1st Lecture on Solutions | Chemistry Part I | 12th Science by Rizwana Mohammad

1. The Malegaon High School & Jr. College
Malegaon, (Nashik), 423203
1st Lecture on Solutions
Chemistry Part I, 12th Science
By
Rizwana Mohammad

2. Mixture: The Mixture is a combination of two or more
substances
e.g. Air, Rock etc.
• Mixtures are homogenous
• Homogeneous mixtures are classified according to the size
of their constituent particles as Colloids or as True
Solutions.
• In this chapter we will deal with True Solutions

3. Solutions: The solution is a homogenous mixture of two or
more pure components.
• A true solution consist of a Solvent and one or more
Solutes.
• Here we will deal with Binary Solutions.

4. No. State of Solute State of Solvent Examples
1 Solid Liquid
Sea water, Benzoic Acid in Benzene, Sugar
in water.
2 Solid Solid Metal alloys such as Brass, Bronze
3 Solid Gas Iodine in air
4 Liquid Liquid Gasoline, Ethanol in water
5 Liquid Solid
Amalgams of Mercury with metals as
Mercury in Silver
6 Liquid Gas Chloroform in Nitrogen
7 Gas Liquid Carbonated water, Oxygen in water
8 Gas Solid H2 in Palladium
9 Gas Gas Air
Types of Solutions:

5. • Saturated Solution: This solution contains maximum
amount of solute dissolved in a given amount of solvent at
a given temperature.
• Super Saturated Solutions: A solution containing greater
than the equilibrium amount of solute is Supersaturated
Solution.
• Such solutions are unstable.
• Solubility: Solubility of a solute is its amount per unit
volume of saturated solution at a specific temperature.

6. Factors affecting solubility:
• Solubility depends on the nature of solute and solvent,
temperature and pressure.
1. Nature of Solute and Solvent:
‘Like dissolves like’ guides to predict the solubility of a
solute in a given solvent.
• Polar solutes dissolve in polar solvents
e.g. NaCl dissolves in water.
• Non polar organic compounds like cholesterol dissolves in
non polar solvent such as Benzene.

7. 2. Effects of Temperature on Solubility:
• How solubility of substance changes with temperature
depends upon enthalpy of solution.
e.g. KCl dissolves in water by endothermic process.
• When temperature is increased by adding heat to the
system, the solubility of substance increases according to
Le-Chatelier principle.
• When substance dissolves in water by an exothermic
process, its solubility decreases with increase in
temperature.

8. • There is no direct correlation between solubility and
exothermicity or endothermicity.
e.g. dissolution of CaCl2 in water is exothermic and that
of NH4NO3 is endothermic. The solubility of these
increases with the temperature.
• Following are some experimental observations
• Solubilities of NaBr, NaCl and KCl change slightly with
temperature.
• Solubilities of KNO3, NaNO3 and KBr increases with
increase in temperature.

9. • Solubility of Na2SO4 decreases with increase of
temperature.
• The solubility of gases in water usually decreases with
increase of temperature.
Fig1 : Variation of Solubilities of some ionic solids
with temperature

10. 3. Effects of Pressure on Solubility:
• Pressure has no effect on the Solubilities of solids and liquids as
they are incompressible.
• Pressure affects solubility of gases in liquids.
Henry’s Law: It states that the solubility of gas in a liquid is
directly proportional to pressure of the gas over the solution,
Thus, S ∝ P or S = KHP
Where, S = Solubility of gas in mol L-1
P = Pressure of gas in Bar
KH = Proportionality constant called Henry’s Constant

11. Unit of KH =
𝑆
𝑃
=
𝑚𝑜𝑙 𝐿
−
1
𝑏𝑎𝑟
= mol L
− 1 bar -1
When P = 1 bar, KH = S
• KH is the solubility of the gas in a liquid when its
pressure over the solution is 1 bar.
• Vapour pressure of solutions of liquids in liquids:
• Raoult’s Law: It states that, “the partial vapour pressure
of any volatile component of a solution is equal to the
vapour pressure of the pure component multiplied by its
mole fraction in the solution.”

12. Therefore, P1 = x1 P1
0 and P2 = x2 P2
0
P1
0 , P2
0 are vapour pressures of pure liquids A1 & A2
P = P1 + P2
= P1
0 x1 + P2
0 x2
x1 = 1 - x2
P = P1
0 (1 - x2) + P2
0 x2
= P1
0 - P1
0 x2 + P2
0 x2
= (P2
0 - P1
0) x2 + P1
0
Fig 2 : Variation of vapour pressure with mole fraction of solute.
Figure shows plots of P1 Vs x1 and P2 Vs x2 are straight
lines passing through origin.

13. Ideal and Non Ideal Solutions:
1. Ideal Solutions:
i. Ideal solutions obey Raoult’s law over entire range of
concentrations,
ii. ΔmixH = 0
iii. Volume of an ideal solution is equal to the sum of volumes of two
components taken for mixing.
ΔmixV = 0
iv. In an ideal solution solvent-solute, solute-solute, solvent-solvent
molecular interactions are comparable.
v. The vapour pressure of ideal solution always lies between vapour
pressures of pure components.

14. 2. Non Ideal Solutions:
i. These solutions do not obey Raoult’s law over entire range of
concentrations.
ii. The vapour pressures of these solutions can be higher or lower
than those of pure components.
iii. These solutions show deviation from the Raoult’s law.

15. A- Positive Deviations from Raoult’s law:
The solutions in which solute-solvent intermolecular attractions are
weaker than those between solute-solute molecules and solvent-solvent
molecules exhibit positive deviations. The vapour pressure of such
solutions are higher than those of pure components.
Fig 3 : Positive deviations from Raoult’s law

16. B- Negative Deviations from Raoult’s law:
The solutions in which the interactions between solvent-solute
molecules are stronger than solute-solute or solvent-solvent molecules,
exhibit negative deviations.
The vapour pressures of such solutions are lower than those of pure
components.
Fig 4 : Negative deviations from Raoult’s law