1. Chemistry

2. Solution - I

3. Session objectives Introduction Solubility Henry’s law
Different concentration terms
Vapour pressure
Raoult’s law and its modification
Relative lowering of vapour pressure
Ideal solutions and non-ideal solutions
Maximum and minimum boiling solutions

4. Introduction Solute: Component of solution present in smaller amount.
Solvent: Component of solution present in the larger amount.
Solution: a homogenous mixture of two or more substances.

5. Solubility
Maximum amount of solute in grams which can be dissolved in a given amount of solvent (generally 100 g) to form a saturated solution at that particular temperature is known as its solubility
For solids
Solubility of ionic compounds in water generally increases with increase in temperature.
For gases
The solubility of gases in water decreases with increase in temperature.
Solubility tends to zero at the boiling point of water.

6. Effect of pressure on solubility of gases
Increase in pressure of the gas above the solution increases the solubility of the gas in the solution.
More dilute solution
More concentrated solution

7. Henry’s law
Solubility of a gas in a liquid is proportional to the pressure of the gas over the solution.
C = kP C: molar concentration, P: pressure,
k: temperature-dependent constant
Carbonated cold drink is an
application of Henry’s law.

8. Different concentration terms
Molarity of a solution changes with temperature due to accompanied change in volume of the solution.
x1=mole fraction of solvent
x2=mole fraction of solute

9. Illustrative Example Determine the molality of a solution prepared
by dissolving 75 g of Ba(NO3)2(s) in 374 g of water at 25oC.
Solution:
Molar mass of Ba(NO3)2 = 261

10. Illustrative Example
Calculate the molality of 1 molar solution of NaOH given density of solution is 1.04 gram/ml.
Solution:
1 molar solution means 1 mole of solute present per litre of solution.
Therefore, mass of 1 litre solution = 1000 x 1.04
= 1040 gram
Mass of solute = 1 x 40 = 40g
Therefore, mass of solvent 1040 – 40 = 1000g

11. Different concentration terms

12. Illustrative Example
Calculate the concentration of 1 molal solution of NaOH in terms of percentage by mass.
Solution:
1 molal solution means 1 mole (or 40g) NaOH present in 1000g of solvent.
Total mass of solution = = 1040g
Therefore, 1040g solution contains 40g NaOH
Therefore, 100g solution contains
= 3.84% by mass.

13. Different Concentration terms
Relation between Molarity (M) and molality (m)
Relation between molality(m) and mol-fraction (x2) of solute

14. Illustrative Example
Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea (molecular mass = 60) in 250 g of water.
Solution:
Mole fraction of urea =
Mole fraction of water = 1 – = 0.996

15. Illustrative Example
Calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass.
Solution:
95% of ethanol by mass means 95 g ethanol present in 100 g of solution.
Hence, mass of water = 100 – 95 = 5 g
Moles of C2H5OH =
= 2.07 moles
Moles of water(H2O)=
Mole fraction of C2H5OH =
Mole fraction of water = 1 – 0.88 = 0.12

16. Vapour pressure of solution
Liquid molecules evaporate from the surface
Vapourised molecules condensed to liquid
vapour pressure of pure liquid
Both processes reach equilibrium
Po=Pressure exerted by the vapour above the liquid surface at eqm.

17. Factors affecting Vapour Pressure
Nature of liquid: More volatile liquids exert more pressure on the liquid surface.
Temperature:
Increase in temperature increases vapour pressure.
Presence of a solute Due to presence of volatile and non-volatile solute, vapour pressure of solution decreases.

18. Vapor Pressure of Solution
Some of the solute particles
will be near the surface.
Block solvent molecules
from entering the gas phase.
Less no. of molecules per
unit surface area are
involved in equilibrium.

19. Raoult’s law for non-volatile solute
For highly dilute solutions
po=vapour pressure of pure liquid x1=mol. fraction of solvent
ps=vapour pressure of solution
ps=x1po

20. Raoult’s law for non-volatile solute
Applicable for ideal solution
Here, solute-solute and solvent-solvent interaction exactly equal in magnitude with solute-solvent interaction.

21. Relative lowering of vapour pressure
From Raoult’s law,
n moles of solute N moles of solvent

22. Modification (two volatile liquids)
According to Raoult’s law,for two volatile miscible liquids pA
Partial vapour pressure of A. xA
Mol fraction of A in liquid phase.

23. Modification (two volatile liquids)

24. Illustrative Example
Vapour pressure of liquids A and B at a particular temperature are 120 mm and 180 mm of Hg. If 2 moles of A and 3 moles of B are mixed to form an ideal solution, what would be the vapour pressure of the solution?
Solution :

25. Illustrative Problem
At 40oC, the vapour pressure in torr of methyl alcohol-ethyl alcohol solution is represented by P = 119Xm where Xm is the mole fraction of methyl alcohol. What are the vapour pressures of pure methyl alcohol & ethyl alcohol ?

26. Solution

27. Illustrative Problem
6g of urea is disolved in 90g water at 25oC ? What is vapour pressure of sol. If vapour pressure of water is 40mmHg.

28. Solution
ps = po x solvent
ps = x 40 = 39.2 mm Hg

29. Modification (two volatile liquids)
From Dalton’s law of partial pressure
yA=mol. fraction of A in vapour phase
ps=vapour pressure of solution.
From (2)

30. Modification (two volatile liquids)

31. Illustrative Problem An unknown compound is immiscible
with water. It is steam distilled at 98.0oC
and P = 737 Torr.poH20 = 707 torr at
98.0oC. This distillate was 75% by weight of water.
Calculate the molecular weight of the unknown

32. Solution Using Dalton’s law of partial pressure
Ptotal = 737 torr PoH2O = 707 torr
Pounknown = 737 – 707 = 30 torr.
If water = 100 g the unknown = 75.0 g

33. Non-ideal solution
Solute-solvent interaction are different than solute-solute and solvent solvent in non ideal solutions.
These do not obey Raoult’s Law.
For non ideal solutions

34. Non ideal solution For solution showing positive
deviation from Raoult's law.
For solution showing negative deviation from Raoult's law.

35. Azeotropic mixtures
Liquid mixtures which distil without any change in composition are called Azeotropes or Azeotropic mixtures.
Solution showing positive deviation from Raoult’s form minimum boiling azeotrope
Interaction between A–B < interaction between A–A or B–B

37. Azeotropic mixtures
Solution showing negative deviation from Raoult’s law form maximum boiling azeotropes
Interaction between A – B > interaction between A – A or B – B

38. Thank you