Chemistry

Solution - I

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

Session objectives

Solute: Component of solution present in smaller amount.

Solvent: Component of solution present in the larger amount.

Introduction

Solution: a homogenous mixture of two or more substances.

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.

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

Effect of pressure on solubility of gases

Increase in pressure of the gas above the solutionincreases 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.

moles of solute (mol)Molarity (M)

volume of solution (L)

moles of solute

Molality (m) Weight of solvent (inkg)

Different concentration terms

i

i

nMole fraction(x)

n x1=mole fraction of solvent

x2=mole fraction of solute

Molarity of a solution changes with temperature dueto accompanied change in volume of the solution.

Illustrative Example

Determine the molality of a solution preparedby dissolving 75 g of Ba(NO3)2(s) in 374 g of water at 25oC.

Solution:

Molar mass of Ba(NO3)2 = 261

275

0 287

0 7670 374

3 -1g

Number of moles of Ba(NO )261 g mol

. mole

0.287 moleMolality = . m

. kg

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 perlitre of solution.

1

m = ×1000 =1 molal solution.1000

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

Mass of solute% by mass = ×100

Mass of solution

w= %

W

Volume of solute% by volume 100

Volume of solution

v%

V

6mass of soluteParts per million (ppm) = ×10

mass of solution

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 contains40

1001040

= 3.84% by mass.

Different Concentration terms

Relation between Molarity (M) and molality (m)

23

mdM

MM1

10

2M Molar mass of solute

d density of solution

Relation between molality(m) and mol-fraction (x2) of solute

2 3

1

1

mx =

10m+

M

Where M = Molar mass of the solvent

12 3

1

mMx

mM 10

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.

Mass of solute 1000

MolalityMolecular mass of solute mass of solvent in gram

31000 0.2

60 250

Mole fraction of urea =Moles of urea 3 / 60

0.003593 250Total moles

60 18

Mole fraction of water = 1 – 0.00359 = 0.996

Solution:

Illustrative Example

Calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass.

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 =9546

= 2.07 moles

Moles of water(H2O)=5

= 0.28mol18

Solution:

Mole fraction of C2H5OH = 0.28

= 0.880.28 +2.07

Mole fraction of water = 1 – 0.88 = 0.12

Vapour pressure of solution

Liquid molecules evaporate from the surface

Vapourised molecules condensed to liquid

Both processes reach equilibrium

Po=Pressure exerted by the vapour above the liquid surface at eqm.

vapour pressure of pure

liquid

Factors affecting Vapour PressureNature of liquid:

More volatile liquids exert more pressure on the liquid surface.

Presence of a solute Due to presence of volatile and non-volatile solute, vapour pressure of solution decreases.

Temperature: Increase in temperature increases vapour pressure.

Vapour pressure Temperature

Vapor Pressure of Solution

Some of the solute particles will be near the surface.

Less no. of molecules per unit surface area are involved in equilibrium.

Block solvent molecules from entering the gas phase.

Raoult’s law for non-volatile solute

For highly dilute solutions

po=vapour pressure of pure liquidx1=mol. fraction of solventps=vapour pressure of solution

ps=x1po

Applicable for ideal solution

mix mixH 0 V 0

Here, solute-solute and solvent-solvent interaction exactly equal in magnitude with solute-solvent interaction.

Raoult’s law for non-volatile solute

Relative lowering of vapour pressure

From Raoult’s law,

o ss solvent soluteo

pp x p , 1 x

p

ssolute o

os

o

px 1

p

p p

p

os

soluteo

Relative lowering of v.p,

p px

p

os

o

p p nn Np

n moles of soluteN moles of solvent

os

o

p p nNp

when n 10%

Modification (two volatile liquids)

According to Raoult’s law,for two volatile miscible liquids

o os A A B B

A B

p p x p x

p p (1)

pA Partial vapour pressure of A.

xA Mol fraction of A in liquid phase.

A Bx x 1

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 :

o oA Bp 120 mmHg p 180 mmHg

A Bn 2 mol n 3mol.

A B2 3

x x5 5

o oS A A B Bp p x p x

2 3120 180

5 5

48 108 156

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

o om m E E

o om m E m

o o om E m E

P = p x +p x

= p x +p 1- x

= p - p x +p

m

oE

o om E

om

Comparing it with

p 119x 135

p 135

p p 119

p 119 135 245 torr

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

solventsolvent

total

nX =

n

90/18=

90/18 +6/60

50.980

5 .1

ps = 0.980 x 40 = 39.2 mm Hg

Modification (two volatile liquids)

From Dalton’s law of partial pressure

A A sp = y p - - - (2)

yA=mol. fraction of A in vapour phase

ps=vapour pressure of solution.

From (2)

AA

S

oA A

o oA A B B

py

p

p x

p x p x

o oA A B B

oA A A

oB A

A BoA A

p x p x1y p x

p (1 x )1 x x 1

p x

o oB Bo o

A A A

p p1

p x p

o oB Bo o

A A A A

p p11

y p x p

Modification (two volatile liquids)

Illustrative Problem

An unknown compound is immiscible with water. It is steam distilled at 98.0oCand P = 737 Torr.po

H20 = 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 pressurePtotal = 737 torr PoH2O = 707 torr

Pounknown = 737 – 707 = 30 torr.

If water = 100 g the unknown = 75.0 g

22

ounknown unknown

oH OH O

P n 7518

n M 100P

30 75 18707 m 100m 318.15 gmol

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 H 0 and V 0

Non ideal solution

For solution showing negative deviation from Raoult's law.

For solution showing positive deviation from Raoult's law.

H 0 and V 0

s idealP >PH 0 and V 0

s idealP P

Azeotropic mixtures

Solution showing positive deviation from Raoult’s form minimum boiling azeotrope

Interaction between A–B < interaction between A–A or B–B

Liquid mixtures which distil without any change in composition are called Azeotropes or Azeotropic mixtures.

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

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