complete chapter solution of 12 cbse

1. RYAN INTERNATIONAL SCHOOL,
DUMAS
STD: XII
SUB: CHEMISTRY CHATER-2 SOLUTION

2. SOLUTIONS
• GENERAL INTRODUCTION
• TYPES OF SOLUTION
• EXPRESSING CONCENTRATION.

3. WHAT IS SOLUTION?
• Solution is homogeneous mixture of two or more chemically non
reacting substances.
• Solute + Solvent = Solution
• Solute = substance present in lesser amount
• Solvent = substance present in larger amount
• Mixture of only two substances one solute and one solvent is called
“Binary Mixture”
• In this chapter we are going to study only binary mixture.

4. WHAT IS SOLUTION?
• Solutions in which water is used as solvent is called aqueous solution.
• Eg. Sugar + water, copper sulphate + water.
• On the other hand solution without water as solvent are called non aqueous
solution.
• Eg. Benzene and naphthalene, petrol and kerosene.
• Solution occurs in all three states of matter i.e. Solid , liquid and gas.
• State of solution depend on State of Solvent.

5. TYPES OF SOLUTIONS
No. Solute Solvent Examples
1. Solid Solid Alloys
2. Liquid Solid Amalgam, hydrated salts
3. Gas Solid H2 in platinum, palladium
4. Solid Liquid Sugar solution
5. Liquid Liquid Methanol or ethanol in water
6. Gas Liquid Aerated drinks, oxygen in
water
7. Solid Gas camphor in air
8. Liquid Gas Humidity
9. Gas Gas All gaseous mixtures.

6. EXPRESSING CONCENTRATION OF SOLUTIONS
• 1. Percentage :
• 2. Strength :
• 3. Molarity:
• 4. Normality:
• 5. Molality
• 6. Mole fraction
• 7. Parts per million(ppm):

7. PERCENTAGE
• The percentage of solution is usually expressed either as ‘ percentage by mass’ or
‘percentage by volume’
• Formula
Mass (w/w)% =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 100
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 + 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒.
• Volume (V/V)% =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 100
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 + 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒.

8. STRENGTH
• It is amount of solute in gm present in one litre of solution.
• It is expressed as gm/l or gm L-1
• Formula :
Strength of solution(gm/l ) =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔𝑚
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝐿

9. MOLARITY
• Molarity is defined as moles of solute dissolved in 1 litre of solution.
• In case of ionic compounds like NaCl, Na2CO3 etc Formality is used instead. In
this case formula weight is calculated instead of molecular weight.
• It is expressed as moles/litre or mol/l or M
• Formula :
Molarity=
𝑤𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔𝑚
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
×
1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝑙
Note : ‘Molarity changes with temperature’ and its inversely proportional.

10. NORMALITY
• It is defined as number of gram equivalents dissolved per litre of solution.
• It is expressed as N or equivalents/ litre.
• In this case equivalent weight has to be calculated first.
• Equivalent weight is ratio of molecular weight to either its valency or its
acidity/basicity or change in oxidation state.
• Formula :
• Normality (N) =
𝑤𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔𝑚
𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑤𝑒𝑖𝑔ℎ𝑡
×
1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝑙

11. MOLALITY
• It is no. of moles of solute dissolved in 1 kg of solvent.
• It is denoted by m or moles/kg
• Molality is not affected by change in temperature.
• Formula:
molality (m) :
𝑤𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔𝑚
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
×
1000
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑔𝑚

12. MOLE FRACTION
• It is ratio of moles of one component to total moles of all component.
• Mole fraction can be determined for both solute and solvent.
• Sum of mole fraction of all components is always equal to 1.
• Mole fraction can never be greater than 1.
• Formula : Let moles of component A be n1 and component B be n2
• Mole fraction of A will be ΧA =
n1
n1+ n2
• Mole fraction of B will be ΧB =
n2
n1+ n2
• XA + XB = 1

13. PARTS PER MILLION(PPM)
• It is used only for very dilute solutions.
• E.g. pollution in atmosphere, dissolved oxygen in water, impurity present in
drinking water etc.
• It is defined as mass of solute present in one million parts by mass of solution.
• It is denoted as ppm
• Formula: ppm =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 106

14. NUMERICALS
• Example1: 15 g of methyl alcohol is dissolved in 35 g of water. The weight
percentage of methyl alcohol in solution is
• Soln. Weight percentage =
Weight of solute
Weight of solution
× 100
Total weight of solution = (15 + 35) g = 50 g
Weight percentage of methyl alcohol =
Weight of methyl alcohol
Weight of solution
× 100
=
15
50
× 100
= 30%

15. NUMERICALS
• Example 2: Sea water contains 5.8 × 10−3 𝑔 of dissolved oxygen per kilogram. The
concentration of oxygen in parts per million is
• Soln : Part per million =
Mass of solute
Mass of solution
× 106
=
5.8 × 10−3 𝑔
103 𝑔
× 106
= 5.8𝑝𝑝𝑚

16. NUMERICALS
• Example 3: Amount of 𝑁𝑎𝑂𝐻 present in 200 𝑚𝑙 of 0.5𝑁 solution is
• Soln: Normality (N) =
𝑤𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔𝑚
𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑤𝑒𝑖𝑔ℎ𝑡
×
1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝑙
Wt. of solute =
𝑁×𝑉× 𝑒𝑞. 𝑤𝑡.
1000
=
0.5×200×40
1000
= 4𝑔

17. NUMERICALS
• Example 4: 100𝑚𝑙 of 0.3𝑁 𝐻𝐶𝑙 is mixed with 200𝑚𝑙 of 0.6𝑁 𝐻2 𝑆𝑂4. The final
normality of the resulting solution will be
• Soln: 𝑁1 𝑉1 + 𝑁2 𝑉2 = 𝑁3 𝑉3
i.e., 0.3 × 100 + 0.6 × 200 = 𝑁3 × 300
0.3 + 1.2 = 3𝑁3
𝑁3 =
1.5
3
= 0.5

18. NUMERICALS
• Example 5: Calculate the molarity of 𝐻2 𝑆𝑂4 solution that has a density of
1.84𝑔/𝑐𝑐 at 35 𝑜 𝐶 and contains 98% by weight ?
• Soln: Vol. of solution =
mass
density
=
100
1.84
= 54.34 𝑚𝑙
• Molarity =
Wt. of solute
Mol. wt.
×
1000
Vol. of solution in 𝑚𝑙.
=
98
98
×
1000
54.34
= 18.4𝑀

19. NUMERICALS
• Example 6: A solution has 25% of water, 25% ethanol and 50% acetic acid by mass.
Find the mole fraction of each component.
• Soln: Since 18𝑔 of water = 1mole, 25𝑔 of water =
25
18
= 1.38 mole
Similarly, 46𝑔 of ethanol = 1 mole, 25𝑔 of ethanol =
25
46
= 0.55 moles
Again, 60𝑔 of acetic acid = 1 mole, 50𝑔 of acetic acid =
50
60
= 0.83 mole
∴ Mole fraction of water =
1.38
1.38+0.55+0.83
= 0.50
Similarly, Mole fraction of ethanol =
0.55
1.38+0.55+0.83
= 0.19
Mole fraction of acetic acid =
0.83
1.38+0.55+0.83
= 0.3

20. SOLUBILITY OF SOLIDS IN LIQUIDS
• Maximum amount of solid which can be dissolved in 100 g
of solvent is called solubility at that particular
temperature.
• Factors affecting Solubility of solids in liquids are
1. Nature of solute and solvent(likes dissolves likes).
2. Temperature.

21. SOLUBILITY OF SOLIDS
• Dissolution
process is either
• Exothermic
• Endothermic

22. SOLUBILITY OF GAS IN LIQUIDS
• It is maximum volume of gas (converted at STP) dissolved
in 1 ml of solvent at a particular temperature.
• Factors affecting solubility of gas in liquids.
• 1. Nature of gas and solvent.
• 2. Temperature.
• 3. Pressure.

23. HENRY’S LAW
• Henry's law is one of the gas laws formulated by William Henry in 1803 and
states:
"At a constant temperature, the amount of a gas that dissolves in a volume of
liquid is directly proportional to the partial pressure of that gas in equilibrium
with that liquid.“
• P ∝ XA (or) PA = KH XA

24. HENRY’S LAW AND RELATION WITH TEMPERATURE

25. AN EXPERIMENT

26. APPLICATION OF HENRY’S LAW
• In the production of beverages.
• In deep sea diving.
• Dissolution of gas in blood stream.
• For climbers at high altitude.

30. QUESTIONS
1. Do Whales and Dolphins suffer from BEND?
2. Sherpas. How it is so easy for them to climb
mountains with carrying Loads of other
mountaineers?

31. LIMITATION OF HENRY’S LAW
• Pressure should be low and temperature should be high.
• The gas should not react with solvent.
• No association or dissociation of gas should occur.

32. NUMERICALS
1. How many grams of carbon dioxide gas is dissolved in a 1 L bottle
of carbonated water if the manufacturer uses a pressure of 2.4 atm
in the bottling process at 25 °C?Given: KH of CO2 in water =
211.67 atm/(mol/L) at 25 °C
• Soln => PA = KH XA
2.5 = 1.67 X XA
XA = 2.5 / 211.67
XA = 0.012

33. • XA =
𝑛 𝐴
𝑛 𝐴+ 𝑛 𝐵
• 0.012 =
𝑛 𝐴
𝑛 𝐴+55.55
• 0.012 =
𝑛 𝐴
55.5.5
• nA = 0.666
• amount of CO2 = 0.666 X 44
• = 29.33 g

34. COLLIGATIVE PROPERTIES.
• Certain properties of dilute solutions depend only upon the concentration i.e.,
the number of particles of the solute present in the solution. Such properties are
called colligative properties. The four well known examples of the colligative
properties are
• (1) Relative Lowering of vapour pressure of the solution.
• (2) Elevation in boiling point of the solvent.
• (3) Depression in freezing point of the solvent.
• (4) Osmotic pressure of the solution.

35. COLLIGATIVE PROPERTIES.
• Since colligative properties depend upon the number of solute particles present
in the solution, the simple case will be that when the solute is a non-electrolyte.
• In case the solute is an electrolyte, it may split to a number of ions each of which
acts as a particle and thus will affect the value of the colligative property.
• The study of colligative properties is very useful in the calculation of molecular
weights of the solutes.

36. VAPOUR PRESSURE
• The pressure exerted by the vapours above the liquid surface in
equilibrium with the liquid at a given temperature is called vapour
pressure of the liquid.

37. VAPOUR PRESSURE
• The vapour pressure of a liquid depends on
• (1) Nature of liquid : Liquids, which have weak intermolecular
forces, are volatile and have greater vapour pressure.
• (2) Temperature : Vapour pressure increases with increase in
temperature.
• (3) Purity of liquid : Pure liquid always has a vapour pressure
greater than its solution.

39. LOWERING OF VAPOUR PRESSURE
• According to Raoult’s law
(1887), at any given
temperature the partial
vapour pressure (pA) of any
component of a solution is
equal to its mole fraction (XA)
multiplied by the vapour
pressure of this component
in the pure state (𝑝 𝐴
0
). That is,
𝑝 𝐴 = 𝑝 𝐴
0
× 𝑋𝐴
Raoult’s law may be stated as “the relative lowering of
vapour pressure of a solution containing a non-
volatile solute is equal to the mole fraction of the
solute in the solution.”

40. RAOULT’S LAW FOR NON- VOLATILE SOLUTE
• Relative lowering of vapour pressure
• Vapour pressure of a solvent present insolution is less than the vapour
pressure of the pure solvent.
• According to Raoult lowering of vapour pressure of a solution depends only on
the concentration of the solute particles and remains independent of their
identity.
• p1 =x1 p1
0

41. RAOULT’S LAW FOR NON- VOLATILE SOLUTE

42. RAOULT’S LAW FOR NON- VOLATILE SOLUTE

43. NUMERICALS
• Question 1. 34.2 g of canesugar is dissolved in 180 g of
water. The relative lowering of vapour pressure will be
• Soln : 𝑷 𝑨
𝟎
−𝑷 𝑨
𝑷 𝑨
𝟎 =
𝑾 𝑩/𝑴 𝑨
𝑾 𝑩/𝑴 𝑩+𝑾 𝑨/𝑴 𝑨
=
𝟑𝟒.𝟐/𝟑𝟒𝟐
𝟑𝟒.𝟐/𝟑𝟒𝟐+𝟏𝟖𝟎/𝟏𝟖
=
𝟎.𝟏
𝟏𝟎.𝟏
= 𝟎. 𝟎𝟎𝟗𝟗

44. NUMERICALS
• Question 2. Lowering in vapour pressure is the highest for
(a) 0.2 m urea
(b) 0.1 m glucose
(c) 0.1m 𝑀𝑔𝑆𝑂4
(d) 0.1m 𝐵𝑎𝐶𝑙2

45. NUMERICALS
• Question 3. Vapour pressure of 𝐶𝐶𝑙4 at 25 𝑜
𝐶 is 143 mm Hg 0.5 g of a non-
volatile solute (mol. wt. 65) is dissolved in 100 ml of 𝐶𝐶𝑙4. Find the vapour
pressure of the solution. (Density of 𝐶𝐶𝑙4 = 1.58𝑔/𝑐𝑚3)
•
𝑷 𝟎−𝑷 𝒔
𝑷 𝟎 =
𝒏 𝟐
𝒏 𝟏
;
•
𝟏𝟒𝟑−𝑷 𝒔
𝟏𝟒𝟑
=
𝟎.𝟓/𝟔𝟓
𝟏𝟓𝟖/𝟏𝟓𝟒
or
• 𝑷 𝒔 = 𝟏𝟒𝟏. 𝟗𝟑𝒎𝒎

46. Solution
Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a
density of 1.26 g/mL at 25 ℃. Calculate the vapor pressure
at 25 ℃ of a solution made by adding 50.0 mL of glycerin
to 500.0 mL of water. The vapor pressure of pure water at
25 ℃ is 23.8 torr , and its density is 1.00 g/mL.
Calculation of Vapor Pressure of a Solution

47. POP UP QUESTION FOR GENIUS
Which method is used
to measure relative
lowering of vapour
pressure?
Hello , I m Dexter.
All smart students
over there I have a
question for you.

48. RAOULT’S LAW FOR VOLATILE SOLUTE

49. MIXTURES OF VOLATILE LIQUIDS

50. RAOULT’S LAW FOR VOLATILE SOLUTE
• When both solute and solvent are volatile (e.g benzene+ toluene) then
vapours of both component will be formed.
• In such cases Raoult’s law is applied for both components.
• Let us take a solution of A and B (both volatile)
• According to Raoult’s Law 𝑃𝐴 = 𝑃𝐴
0
𝑋𝐴 and 𝑃𝐵 = 𝑃𝐵
0
𝑋 𝐵.
• And by Daltons Law of Partial pressure 𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑝 𝐴 + 𝑝 𝐵
= (𝑝 𝐴
0
× 𝑋𝐴) + (𝑝 𝐵
0
× 𝑋 𝐵)

51. RELATIONSHIP BETWEEN VAPOUR PRESSURE AND
MOLE FRACTION OF AN IDEAL SOLUTION

52. MOLE FRACTION OF SOLUTE AND SOLVENT IN
VAPOUR PHASE
PA = YA X P TOTAL
PB = YB X P TOTAL
PA = P°A × XA
PB = P°B × XB

53. 53
RAOULT'S LAW
One last point about Raoult's Law….
It is for the ideal solutions.
BUT, even deviations from the law gives us information on the IMF of the
solute to the solvent and IMF of the solvent molecules to solvent
molecules.
If the solute-solvent IMF is strong, vp of solution would be lowered more
than expected.
If the solvent-solvent IMF is strong, vp of the solution would not be
lowered as much as expected (adding solute disrupts the IMF of the
solvent and allows more solvent to vaporize).

54. IDEAL AND NON-IDEAL SOLUTION

55. COMPARISON

56. DIFFERENCE BETWEEN

57. EXAMPLES OF IDEAL SOLUTIONS
• Dilute solutions;
• benzene + toluene:
• n-hexane + n-heptane;
• chlorobenzene + bromobenzene;
• ethyl bromide + ethyl iodide;
• n-butyl chloride + n-butyl bromide

58. EXAMPLES OF NON-IDEAL SOLUTIONS SHOWING
POSITIVE DEVIATION
• acetone +ethanol
• acetone +CS2:
• water + methanol;
• water + ethanol;
• CCl4 + toluene;
• CCl4 + CHCl3;
• acetone + benzene;
• CCl4+CH3OH;
• cyclohexane + ethanol

59. EXAMPLES OF NON-IDEAL SOLUTIONS SHOWING
NEGATIVE DEVIATION
• Acetone + aniline;
• acetone + chloroform;
• CH3OH+CH3COOH;
• H2O+HNO3
• chloroform + diethyl ether;
• water + HCl;
• acetic acid + pyridine;
• chloroform + benzene

60. A mixture of benzene (C6H6) and toluene (C7H8) containing
1.0 mol of benzene and 2.0 mol of toluene. At 20 °C the vapor
pressures of pure benzene and toluene are 75 torr and 22 torr,
respectively. What is the vapor pressure of the mixture?
What is the mole fraction of benzene in the vapor?
Numerical

61. A solution is prepared by mixing 5.81 g acetone (molar mass =
58.1 g/mol) and 11.0 g chloroform (molar mass = 119.4 g/mol).
At 35 °C, this solution has a total vapor pressure of 260. torr.
Is this an ideal solution? The vapor pressure of pure acetone
and pure chloroform at 35 °C are 345 torr and 293 torr,
respectively.
Numericals

62. WHAT IS HAPPENING?
What Is
Happening?

63. AZEOTROPES
• These are mixtures of two or more liquids with definite composition and boils like a
pure liquids.
• They are also called constant boiling mixture.
• There composition cannot be altered and they can’t be separated by fractional
distillation in pure form.

64. AZEOTROPE

65. MINIMUM BOILING AZEOTROPES
 These are liquid mixtures which shows positive deviation from Raoult’s Law.
 E.g water and ethanol (5:95) B.P = 78 C
 Acetone and CS2 (33: 67) B.P = 39 C

66. MAXIMUM BOILING AZEOTROPES
 These are liquid mixtures which shows negative deviation from Raoult’s Law.
 E.g Water and HCl (80:20) B.P = 110 C
 E.g Water and HNO3 (32: 68) B.P = 120 C

67. ELEVATION IN BOILING POINT
• Boiling point of liquid is temperature at which the vapour pressure of liquid
becomes equal to atmospheric pressure.
• Boiling point of solution is always higher than pure solvent.
• This increase in boiling point is directly proportional to molality of solution.
• 𝑻 − 𝑻 𝒃 = 𝜟𝑻 𝒃
• 𝜟𝑻 𝒃 = 𝑲 𝒃 × 𝒎
• 𝜟𝑻 𝒃 =
𝟏𝟎𝟎𝟎×𝑲 𝒃×𝒘
𝒎×𝑾
or 𝒎 =
𝟏𝟎𝟎𝟎×𝑲 𝒃×𝒘
𝜟𝑻 𝒃×𝑾
• 𝑲 𝒃 =
𝑹(𝑻 𝟎) 𝟐
𝟏𝟎𝟎𝟎 𝒍 𝑽
𝑲 𝒃 =
𝑴 𝑨
𝑹 (𝑻 𝟎) 𝟐
𝟏𝟎𝟎𝟎 𝜟𝑯 𝑽𝒂𝒑

68. Phase diagram illustrating boiling-point elevation

69. MOLAL ELEVATION CONSTANT OR EBULLIOSCOPIC
CONSTANT(𝑲 𝒃)
solvent normal boiling Kb, oC m-1
water 100.0 0.512
acetic acid 118.1 3.07
benzene 80.1 2.53
chloroform 61.3 3.63
nitrobenzene 210.9 5.24

70. Freezing Point Depression
• At freezing point, liquid solvent molecules are solidified
to an ordered crystal form.
• The freezing point depression calculation:
ΔTf = Tf(solution) – Tf(solvent) = Kf m
m: molality
Kf: molal freezing-point depression constant
For water, Kf is 1.86 oC/m
• 𝜟𝑻 𝒇 =
𝟏𝟎𝟎𝟎×𝑲 𝒇×𝒘
𝒎×𝑾
or 𝒎 =
𝟏𝟎𝟎𝟎×𝑲 𝒇×𝒘
𝜟𝑻 𝒇×𝑾
• 𝑲 𝒇 =
𝑹(𝑻 𝟎) 𝟐
𝟏𝟎𝟎𝟎 𝒍 𝒇
𝑲 𝒇 =
𝑴 𝑨
𝑹 (𝑻 𝟎) 𝟐
𝟏𝟎𝟎𝟎 𝜟𝑯 𝒇𝒖𝒔

71. Phase diagram illustrating freezing-point depression

72. VALUES OF 𝑲 𝒃 AND KF

73. Ascorbic acid
Detergent
POP UP QUESTION

74. Why salt is sprinkled on
roads covered with snow?
Why antifreeze is mixed
with water and added in a
car’s radiator?

75. OSMOTIC PRESSURE
• Osmosis is flow of solvent molecules from a region of lower concentration
to region of higher concentration across a semi-permeable membrane.
• Solute particles in solvent act like a gas molecules and hence they exert
pressure.
• So osmotic pressure can be defined as “ minimum pressure that has to
be applied on the solution to prevent the entry of solvent molecules
through a semi permeable membrane.

76. OSMOTIC PRESSURE
The pressure required to prevent
osmosis is known as osmotic pressure
(p) of the solution.
p = MRT
M = Molarity
R = 0.082 l.atm/mol.K
T = Temp. in K.

77.  Osmotic Pressure
 Osmosis: The flow of solvent from a solution of
lower solute concentration to one of higher solute
concentration.

78. Types of Solutions and Osmosis
• Isotonic solution: equal concentrations
• Hypertonic solution: the more concentrated solution
• Hypotonic solution: the more dilute solution

79. • Various relations derived above about colligative properties are only applicable if
solute is non electrolyte which neither undergoes dissociation or association.
• In case of electrolytes like NaCl, K2SO4, MgSO4 etc they undergo dissociation in
aqueous solution hence number of particles in solution increases than expected.
• In case of solutes like benzoic acid, formic acid, acetic acid etc they undergo
association hence number of particles in solution decreases than expected.
• If molecular mass is determined for such solute then EXPERIMENTALLY measured
mass will be different from expected ; called as ‘ABNORMAL MOLECULAR
MASS’

81. VAN’T HOFF FACTOR

82. VAN’T HOFF FACTOR
• Van’t Hoff factor is defined as ratio of experimental colligative property to
calculated one.
• ASSOCIATION : polar solute particles will undergo association in non polar
solvent
• DISSOCIATION : ionic solute will undergo dissociation in polar solvent.

83. As concentration increases, deviation increases.
Dilution leads towards the theoretical i value.
Compare NaCl with MgSO4. Both have theoretical i = 2.
Why does MgSO4 deviate much more than NaCl?

84. VAN’T HOFF FACTOR IN CASE OF DISSOCIATION

85. VAN’T HOFF FACTOR IN CASE OF ASSOCIATION

86. Vodka = ethanol + water Brass = copper + zinc

87. Note – Only few days until all Missing
and Late Work is due by 25/04/2020
unless previous arrangements have been
made with me.
Thank You

88. Molar Mass determination
 A compound is composed of 32.0 % carbon, 4.0%
hydrogen and 64.0 % oxygen. It is found that 15.0 g of
the substance added to 1.00 kg of water lowers the
freezing point of the water to -0.186 °C . What are the
empirical formula, molar mass and molecular formula?

89. TO BE CONTINUED……
STAY SAFE….