Chemistry

Solution - I

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

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.

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.

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

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.

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

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

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

Different concentration terms

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 = 1000 + 40 = 1040g Therefore, 1040g solution contains 40g NaOH Therefore, 100g solution contains = 3.84% by mass.

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

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 – 0.00359 = 0.996

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

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.

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.

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.

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

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.

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

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.

Modification (two volatile liquids)

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 :

Illustrative Problem At 40oC, the vapour pressure in torr of methyl alcohol-ethyl alcohol solution is represented by P = 119Xm + 135 where Xm is the mole fraction of methyl alcohol. What are the vapour pressures of pure methyl alcohol & ethyl alcohol ?

Solution

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.

Solution ps = po x solvent ps = 0.980 x 40 = 39.2 mm Hg

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)

Modification (two volatile liquids)

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

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

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

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

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

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

Thank you