[Ans. 142.6 g]
(7) Battery acid is 4.27 M H2SO4 aqueous solution of sulphuric acid of density 1.25 g/mL. Find its molality. [Ans 5.135 m]
HOTS
(8) How many mL of sulphuric acid of density 1.84 g/mL containing 95.6 mass % of H2SO4 should be added to one litre of
40 mass % solution of H2SO4 of density 1.31 g/mL in order to prepare 50 mass % solution of sulphuric acid of density 1.40
g/mL. [Ans : 166.2 mL of 95.6 mass%]
(9) Calculate the density of the sulphuric acid solution whose molarity is 10.8 M & molality is 92.6 m. [Ans: 1.17 g/mL]
(10)How many mL of a 0.1M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing
equimolar amount of both? [Ans. 157.8 ml]
(11) When 10.0 mL of ethanol of density 0.7893 g/mL is mixed with 20.0 mL of water if density 0.997 g/mL at 25°C, the
final solution has the density of 0.9571 g/mL. Calculate the percentage change in volume on mixing. Also calculate the
molarity of the final solution. [Ans 3.1% contraction, 5.89 M]

12. Solutions of solids in a liquid as solvent:
Ionic compounds like NaCl, KNO3 etc are soluble in polar solvent like water.
Molecular solids like iodine, naphthalene dissolve in non polar solvents like benzene, CCl4,
CS2 etc.
Certain non-ionic solids like urea, glucose, fructose, sucrose etc are soluble in water due to
the formation in hydrogen bonding.
Solubility of a solid in a liquid at any temperature is defined as the maximum amount of solid
solute in grams that saturate 100 g of a liquid solvent at a given temperature.
Solubility of a solid solute is the maximum no. of moles of that solute that saturate 1 L of
solution at a given temperature. It is the maximum molarity of a solution.
An ionic solid is soluble in water if hydration enthalpy of salt in solution > lattice enthalpy of
ionic solid.

13. Effect of temperature on solubility of a salt:
(i) If dissolution of an ionic solute in water is endothermic (heat is absorbed) , the solubility of
salt increases with the increase in temperature according to Le Chatelier’s principle. E.g.
NaNO3, NaCl. KCl, KNO3 etc.
(ii) If dissolution of an ionic solute in water is exothermic (heat is released) , the solubility of
salt decreases with the increase in temperature. E.g. Li2CO3, Na2CO3.H2O, Ce2(SO4)3 etc.
(iii) In case of Na2SO4.10H2O solubility increases with the increase in temperature but if on
heating at 32°C and above it changes into Na2SO4 (anhydrous) & solubility decreases with the
increase in temperature.

14. Solution of gases in a liquid solvent:
Solubility of a gas in a liquid in the no. of moles of that gas that saturate one litre of solution at a given
temperature. It can be expressed in mole fraction (χgas) or M or m.
Factors affecting solubility of a gas in a liquid:
(i) Non polar gases (lower critical temperature value) like He, H2, O2, N2 etc are less soluble than polar gases
(or non polar gases with polar bonds) having high critical temperature like, CO2, H2S, NH3, HCl etc.
(ii) Temperature: The dissolution of a gas in a liquid is an exothermic process, so the solubility of a gas
decreases with the increase in temperature according to Le Chatelier’s principle.
(iii) Pressure: With the increase in pressure solubility of a gas increases.
Henry’s law: At a particular temperature, the solubility of a non-reacting gas ( in terms of mass or mole
fraction) in a given solvent is directly proportional to the pressure of the gas above the liquid.
Mass of a gas (m) α pressure of the gas (p) above the solution (Temperature is constant)
or m = k pgas
or, mole fraction of a gas: (χgas) = K pgas. Here KH = 1/K
It can be written as pgas = KH. Χgas where KH is called Henry’s law constant.
Unit of KH is same with the unit of pressure of the gas.

15. Henry’s law: pgas = KH. Χgas
The graph is plotted between pressure of the gas & mole fraction of the gas, the
slope = Henry’s law constant (KH)
Thus: (i) KH is the function of nature of a gas.
(ii) At same pressure, higher value of KH of a gas means it has lower solubility.
(iii) KH increases with the increase in temperature means solubility of a gas decreases with the
increase in temperature.
Remember: (i) Aquatic life becomes difficult in summer than winter.
(ii) According to Henry’s law, mass of a gas dissolved is directly proportional to the pressure of
the gas above the liquid but volume of the gas dissolved measured at a pressure used is
independent of pressure.
(iii) For solubility of a gas in a liquid,
ΔS = -ve ( gas dissolved in solution is less random, entropy decreases)
ΔG = -ve (dissolution of a gas is spontaneous in a liquid)
ΔH = -ve (as only then ΔG can be -ve according to the equation:
ΔG = ΔH – TΔS), Hence the dissolution of a gas in water is always exothermic.

16. Application of Henry’s law:
(i) CO2 is dissolved in cold drinks at high pressure to increase its solubility. If a soda bottle is opened,, the
pressure inside the bottle equals to the atmospheric pressure, the solubility of CO2 decreases, the undissolved
CO2 escapes out producing bubbles with fizzing sound.
(ii) In deep sea divers, the oxygen cylinder contains helium having high KH value, in place of N2 to save the scuba
divers from the disease called bends or decompression disease.
(iii) Atmospheric pressure at high altitude is low so, O2 pressure in human blood & tissue are less so, people at
high altitude feel weak & are unable to think properly, this is a disease called anoxia.
(iv) In the lungs, partial pressure of O2 is high. It combines with haemoglobin of blood to form oxyhaemoglobin. In
the tissues, partial pressure of O2 is low, so O2 is released from oxyhaemoglobin for functions of cells.
Examples:
1: Give reasons: At higher altitudes, people suffer from a disease called anoxia. In this disease, they
become weak and cannot think clearly.
Ans: At higher altitudes, the partial pressure of oxygen is less than at ground level. This leads to low concentration
of oxygen in blood and the tissues of the people living at higher altitudes. As a result of low oxygen in the blood,
the people become weak and unable to think clearly. These are the symptoms of a condition known as anoxia.

17. Numerical problems:
1) If N2 gas is bubbled through the water at 293 K, how many millimoles of N2 gas would dissolve in 1 L of
water? (Assume that N2 exerts a partial pressure of 0.987 bar. The KH for N2 at 293 is 76.48 K bar).
Ans: The solubility of gas is related to its mole fraction in aqueous solution.
The mole fraction of gas in solution,
ꭓN2
=
pN2
KH
=
0.987 bar
76480 bar
= 1.29 X 10-5
If n is the number of moles of N2 in solution & 1 L of water contains 55.5 mol, then
ꭓN2
=
n
n+55.5
≈ 1.29 X 10-5 (n in the denominator is neglected because it is << 55.5)
Therefore, n= 55.5 X 1.29 X 10-5 =7.16 X 10-4 mol =7.16 X 10-1 =0.716 m mol (Ans)
2) Dry air contains 79% N2 and 21% O2. Determine the proportion of N2 and O2 (in terms of mole fractions)
dissolved in water at 1 atm pressure. Henry’s law constant N2 and O2 in H2O are 8.54 X 104 atm & 4.56 X 104
atm respectively.
Ans: Total pressure of air over water = 1 atm partial pressure of N2 & O2 are:
pN2
=
1 X 79
100
= 0.79 atm and pO2
=
1 X 21
100
= 0.21 atm
Applying, Henry’s Law: pN2
= KH . ꭓN2
or,
ꭓN2
=
pN2
KN2
=
0.79
8.54 X 104 = 9.25 X 10-6 & ꭓO2
=
pO2
KO2
=
0.21
4.56 X 104 = 4.60 X 10-6
Ratio of N2 & O2 = 9.25 X 10-6 : 4.60 X 10-6 = 2:1 (Ans.)

18. Numerical Problems to practice:
Q.1. One litre of water under a nitrogen pressure 1 atm dissolves 0.02g of nitrogen at 293 K.
Calculate Henry’s law constant. [Ans. 7.75 x 104 atm]
Q.2. Calculate the amount of CO2 dissolved at 4 atm in 1 dm3 of water at 298 K. The Henry’s law
constant for CO2 at 298 K is 1.67 K bar. [Ans. 5.85 g]
Q.3. The mole fraction of He gas in a saturated solution at 20° C is 1.25 X 10-6. Calculate the
pressure of He gas above the solution. (KH of Helium at 20° C = 144.98 K bar). [Ans. 0.181 bar]
Q.4 Henrys law constant for the molality of methane in benzene at 298 K is 4.27 x 105 mm Hg.
Calculate the solubility of methane in benzene at 298 K under 760 mm Hg. [Ans. 178 x 10-5]
Think: At same temperature hydrogen is more soluble than helium gas in water. Which of these
gases will have more KH value & why?

19. Liquid in liquid solutions:
(i) Vapour pressure: of a liquid of a solution is the
pressure exerted by the vapour in equilibrium with the
liquid or solution at a particular temperature.
(ii) Raoult’s law: In a solution, the vapour pressure of a
component of a given temperature is directly
proportional to the mole fraction of that component in
the solution.
For a binary liquid solution of liquids A & B,
For liquid A; pA α χA or pA = 𝐩𝐀
𝐨
. 𝛘𝐀
& for liquid B: pB = 𝐩𝐁
𝐨
. 𝛘𝐁
So, total pressure: Ptotal = pA + pB = 𝐩𝐀
𝐨
. 𝛘𝐀 + 𝐩𝐁
𝐨
. 𝛘𝐁
= (1 - 𝛘𝐁) 𝐩𝐀
𝐨
+ 𝐩𝐁
𝐨
. 𝛘𝐁
(where pA
o
& pB
o
are vapour pressures of pure liquids A &
B, pA & pB are vapour pressure of liquids A & B in
solution & χA and χB are mole fractions of liquids A & B
in solution.
Graphically it can be explained as

20. REMEMBER:
Raoult’s law is applicable to two completely miscible liquids having nearly similar structures forming ideal solution. E.g. benzene +
toluene, n-hexane + n- heptane.
The vapour phase contains vapour molecules of both the components of binary liquid solution, but the vapour pressure of more
volatile component (liquid with less b.p.) will be more.
The mole fraction in the vapour phase are calculated as follows:
Mole fraction of liquid A in vapour phase yA =
𝐩𝐀
𝐩𝐀+ 𝐩𝐁
=
𝐩𝐀
𝐩𝐭𝐨𝐭𝐚𝐥
and
Mole fraction of liquid B in vapour phase yB =
pB
pA+ pB
=
pB
ptotal
Total vapour pressure of solution increases when a volatile component is added & decreases if less volatile component is added.
Henry’s law is the special case of Raoult’s law .
Raoult’s law: pA = pA
o
. χA
Henry’s law: pgas = KH. χgas Hence, on comparing pA
o
= KH . so for a solution, vapour pressure of pure liquid & Henry’s law
constant of a gas is same. & mole fraction of gas above the solution is compared with mole fraction of solvent in solution.
Raoult’s law can be applied to a solution of non-volatile, non electrolytic solute.
For a solution of non-volatile solute(B) in solution of solvent (A), pA = pA
o
. χA and χA + χB = 1
pA = pA
o
. (1 − χB ) = pA
o
−χB pA
o
Hence,
pA
o
− pA
pA
o = χB or,
Δp
pA
o = χB or, Δp = pA
o
. χB
(Δp = lowering of vp of solution &
pA
o
− pA
pA
o =
Δp
pA
o = relative lowering of v.p. of solution of non- volatile solute).
Hence, Raoult’s law of a solution of non-volatile solute can be stated as “ the relative lowering of vapour pressure
of a solution of non-volatile solute is equal to the mole fraction of solute in solution at a particular temperature”.

21. To, explain Raoult’s law of solution of non-volatile solute, we can find that vapour pressure of a solution of
non-volatile solute is less than the vapour pressure of pure liquid solvent. i.e. pA < pA
o
. This is due to the
presence of non volatile particles occupying a part of liquid surface & decreases the surface area of the
solvent to evaporate on the surface & thus decreasing the vapour pressure of solution. Diagrammatically,
Pure Solvent (vapour pressure = pA
o) Solution of non-volatile solute (vap. pressure = (pA)
pA
o
pA
Solved example:
An aqueous solution of glucose (molar mass 180g/mol) is made by dissolving 10 g of glucose in 90 g of
water at 303 K. If the vapour pressure of pure water at 303K be 32.8 mm Hg, what would be the vapour
pressure of the solution ?
Ans. According to Raoult’s law, vapour pressure of the solution, pA = pA
0xA. Given: pA
0 = 32.8 mm at 303K
No. of moles of water = 90/18 = 5 and no. of moles of glucose = 10/180 = 0.0556
Mole fraction of water = xA =
5.0
5.0+0.0556
= 0.989, Hence v.p. of solution = 32.8x0.989 = 32.44 mm Hg (Ans)

22. Numericals for practice:
(i) Vapour pressure of chloroform (molar mass = 119.5 g/mol) & dichloromethane (molar mass = 55 g/mol) at 298
K are 200 mm & 415 mm Hg respectively. (i) calculate the vapour pressure of solution prepared by mixing 25.2 g
of chloroform & 40 g of dichloromethane at 298K & (ii) mole fraction of each component in solution & vapour
phase? (Ans. P = 347.9 mm Hg, Mole fraction in vapour phase of CHCl3 = 0.18 & CH2Cl2 = 0.82)
(ii) An aqueous solution containing 28 % by weight of a liquid A (molar mass = 140 g/mol) has vapour pressure of
0.210 bar at 300K. Calculate the vapour pressure of pure liquid.
(Vapour pressure of pure water at 300 K = 0.198 bar) [Ans.0.448 bar]
(iii) At 20℃, the vapour pressure of pure liquid A is 22 mm Hg & that of pure liquid B is 75 mm Hg. What is the
composition of the solution of these two components that has a vapour pressure of 48.5 mm Hg at this
temperature(assume ideal behaviour). [Ans: xA = xB = 0.5]
(iv) Two liquids A & B have vapour pressures of 0.658 bar & 0.264 bar respectively. In an ideal solution of the two,
calculate the mole fraction of A at which the two liquids have equal partial pressures. [Ans.0.286]
(v) The liquids X & Y form ideal solution having vapour pressures 200 & 100 mm Hg respectively. Calculate the
mole fraction of component X in vapour phase in equilibrium with an equimolar solution of the two. [Ans: 0.67]

23. Ideal solution & non ideal solution:
Ideal Solution:
A binary liquid solution is said to be an ideal solution when the solution
obeys Raoult’s law under all conditions.
Conditions are:-
(i) For solutions of two liquids A & B, pA = 𝐩𝐀
𝐨
. 𝛘𝐀 , pB = 𝐩𝐁
𝐨
. 𝛘𝐁
Ptotal = 𝐩𝐀
𝐨
. 𝛘𝐀 + 𝐩𝐁
𝐨
. 𝛘𝐁
(ii) ΔHmixing = 0 ( there is no change in enthalpy on mixing the liquids in
solution).
(iii) ΔVmixing = 0 (there in no volume change on mixing the liquids in
solution).
(iv) A-A- bonds of liquid A & B-B- bonds of liquid B are as strong as A-B-
A-B- bonds in solution.
Examples: Benzene & toluene, n-hexane & n- heptane, chlorobenzene &
bromobenzene etc.

24. Non-ideal solution:
A binary liquid solution is said to be an non-ideal solution when the solution does not obey Raoult’s law
under all conditions. Conditions are:-
(i) For solutions of two liquids A & B, pA ≠ 𝐩𝐀
𝐨
. 𝛘𝐀 , pB ≠ 𝐩𝐁
𝐨
. 𝛘𝐁 Ptotal≠ 𝐩𝐀
𝐨
. 𝛘𝐀 + 𝐩𝐁
𝐨
. 𝛘𝐁
(ii) ΔHmixing ≠ 0 ( there is no change in enthalpy on mixing the liquids in solution).
(iii) ΔVmixing ≠ 0 (there in no volume change on mixing the liquids in solution).
(iv) A-A- bonds of liquid A & B-B- bonds of liquid B are not as strong as A-B-A-B- bonds in solution
• Ideal Solution
• A binary liquid solution is said to be an ideal solution when
the solution obeys Raoult’s law under all conditions.
• Conditions are:-
(i) For solutions of two liquids A & B, pA = pA
o
. χA , pB = pB
o
. χB
Ptotal = pA
o
. χA + pB
o
. χB
(ii) ΔHmixing = 0 ( there is no change in enthalpy on mixing the
liquids in solution).
(iii) ΔVmixing = 0 (there in no volume change on mixing the
liquids in solution).
(iv) A-A- bonds of liquid A & B-B- bonds of liquid B are as strong
as A-B-A-B- bonds in solution. On change in bond interaction
before & after mixing liquids.
Non ideal Solution
A binary liquid solution is said to be an non-ideal solution
when the solution does not obey Raoult’s law under all
conditions.
• Conditions are:-
(i) For solutions of two liquids A & B, pA ≠ pA
o
. χA , pB ≠
pB
o
. χB Ptotal≠ pA
o
. χA + pB
o
. χB
(ii) ΔHmixing ≠ 0 ( there is no change in enthalpy on
mixing the liquids in solution).
(iii) ΔVmixing ≠ 0 (there in no volume change on mixing
the liquids in solution).
(iv) A-A- bonds of liquid A & B-B- bonds of liquid B are
not as strong as A-B-A-B- bonds in solution

25. Types of non-ideal solution:
Non ideal solution with positive deviation Non ideal solution with negative deviation
(i) For solutions of two liquids A & B, pA> 𝐩𝐀
𝐨
. 𝛘𝐀 ,
pB >
𝐩𝐁
𝐨
. 𝛘𝐁
So, Ptotal > 𝐩𝐀
𝐨
. 𝛘𝐀 + 𝐩𝐁
𝐨
. 𝛘𝐁
(ii) ΔHmixing > 0 ( there is no change in enthalpy on
mixing the liquids in solution).
(iii) ΔVmixing > 0 (there in no volume change on mixing
the liquids in solution).
(iv) A-A- bonds of liquid A & B-B- bonds of liquid B are
stronger than A-B-A-B- bonds in solution.
(i) For solutions of two liquids A & B, pA< 𝐩𝐀
𝐨
. 𝛘𝐀 ,
pB < 𝐩𝐁
𝐨
. 𝛘𝐁
So, Ptotal < 𝐩𝐀
𝐨
. 𝛘𝐀 + 𝐩𝐁
𝐨
. 𝛘𝐁
(ii) ΔHmixing < 0 ( there is no change in enthalpy on
mixing the liquids in solution).
(iii) ΔVmixing < 0 (there in no volume change on mixing
the liquids in solution).
(iv) A-A- bonds of liquid A & B-B- bonds of liquid B are
weaker than A-B-A-B- bonds in solution.
Example: Ethanol & cyclohexane,
Acetone& CCl4, Acetone & ethyl alcohol,
Acetone & benzene, Methanol & water,
ethanol & water, CCl4 & CHCl3, CCl4 & C6H6,
CCl4 & toluene etc.
Example: chloroform & acetone, CHCl3 & C6H6,
Chloroform & diethyl ether, Acetone & aniline,
HCl & water, HNO3 & Water, Acetic acid & pyridine etc

26. NON IDEAL SOLUTIONS : GRAPHICALLY
Ptotal > 𝐩𝐀
𝐨
. 𝛘𝐀 + 𝐩𝐁
𝐨
. 𝛘𝐁 Ptotal < 𝐩𝐀
𝐨
. 𝛘𝐀 + 𝐩𝐁
𝐨
. 𝛘𝐁

27. GRAPHICALLY THE COMPARISON OF IDEAL AND NON IDEAL
SOLUTIONS ARE:
Ideal solution Non ideal solution with +ve deviation solution with- ve
deviation

28. Azeotropic Mixture: It is a binary liquid solution which boils at a particular temperature like a pure liquid.
It distils over in the same composition at this temperature. It is also called as constant boiling mixture.
It is of two types:-
(a) Minimum boiling azeotrope: It is the azeotropic mixture whose b.p. is less than the boiling point of
individual liquid components. It is shown by the non-ideal solution with positive deviation (vapour pressure
of solution is maximum).
Example: Azeotropic Mixture of ethanol (bp 351.3K ) & water (bp 373.15K) has 95.4% ethanol by volume
having bp = 351.5K.
(b) Maximum boiling azeotrope: It is the azeotropic mixture whose b.p. is more than the boiling point of
individual liquid components. It is shown by the non-ideal solution with negative deviation (vapour pressure
of solution is minimum).
Example: Azeotropic Mixture of HNO3 (bp 359K ) & water (bp 373.15K) has 68.0% nitric acid by mass having
bp = 393.5K.
Think:
We cannot separate completely water & nitric acid from its dilute solution by fractional distillation. Why?

29. Numerical problems to practice:
(i) Vapour pressure of chloroform (molar mass = 119.5 g /mol) & dichloromethane (molar mass = 55 g/mol) at 298
K are 200 mm & 415 mm Hg respectively. (i) calculate the vapour pressure of solution prepared by mixing 25.2 g
of chloroform & 40 g of dichloromethane at 298K & (ii) mole fraction of each component in solution & vapour
phase? (Ans. P = 347.9 mm Hg, Mole fraction in vapour phase of CHCl3 = 0.18 & CH2Cl2 = 0.82)
(ii) An aqueous solution containing 28% by weight of a liquid A (molar mass = 140g/mol) has vapour pressure of
0.210 bar at 300K. Calculate the vapour pressure of pure liquid.
(Vapour pressure of pure water at 300 K = 0.198 bar. [Ans.0.448 bar]
(iii) At 20℃, the vapour pressure of pure liquid A is 22 mm Hg & that of pure liquid B is 75 mm Hg. What is the
composition of the solution of these two components that has a vapour pressure of 48.5 mm Hg at this
temperature. (assume ideal behaviour). [Ans: xA = xB = 0.5]
(iv) Two liquids A & B have vapour pressures of 0.658 bar & 0.264 bar respectively. In an ideal solution of the two,
calculate the mole fraction of A at which the two liquids have equal partial pressures. [Ans.0.286]
(v) The liquids X & Y form ideal solution having vapour pressures 200 & 100 mm Hg respectively. Calculate the
mole fraction of component X in vapour phase in equilibrium with an equimolar solution of the two. [Ans: 0.67]

30. Numerical Problems : (continued)
(vi) An aqueous solution of is made by dissolving 10 g of glucose (molar mass = 180 ) in 90 g of
water at 303 K. If the vapour pressure of pure water at 303 K be 32.8 mm Hg, what would be the
vapour pressure of the solution? [Ans 32.44 mm]
(vii) At 298 K the vapour pressure of pure benzene is 0.256 bar & vapour pressure of pure toluene is
0.0925 bar. If the mole fraction of benzene in solutions 0.40, (i) find the total vapour pressure of
solution (ii) calculate the composition of the vapour in terms of male fraction.
[Ans: Ptotal = 0.158 bar, ybenzene= 0.648 & ytoluene = 0.352]
(viii) Two liquids X & Y on mixing form an ideal solution. The vapour pressure of this solution
containing 3 mol of X & 1 mol of Y is 550 mm of Hg. But when 4 mol of X & I mol of Y are mixed the
vapour pressure of solution becomes 560mm of Hg. What will be he vapour pressure of pure liquids
X and Y at this temperature? [Ans: X = 600 mm Y = 400mmof Hg].
HOTS
(ix) At a certain temperature, the vapour pressure (in mm of Hg) of CH3OH & C2H5OH solution is
represented by P= 119 x +135 where x is the mole fraction of CH3OH. Find the vapour pressures of
pure components at this temperature? [Ans: pmethanol=254 mm, pethanol=135mm]

31. Colligative Properties: These are the properties which
depends only on the number of the solute particles (ions or molecules)
in solution but not on the nature of the solute particles or solvent
molecules. Colligative properties depends on no. of moles solute
particles.
The colligative properties are observed in dilute solutions of non-volatile
solute & are of four types:-
(i) relative lowering of vapour pressure of solution (
pA
o
− pA
pA
o =
Δp
pA
o )
(ii) elevation of boiling point of solution (ΔTb)
(iii) depression of freezing point of solution (ΔTf)
(iv) osmotic pressure of solution (π)

32. (i) Relative lowering of vapour pressure of a solution of non-volatile solute (
pA
o
− pA
pA
o =
Δp
pA
o )
From the Raoult’s law of relative lowering of vapour pressure:
Δp
pA
o = χB =
nB
nA+nB
or,
Δp
pA
o =
wB
mB
wA
mA
+
wB
mB
For a dilute solution of non-volatile solute (B), nA >> nB so, nA + nB ≈ nA ,
Hence
Δp
pA
o =
nB
nA
or,
Δp
pA
o =
wB
mB
wA
mA
Therefore, mB =
𝐩𝐀
𝐨
∆𝐩
𝐰𝐁
𝐰𝐀
𝐦𝐀 Thus, we can calculate the molar mass of a solute in solution.
Think:
Lowering of vapour is not a colligative property whereas relative lowering of vapour pressure is a colligative property. Why?

33. RELATIVE LOWERING OF VAPOUR PRESSURE:
Example:
(1) The vapour pressure of water at 20℃ is 17.5 mm Hg & lowering of vapour pressure of a sugar solution is 0.061
mm Hg. Calculate (i) relative lowering of vapour pressure (ii) vapour pressure of the solution (iii) mole fraction of
sugar & water.
Solution: (i) V.p. of pure water pA
o = 17.5 mm, lowering of v.p. = pA
o – pA = Δp = 0.061 mm
Therefore, Relative lowering of v. p. =
𝛥𝑝
𝑝𝐴
𝑜 = 0.061/17.5 = 0.00348 (Ans)
(ii) vapour pressure of solution = pA = vp of solvent – lowering of vp of solution = 17.5 – 0.061 = 17.439 mm Hg
(iii) We know for Raoult’s law:
𝛥𝑝
𝑝𝐴
𝑜 = 𝜒𝐵 = 0.00348 (Ans)
Hence the mole fraction of water = 1 - 𝜒𝐵 = 1 – 0.00348 = 0.996 (Ans)
(2) Calculate the vapour pressure of an aqueous solution containing 5% by mass of urea (Molar mass = 60g/mol) at
298K (Vapour pressure of water at 298 K is 23.75 mm Hg)
Solution: 5% urea solution by mass means 5 g urea in 95 g water
Mass of solute urea (wB)= 5 g, Mass of water as solvent (wA) – 95 g,
molar mass of urea (mB) = 60, Molar mass of water (wA) = 18, v.p. of water pA
o = 23.75 mm to find pA = ?
We know, mB =
𝒑𝑨
𝒐
∆𝒑
𝒘𝑩
𝒘𝑨
𝒎𝑨 or 𝛥𝑝 =
𝒑𝑨
𝒐
𝒎𝑩
𝒘𝑩𝒎𝑨
𝒘𝑨
, ∆𝒑 =
𝟐𝟑.𝟕𝟓𝒙𝟓𝒙𝟏𝟖
𝟔𝟎𝒙𝟗𝟓
= 0.375
pA = pA
o - ∆p = 23.75-0.375 = 23.375 mm Hg (Ans.)

34. Problems to practice:
(1) The vp of water is 12.3 kPa at 300K. Calculate the v.p. of 1 molal solution of a solute in it. [Ans. 12.08kPa]
(2) The vapour pressure of pure benzene (molar mass = 78 g/mol) at a certain temperature is 262 bar. At the
same temperature, the vapour pressure of a solution containing 2 g of non volatile , non electrolytic solid in 100
g of benzene is 256 bar. What is the molecular mass of the solid? [Ans: 68.12 g/mol]
(3) The vapour pressure of a 5% aqueous solution of a non volatile organic substance at 373K is 745 mm Hg.
Calculate the molar mass of the solute. (Vapour pressure of water at 373 K is = 760 mm Hg) [Ans: 48 g/mol]
(4) What mass of a non-volatile solute urea (molar mass 60 g/mol) needed to be dissolved in 100 g of water in
order to decrease the vapour pressure of water by 25%? Calculate the molality of solution?
[Ans: 111 g, 18.5 m]
(5) The vapour pressure of water at 293 K is 0.0231 bar & the vapour pressure of solution of 108.24 g of a
compound in 1000 g of water at the same temperature is 0.0228 bar. Calculate the molar mass of the solute.
[Ans: 150 g/mol]

35. (II) ELEVATION OF BOILING POINT OF A SOLUTION OF NON-VOLATILE
SOLUTE (𝜟𝑻𝒃)
Boiling point of a solution is the temperature at which its vapour pressure
equals to atmospheric pressure.
Example: boiling point of water is 373.05 K (100°C) means vapour pressure
of water at 373.15 K is 1 atmospheric pressure (or 1.013 bar). We know that
the vapour pressure of a pure liquid or solution is a function of temperature.
Here Tb
0
and Tb are the boiling points of pure liquid & solution. Here, at 1
atm. Tb
0
< Tb , so there is elevation of boiling point of solution due to the
presence of non-volatile solute in solution (having lower vapour pressure
than that of pure liquid) so, Δ𝐓𝐛 = 𝐓b - 𝐓𝐛
𝟎
.
Solvent Solutio
n
<—
>
ΔТb

36. Elevation of boiling point of a solution of non-volatile solute (𝜟𝑻𝒃)
Experiments show that for a dilute solution the elevation of boiling point (ΔTb) is
directly proportional to the molality (m) of solution of non-volatile solute Thus,
ΔTb α m or 𝚫𝐓𝐛 = Kb. m
where Kb is called boiling point elevation constant or molal elevation constant or
Ebullioscopic constant.. Its unit is K kg mol-1
. For a solution, Kb is defined as the
elevation of boiling point of one molal solution.
We know, molality of a solution (m) =
No.of moles of solute
mass of solvent in kg
=
1000.wB
wAmB
Therefore, 𝚫𝐓𝐛 = Kb. m 𝐨𝐫, ΔTb = Kb.
1000.wB
wAmB
And we can calculate the molar mass of non-volatile solute in solution.
mB =
𝟏𝟎𝟎𝟎.𝐊𝐛 .𝐰𝐁
∆𝐓𝐛 .𝐰𝐀
Think: When salt is added in water its boiling point increases. Why?

37. Elevation of boiling point of a solution of non-volatile solute (𝜟𝑻𝒃)
Example:
(1) The boiling point of benzene is 353.23 K. When 1..80 g of a non volatile solute was dissolved in 90 g of benzene,
the boiling point of solution is raised to 354.11 K. Calculate the molar mass of solute. (Kb of benzene is 2.53 K kg/mol)
Solution: Given: Kb = 2.53 K kg/mol, wB = 1.80 g, wA = 90 g and 𝛥𝑇𝑏 = 354.11 – 353.23 = 0.88 K
Molar mass can be calculated by the formula: mB =
1000.𝐾𝑏 𝑤𝐵
∆𝑇𝑏 𝑤𝐴
=
1000𝑥 2.53 𝑥 1.80
90 𝑥 0.88
= 57.5 g/mol (Ans.)
(2) 18 g of glucose (molar mass=180 g/mol) is dissolved in 1000 g of water in a saucepan. At what temperature will
water boil at 1.013 bar? (Kb for water = 0.52 K kg mol-1)
Solution: Given: wB = 18 g, wA = 1000 g mB = 180 g/mol Kb = 0.52 K kg/mol
We know, 𝛥𝑇𝑏 = Kb.
1000.𝑤𝐵
𝑤𝐴𝑚𝐵
=
0.52 𝑥 1000 𝑥 18
1000 𝑥 180
= 0.052 K
Therefore the boiling point of solution = Tb
o + 𝛥𝑇𝑏 = 373.15 + 0.052 = 373.202 K (Ans.)

38. Elevation of boiling point of a solution of non-volatile solute (𝜟𝑻𝒃)
PROBLEMS for practice:
(1) On dissolving 3.24 g of sulphur in 40 g of benzene , the boiling point of solution was higher than that of
benzene by 0.81 K. (Kb of benzene is 2.53 K kg/mol). What is the molecular formula of sulphur?
[Ans. S8]
(2) A solution of glycerol (molar mass = 92 g/mol) in water is prepared by dissolving some glycerol in 500 g of
water. This solution had a boiling point of 100.42℃ while pure water boils at 100℃ Find the mass of glycerol
dissolved in the solution? [Ans. 37.73 g]
(3) What is the elevation in boiling point is to be expected when 5 g of urea (molar mass = 60) are dissolved
in 75 g of it? (Kb of alcohol= 1.15 K kg/mol) [Ans. 1.28K)
(4) A solution of 12.5 g of a non electrolyte solid in 175 g of water gave the boiling point elevation of 0.70 K.
Calculate the molar mass of the substance (Kb of water = 0.52K kg/mol) [Ans. 53.06 ]

39. (iii) Depression of freezing point of a solution of non volatile solute (𝜟𝑻𝒇)
Freezing point of a solution is the temperature at which the
vapour pressure of solution in liquid phase is in dynamic
equilibrium with vapour pressure of its solid phase.
According to Raoult’s law when a non-volatile solid is
added to a solvent, its vapour pressure decreases & now it
becomes equal to that of the solid solvent at lower temperature.
Thus, the freezing point decreases as shown in the graph.
Here Tf
0
and Tf are freezing points of pure liquid & solution.
Here, when vapour pressures in liquid phase & solid phase
are equal 𝐓𝐟
𝟎
> Tf , so there is depression of freezing point of
solution due to the presence of non volatile solute in solution
(having lower vapour pressure than that of pure liquid) so,
𝚫𝐓𝐟 = 𝐓𝐟
𝟎
− 𝐓𝐟.

40. (iii) Depression of freezing point of a solution of non volatile solute (𝚫𝐓𝐟)
Experiments show that for a dilute solution the depression of freezing point (ΔTf) is
directly proportional to the molality (m) of solution of non volatile solute.
Thus, ΔTf α m or Δ𝐓𝐟 = Kf. m
where Kf is called freezing point depression constant or molal depression constant or
cryoscopic constant. Its unit is K kg mol-1
.
For a solution, KF is defined as the depression of freezing point of one molal solution.
We know, molality of a solution m =
No.of moles of solute
mass of solvent in kg
=
1000.wB
wAmB
Therefore, Δ𝐓𝐟 = Kf m
or, 𝚫𝐓𝐟 = Kf.
𝟏𝟎𝟎𝟎.𝐰𝐁
𝐰𝐀𝐦𝐁
We can calculate the molar mass of non volatile solute in solution as
mB =
𝟏𝟎𝟎𝟎.𝐊𝐟 𝐰𝐁
∆𝐓𝐟 𝐰𝐀

41. (iii) Depression of freezing point of a solution of non volatile solute (𝚫𝐓𝐟)
1.ΔTb orΔTf valuesinkelvinscale isequaltodegreecentigradescale.Itisnotrequiredtochange the unit.
2.Touse formula,wA & wB should bein g.
3.Kb & Kf valuesdependupon the nature of the solvent&hasafixedvalue.These canbe calculatedas
Kf =
R mA.Tf
2
1000 ∆fus H
and Kb =
R mA.Tb
2
1000 ∆vap H
where Tf &Tb are f.p.& b.p.of puresolvent& ΔfusH &ΔvapHare enthalpiesof fusion&
vaporisationofsolventrespectively.
4. We cancalculate b.p fromf.p.of solution andvice− versaifKb & Kf ofsolventisknown
ΔTb =Kf.m & ΔTf =Kf.m Hence,
∆Tb
∆Tf
=
Kb
Kf
or
Tb−Tb
0
Tf
0
−Tf
=
Kb
Kf
5.Anti-freeze issolutionofethylene glycolinwater,used incarradiatorsincold countries.
6.ByRastmethod, depressioninfreezingpointiscalculated experimentallyusingcaophorassolventforsolid soluteslike
naphthaleneetcascamphorhashigh Kf value =40Kkgmol-1 at39.7°C.
7.Common saltoranhydrousCaCl2 isare usedtoclearsnowonroadsofsnowy mountains.Thisisbecausetheydepressthe f.p.of
watertosuchasextentthatitcannotfreezetoformice.Hence,itmeltsoffeasilyatthattemperature & clearsthe road.

42. (iii) Depression of freezing point of a solution of non volatile solute (𝜟𝑻𝒇)
Example:
(1) 45 g of ethylene glycol (molar mass = 62 g/mol) is mixed wih 600 g of water. Calculate (i) the depression in
freezing point of solution (ii) the freezing point of solution. (Kf of water = 1.86K kg/mol)
Solution: Given: wB = 45 g, wA = 600 g , mB = 62 g/mol, Kf = 1.86 K kg/mol & we know ΔTf = Kf.
1000.wB
wAmB
ΔTf =
1.86 x 1000 x 45
600 x 62
= 2.25 K
So depression in freezing point = 2.25 K & freezing point Tf = Tf
o – ΔTf = 273.15 – 2.25 = 270.90 K (Ans.)
(2) Two elements A & B form compounds having molecular formula AB2 and AB4.When dissolved in 20 g of
benzene, 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers the freezing point by 1.3 K (Kf
of benzene is 5.1 K) Calculate the atomic masses of A and B.
Solution: Let atomic masses of elements A & B are a & b respectively. Molar mass of AB2 m AB2= a + 2b & for AB4
mAB4 = a + 4b and for compound AB2, wB = 1.0 g wA = 20.0 g Kf = 5.1 K kg/mol mAB2 = a + 2b & ΔTf = 2.3 K
We know, mB =
1000.Kf wB
∆Tf wA
or a + 2b =
1000 x 5.1 x 1
2.3 x 20
or a + 2b = 110.87 …………….. (i)
Similarly for compound AB4, mB =
1000.Kf wB
∆Tf wA
or a + 4b =
1000 x 5.1 x 1
1.3 x 20
= 296.15….………… (ii)
Solving (i) & (ii), we get, a = 25.59 & b = 42.64, so, atomic mass of A = 25.59 u & atomic mass of B = 42.64 u (Ans.)

43. (iii) Depression of freezing point of a solution of non volatile solute (𝚫𝐓𝐟)
Problems for practice:
(1) Calculate the mass of a compound (molar mass = 256 g /mol) to be dissolved in 75 g of benzene to lower its
freezing point by 0.48 K (Kf of benzene = 5.12 K kg /mol). [Ans. 1.8 g]
(2) 1.0 g of a non electrolyte solute dissolved in 50 g of benzene lowered the freezing pont of benzene by 0.40K. Find
the molar mass of solute (Kf of benzene = 5.12 K kg/mol). [Ans. 256 g/mol]
(3) Two aqueous solutions containing 7.5 g of urea (molar mass = 60) & 42.75 g substance X in 100 g water freeze at
same temperature. Calculate the molar mass of X. [Ans. 342 g/mol]
(4) An aqueous solution freezes at 272.07 K while pure water freezes at 273 K. Determine the molality & boiling
point of the solution. (Kb & Kf of water are 0.52 & 1.86 K kg/mol respectively) [Ans, 0.5 m, 373.256 K]
(5) In cold countries, water gets frozen causing damage to the car radiator. Ethylene glycol is used as an antifreeze.
Calculate the amount of ethylene glycol to be added to 4 kg of water to prevent freezing ̶6℃.
(Kf for water = 1.85 K/m) [Ans. 804.32 g]

44. (iv) Osmotic Pressure of solution of non-volatile solute. (π)
• Osmosis is the phenomenon of net movement of
solvent particles through from a region of higher
concentration of solvent (lower concentration of
solute) in solution to lower concentration of solvent
(higher concentration of solute) in solution. It is
temperature dependent. A semi permeable
membrane may be animal bladder like pig’s
bladder, parchment paper etc.
• Synthetic semi permeable is gelatinous
Cu2[Fe(CN)6].
• Osmotic pressure (π) of a solution is the minimum
pressure that must be applied on the solution of
higher concentration of solute just to prevent the
net flow of the solvent molecules into the solution
through the semipermeable membrane at a given
temperature. Osmotic pressure depends upon the
concentration of solution & its temperature.

45. (iv) Osmotic Pressure of solution of non-volatile solute. (π)

46. (iv) Osmotic Pressure of solution of non-volatile solute. (π)
[Osmotic pressure of a solution is the excess pressure that must be applied to a solution of higher concentration of
solute to prevent osmosis].
Experimentally, it is observed for a dilute solution of non-volatile solute osmotic pressure is proportional to the molarity
(C in mol L-1) of solution at a given temperature T in kelvin scale.
Hence, π = CRT =
𝐧𝐁
𝐕
𝐑𝐓 =
𝐰𝐁
𝐦𝐁
𝐑𝐓
𝐕
where C =
𝐧𝐁
𝐕
𝐦𝐨𝐥/𝐋
Thus we can calculate the molar mass of solute in solution as mB =
𝐖𝐁𝐑𝐓
𝛑 𝐕
Remember:
1 atm = 760 torr = 760 mm of Hg = 76 cm of Hg = 1.013 bar = 1.013 x 105 Pa & 1 bar = 0.987 atm).
Osmotic pressure method is the best method among other colligative property methods to calculate the
molar mass of macromolecules proteins, polymers etc because, π is calculated at room temperature &
molarity of solution is used in place of molality. So error in measurement is less. Also, for a dilute solution,
the magnitude of π value is appreciably large.

47. (iv) Osmotic Pressure of solution of non-volatile solute. (π)
• Reverse Osmosis:
If the pressure more than the osmotic pressure is
applied on the solution of higher concentration of solute
separated from the solvent by semipermeable membrane,
there is net flow by the solvent through semipermeable
membrane towards the solution of lower concentration of
solute or pure solvent. It is called reverse osmosis (RO).
Application: It is used in desalination of sea water (removal
of salt from sea water). Semi permeable membrane used is
cellulose acetate. Many countries use desalination plants to
meet their potable water requirements.
Isotonic solutions are the solutions having the equal
osmotic pressures.

48. Osmosis
Example: A 0.91 % solution(w/v) of NaCl solution is osmotic with human red blood corpuscles (RBC).
Let solutions X & Y are isotonic solutions having osmotic pressures πX & πY having molar concentrations CX & CY at
T temperature. Here, πX = πY or, CXRT = CYRT or,
𝐧𝐗𝐑𝐓
𝐕
=
𝐧𝐘𝐑𝐓
𝐕
or, nX = nY
i.e. no. of moles of solutes nX & nY are equal in isotonic solutions. Or
𝐰𝐗
𝐌𝐗
=
𝐰𝐘
𝐌𝐘
(Volume & temperature constant)
In general,
𝐰𝐱
𝐌𝐗𝐕𝐗
=
𝐰𝐲
𝐌𝐘𝐕𝐘
Where wX & wY are masses of X & Y solutes of molar concentration Mx & My in volumes VX & VY.
Hypotonic solutions and hypertonic solutions: A solution with less osmotic pressure (lower
concentration) than the other solution in called hypotonic solution. A solution with higher osmotic
pressure (lower concentration) than the other solution is called hypertonic solution.
For solutions X & Y having osmotic pressures πX and πY.
If πX < πY , solution X is hypotonic & solution Y is hypertonic.
A solution of pure NaCl solution less than 0.91% by mass is called hypotonic & RBC will swell & brust
when placed in this solution.
A solution of pure NaCl solution more than 0.91% by mass is called hypertonic & RBC will shrink when
placed in this solution.

49. Osmotic Pressure of solution of non-volatile solute (π).
Some terms related to osmosis:
Turgor: is the pressure developed inside the cell due to inflow of water.
Plasmolysis or crenation: in the shrinking of cell in hypertonic solution.
Homolysis: is the swelling of cell followed by rupture of cell in a hypotonic solution.
The flow of water from a cell to outside is called exosmosis but the flow of water inside a cell
is called endosmosis.
Osmotic pressure is experimentally determined by Berkeley & Hartley method.
The puffiness or swelling due to water retention in tissue cells & intercellular spaces due to
osmosis when people eat more salt in food is called Edema.
The preservation of meat by salting & of fruits by adding sugar protects against bacterial
action. Through the process of osmosis, bacterium on salt meat or candid fruit loses water,
shrivels & dies.
Try: (1) Why it is necessary to administer the intravenous injection with proper concentration to old
people or infants?
(2) A person suffering from sore throat is advised to gargle with lukewarm salty water.

50. Osmotic Pressure:
For numerical problems:
(1) Let a % solution of A is isotonic with b % solution of B at T temperature.
So, molar concentration of solutions of A & B are equal.
𝐚𝐱𝟏𝟎𝟎𝟎
𝟏𝟎𝟎 𝐦𝐀
=
𝐛𝐱𝟏𝟎𝟎𝟎
𝟏𝟎𝟎 𝐦𝐁
or,
𝐚
𝐦𝐀
=
𝐛
𝐦𝐁
(2) If two solutions X & Y of osmotic pressures, πX and πY are mixed of molar concentrations CX & CY in solutions, the
final osmotic pressure (π) of solution mixture of total volume V in litre is calculated as
πX = CXRT and πY = CYRT Hence π = πX + πY = (CX + CY) RT =
𝐧𝐗+𝐧𝐘 ) 𝐑𝐓
𝐕
where nX & nY are no. of moles of X & Y.
Example:
(1) If 1.71 g of sugar (molar mass = 342) are dissolved in 500 cm3 of a solution at 27℃, what will be the osmotic
pressure of the solution? (R = 0.083 L bar /K/mol).
Solution: Given: wB = 1.71 g, V = 500/100 = 0.5 L , T = 27℃ = 300K & R = 0.083 L bar /K/mol
We know, osmotic pressure, π =
WBRT
mB V
=
1.71 x 0.083 x 300
342 x 0.5
= 0.249 bar (Ans.)

51. Osmotic Pressure: (Numerical problems)
(2) 36 g of glucose dissolved in 1 L of solution has an osmotic pressure of 4.98 bar at 300 K. If the osmotic pressure
of the solution is 1.52 bar at the same temperature, what would be its concentration?
Solution: According to Van’t Hoff equation π = CRT
Given concentration of glucose C1 = 36/180 = 0.2 M at π1 = 4.98 bar
And concentration (C2) of solution = x at osmotic pressure π2 = 1.52 bar
We have π1 = C1RT & π2 = C2RT So,
π1
π2
=
C1
C2
or, C2 =
μ2C1
π1
=
1.52 x 0.2
4.98
= 0.061 M
(3) A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of
0.335 torr at 298 K. Assuming the gene fragment is a non-electrolyte, determine its molar mass.
(R = 0.0821 L atm /mol/K)
Solution:
Mass of gene fragment wB = 8.95 x 10-3 g, V = 35 mL = 35 x 10-3 L, T = 298K,
Osmotic pressure (π) = 0.335 torr =
0.335
760
atm
Molecular mass of gene mB =
WBRT
π V
=
8.95 x10−3x 0.0821 x 298
0.335 x 35 x 10−3
= 14193.3 g/mol = 1.42 x 104 g/mol (Ans.)

52. OSMOTIC PRESSURE (Problems for practice):
(1) Calculate the osmotic pressure in Pa exerted by a solution prepared by dissolving 1.0 g of polymer of
molar mass 185,000 in 450 mL of water at 37℃. R = 8.314 J/mol/K. [Ans. 30.96 Pa]
(2) A 5% solution of cane sugar (malar mass = 342) is isotonic with 0.877% solution of urea. Calculate the
molar mass of urea. [Ans.59.99 g/mol]
(3) The osmotic pressure of human blood is 7.7 atm at 40℃. (a) What is the total concentration of all solute
in blood? (b) Assuming the concentration to be essentially the same as the molality, calculate the freezing
point of blood. (Kf = 1.86 K kg/mol) [ Ans.0.3 M , - 0.558℃]
(4) 600 mL of aqueous solution containing 2.5 g of a protein shows an osmotic pressure of 25 mm Hg at
27℃. Calculate the relative molar mass of protein. [Ans, 3119.8 g/mol]
HOTS
(5) At 10℃, the osmotic pressure of urea solution is 500 mm Hg. The solution is diluted & the temperature
is raised to 25℃, when the osmotic pressure becomes 105.3mm Hg. Calculate the extent of dilution.
[Ans. 5 times]

53. Abnormal molar mass of solute in a solution is sometimes calculated by colligative property methods.
We know that colligative property is directly proportional to the number of solute particles in solution & inversely
proportional to the molar mass of solute in solution, Hence, Molar mass of solute is inversely proportional to the
number of solute particles in solution.
The abnormal molar mass of solute is observes when solute undergoes dissociation or association in
solution.
In case of dissociation of ionic solid as solute in solution, more ions are forms, no. of particles
(ions) increases & hence molar mass of ionic solid decreases as calculated by colligative property
method.
e.g. KCl(s) (1mole) + H2O (solvent)  K+
(aq) + Cl-
(aq) (2 moles of ions)
Molar mass of KCl will decrease, calculated by colligative property method.
In case of association of solute in solution, the associated molecules are formed, no. of particles
decrease, hence colligative property increases & calculated molar mass of solute particles increases.
Example: benzoic acid or ethanoic acid in benzene undergoes association to form dimer.
So, molar mass of solute benzoic acid or ethanoic acid increases as calculated by colligative property
method.
Abnormal molar mass of solute in a solution

54. In 1880, Van’t Hoff introduced a factor ‘і’ known as Van’t Hoff factor to account the extent of
dissociation or association of solute in solution. ‘і’ is defined as
i =
𝐍𝐨𝐫𝐦𝐚𝐥 𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
𝐚𝐛𝐧𝐨𝐫𝐦𝐚𝐥 𝐨𝐫 𝐎𝐛𝐬𝐞𝐫𝐯𝐞𝐝 𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 𝐢𝐧 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
i =
𝐎𝐛𝐬𝐞𝐫𝐯𝐞𝐝 𝐜𝐨𝐥𝐥𝐨𝐠𝐚𝐭𝐢𝐯𝐞 𝐩𝐫𝐨𝐩𝐞𝐫𝐭𝐲
𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞𝐝 𝐜𝐨𝐥𝐥𝐢𝐠𝐚𝐭𝐢𝐯𝐞 𝐩𝐫𝐨𝐩𝐞𝐫𝐭𝐲 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
i =
𝐍𝐨.𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟𝐬𝐨𝐥𝐮𝐭𝐞 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 𝐚𝐟𝐭𝐞𝐫 𝐝𝐢𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐨𝐫 𝐚𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐢𝐧 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
𝐍𝐨.𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 𝐛𝐞𝐟𝐨𝐫𝐞 𝐚𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐨𝐫 𝐝𝐢𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧
(a) If i = 1 means solute does not undergo dissociation or association in solution.
(b) If i > 1 means solute undergoes dissociation in solution as ions are formed & no. of
particles increases.
(c) If i < 1, means solute undergoes association in solution as associated molecules
are formed, no. of particles decreases.
Van’t Hoff factor (i) is used in the formula as to calculate colligative property or molar
mass of solute;

55. Colligative property General formula To calculate molar
mass of solute
Relative lowering of
vapour pressure
𝜟𝒑
𝒑𝑨
𝒐 = 𝒊𝝌𝑩 = 𝒊
𝒏𝑩
𝒏𝑨+𝒏𝑩
mB = i
𝒑𝑨
𝒐
∆𝒑
𝒘𝑩
𝒘𝑨
𝒎𝑨
Elevation of boiling point 𝜟𝑻𝒃 = 𝒊 Kb. m mB = 𝒊
𝟏𝟎𝟎𝟎.𝑲𝒃 𝒘𝑩
∆𝑻𝒃 𝒘𝑨
Depression of freezing
point
𝜟𝑻𝒇 = 𝒊 Kf. m mB = 𝒊
𝟏𝟎𝟎𝟎.𝑲𝒇 𝒘𝑩
∆𝑻𝒇 𝒘𝑨
Osmotic pressure π = i CRT = i
𝒏𝑩
𝑽
𝑹𝑻 mB = 𝒊
𝑾𝑩𝑹𝑻
𝝅 𝑽

56. Van’t Hoff factor (i) and Extent of dissociation or association in an electrolytic solution:
(i) Dissociation: Let 𝝰 is the degree of dissociation of an electrolyte forming n ions in
solution. For a reaction: An ⇌ nA (ions)
Initially, t = 0 1 mole 0 mole
At equilibrium, no. of moles (1- 𝝰) n𝝰 moles
Hence total no. of moles = 1 – 𝝰 + n 𝝰
Van’t Hoff factor (i) =
𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑓𝑡𝑒𝑟 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛
𝑁𝑜.𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑡𝑎𝑘𝑒𝑛
=
1 – α + n α
1
or, 𝝰 =
𝑖−1
𝑛−1
For 100% dissociation of electrolyte, 𝝰 = 100 % = 1, so, i = n,
Example: For a strong electrolyte: Al2(SO4)3(s) +H2O ⇌ 2Al3+
(aq) + 3SO4
2-
(aq)
Here n = 5, 𝝰 = 1
So, i = n = 5

57. (ii) Association:
Suppose a solute molecules associate in solution to form the associated molecules An &
𝝰 is the degree of association of solute molecule A.
nA ⇌ An (associated molecule)
Initially, t = o 1 mole 0 mole
At equilibrium, No. of moles (1- 𝝰)
𝛂
𝒏
moles
Hence, total no. of moles after association = (1- 𝝰) +
𝛂
𝒏
Van’t Hoff factor (i) =
𝑵𝒐.𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔 𝒂𝒇𝒕𝒆𝒓 𝒂𝒔𝒔𝒐𝒄𝒊𝒂𝒕𝒊𝒐𝒏
𝑵𝒐.𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔 𝒕𝒂𝒌𝒆𝒏 𝒊𝒏 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
i =
𝟏 – 𝛂 +
𝜶
𝒏
𝟏
or, 𝝰 =
𝒊−𝟏
𝟏
𝒏
−𝟏
For 100% association of solute in a solvent, α = 100 % = 1 Hence, i =
𝟏
𝒏
.
e.g. (a) For a dimer, n = 2 so, i = ½
(b) For a trimer, n = 3, so, i =
𝟏
𝟑
(c) For tetramerization of a solute in solution, n = 4 so, i =
𝟏
𝟒
Van’t Hoff factor (i) and Extent of dissociation or association in an electrolytic solution:

58. Numerical problems using van’t Hoff’s factor :
(1) Calculate the boiling point of solution when 2 g of Na2SO4 (molar mass = 142) was dissolved in 50 g of water, assuming Na2SO4
undergoes complete ionisation. (Kb for water = 0.52 K kg/mol)
Solution: Weight of solute, wB = 2 g, molar mass mB = 142 g/mol, wA = 50 g, Kb = 0.52 K kg/mol
And Na2SO4 → 2Na+ + SO4
2- Here i = 3
We know, ΔTb = 𝐢
𝟏𝟎𝟎𝟎.𝐊𝐛 𝐰𝐁
𝐦𝐁 𝐰𝐀
=
𝟑 𝐱 𝟏𝟎𝟎𝟎 𝐱 𝟎.𝟓𝟐 𝐱 𝟐
𝟏𝟒𝟐 𝐱 𝟓𝟎
= 0.439 Boiling point of solution] Tb = 373 + 0.439 = 373.439 K (Ans.)
(2) A solution containing 3.1 g of BaCl2 (Molar mass = 208.3) in 250 g of water boils at 100.083℃. Calculate the Van’t Hoff’s factor and
molality of BaCl2 in this solution. (Kb for water = 0.52 K/m)
Solution: wB = 3.1 g, wA = 250 g, mB = 208.3, Kb = 0.52 K/m & ΔTb =100.083 – 100 = 0.083℃ = 0.083 K
We know, ΔTb = 𝐢
𝟏𝟎𝟎𝟎.𝐊𝐛 𝐰𝐁
𝐦𝐁 𝐰𝐀
𝐨𝐫, 0.083 =
𝐢 𝐱 𝟏𝟎𝟎𝟎 𝐱 𝟎.𝟓𝟐 𝐱 𝟑.𝟏
𝟐𝟎𝟖.𝟑 𝐱 𝟐𝟓𝟎
On calculation I = 2.68 (Ans.)
(3) 3.9 g of benzoic acid (molar mass = 122 g/mol) dissolved in 49 g of benzene shows a depression in freezing point of 1.62K. Calculate
the van’t Hoff factor & predict the nature of solute (associated or dissociated) ( Given: Kf of benzene = 4.9 K kg/mol)
Solution: Given : wB = 3.9 g, wA = 49 g, mB = 122, ΔTf = 1.62 K, Kf = 4.9K kg / mol
We know, ΔTf = 𝐢
𝟏𝟎𝟎𝟎.𝐊𝐟 𝐰𝐁
𝐦𝐁 𝐰𝐀
or 1.62 =
𝐢 𝐱 𝟏𝟎𝟎𝟎 𝐱 𝟒.𝟗 𝐱 𝟑.𝟗
𝟏𝟐𝟐 𝐱 𝟒𝟗
On calculation, value of i = 0.506
Since ’i’ < 1, solute benzoic acid is associated in benzene. (Ans.)

59. Some Typical questions:
(i). Why are aquatic species more comfortable in cold water in comparison to warm water ?
(ii). Why is the vapour pressure of an aqueous solution of glucose lower than that of water ?
(iii) Why is ethylene glycol added in car radiator water in cold countries?
Problems for practice:
(1) A 1.2% (w/V) solution of NaCl (molar mass = 58.5) is isotonic with 7.2% (w/V) of glucose (molar mass = 180)
solution Calculate the degree of dissociation of NaCl solution. (Ans. degree of dissociation = 0.96
(2)0.6 mL of acetic acid (molar mass = 60) having density 1.06 g/mL is dissolved in 1 L of water. The depression in
freezing point observed for this strength of acid was 0.0205℃. Calculate the Van’t Hoff factor & dissociation
constant of the acid. (Kf = 1.86K kg/mol,) [Ans i = 1.041 & Ka = 1.86 x 10-5]
(3) Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depresses by 2 K
(Kf = 1.86 K kg/mol & atomic masses: K = 39, Cl = 35.5) [ Ans.40.05 g]
(4) 0.01 m aqueous solution of K4[Fe(CN)6] freezes at – 0.062℃. What is the apparent %age of dissociation
(Kf for water = 1.86K kg/mol) [Ans. 77.7%]
(5) Calculate the normal freezing point of sea water sample containing 3.8 % NaCl & 0.12 & MgCl2 by mass.
(Kf for water = 1.86 K/m & atomic masses Na = 23 Cl = 35.5, Mg = 24) [Ans. – 2.59℃]

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