41. solution 1

7/31/2019 41. Solution 1

1/38

Chemistry

7/31/2019 41. Solution 1

2/38

Solution - I

7/31/2019 41. Solution 1

3/38

Introduction

Solubility

Henrys law

Different concentration terms

Vapour pressure

Raoults law and its modification

Relative lowering of vapour pressure

Ideal solutions and non-ideal solutions

Maximum and minimum boiling solutions

Session objectives

7/31/2019 41. Solution 1

4/38

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.

7/31/2019 41. Solution 1

5/38

Solubility

For solids Solubility of ionic compounds

in water generally increaseswith increase in temperature.

For gases

The solubility of gases in

water decreases withincrease in temperature.

Solubility tends to zero at theboiling 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 particulartemperature is known as its solubility

7/31/2019 41. Solution 1

6/38

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

7/31/2019 41. Solution 1

7/38

Henrys law

Solubility of a gas in a liquid is proportional tothe 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 Henrys law.

7/31/2019 41. Solution 1

8/38

moles of solute (mol)Molarity (M)

volume of solution (L)

moles of solute

Molality (m)Weight of solvent (inkg)


i

i

nMole fraction(x)

n

x1=mole fraction of solventx2=mole fraction of solute

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

7/31/2019 41. Solution 1

9/38

Illustrative Example

Determine the molality of a solution prepared

by dissolving 75 g of Ba(NO3)2(s) in 374 g ofwater at 25oC.

Solution:

Molar mass of Ba(NO3)2 = 261

2

75

0 287

0 767

0 374

3 -1

gNumberofmoles of Ba(NO )

261 g mol

. mole

0.287 moleMolality = . m

. kg

7/31/2019 41. Solution 1

10/38


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

7/31/2019 41. Solution 1

11/38


Mass of solute% by mass = 100Mass of solution w= %W

Volume of solute% by volume 100

Volume of solution

v%

V

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

7/31/2019 41. Solution 1

12/38


Calculate the concentration of 1 molal solution

of NaOH in terms of percentage by mass.

Solution:

1 molal solution means 1 mole (or 40g) NaOHpresent 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.

7/31/2019 41. Solution 1

13/38

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

7/31/2019 41. Solution 1

14/38


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 1000Molality

Molecular mass of solute mass of solvent in gram

31000 0.2

60 250

Mole fraction of urea =Moles of urea 3/60

0.00359

3 250Total moles60 18

Mole fraction of water = 1 0.00359 = 0.996

Solution:

7/31/2019 41. Solution 1

15/38


Calculate the mol fraction of ethanol and waterin a sample of rectified spirit which contains

95% of ethanol by mass.

95% of ethanol by mass means 95 g ethanolpresent in 100 g of solution.

Hence, mass of water = 100 95 = 5 g

Moles of C2H5OH =95

46= 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

7/31/2019 41. Solution 1

16/38

Vapour pressure of solution

Liquid molecules evaporatefrom the surface

Vapourised molecules condensed to liquid

Both processes reach equilibrium

Po=Pressure exerted by the vapourabove the liquid surface at eqm.

vapour pressure ofpure liquid

7/31/2019 41. Solution 1

17/38

Factors affecting Vapour Pressure

Nature of liquid:More volatile liquids exert more

pressure on the liquid surface.

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

Temperature:Increase in temperature

increases vapour pressure.

Vapour pressure Temperature

7/31/2019 41. Solution 1

18/38

Vapor Pressure of Solution

Some of the solute particles

will be near the surface.

Less no. of molecules perunit surface area are

involved in equilibrium.

Block solvent molecules

from entering the gas phase.

7/31/2019 41. Solution 1

19/38

Raoults law for non-volatile solute

For highly dilute solutions

po=vapour pressure of pureliquid

x1=mol. fraction of solventps=vapour pressure of

solution

ps=x1po

7/31/2019 41. Solution 1

20/38

Applicable for ideal solution

mix mixH 0 V 0

Here, solute-solute and solvent-solventinteraction exactly equal in magnitudewith solute-solvent interaction.

Raoults law for non-volatile solute

7/31/2019 41. Solution 1

21/38

Relative lowering of vapour pressure

From Raoults law,

o s

s solvent soluteo

pp x p , 1 x

p

ssolute o

os

o

px 1

pp p

p

os

soluteo

Relative lowering of v.p,

p px

p

os

o

p p n

n Np

n moles of solute

N moles of solvent

os

o

p p n

Np

when n 10%

7/31/2019 41. Solution 1

22/38

Modification (two volatile liquids)

According to Raoults law,for twovolatile 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

7/31/2019 41. Solution 1

23/38


7/31/2019 41. Solution 1

24/38


Vapour pressure of liquids A and B at a particulartemperature 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 o

A Bp 120 mmHg p 180 mmHg

A Bn 2 mol n 3mol.

A B

2 3x x

5 5

o oS A A B Bp p x p x

2 3120 180

5 5

48 108 156

7/31/2019 41. Solution 1

25/38

Illustrative Problem

At 40oC, the vapour pressure in torr of methylalcohol-ethyl alcohol solution is represented byP = 119Xm + 135 where Xm is the mole fractionof methyl alcohol. What are the vapour pressuresof pure methyl alcohol & ethyl alcohol ?

7/31/2019 41. Solution 1

26/38

Solution

o o

m m E E

o om m E m

o o o

m E m E

P =p x +p x

= p x +p 1- x

= p - p x +p

m

o

E

o o

m E

o

m

Comparing it with

p 119x 135p 135

p p 119

p 119 135 245 torr

7/31/2019 41. Solution 1

27/38


6g of urea is disolved in 90g water at 25oC ?

What is vapour pressure of sol. If vapourpressure of water is 40mmHg.

7/31/2019 41. Solution 1

28/38

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

7/31/2019 41. Solution 1

29/38


From Daltons 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

So

A Ao o

A A B B

py

pp x

p x p x

7/31/2019 41. Solution 1

30/38

o o

A A B BoA A A

oB A

A BoA A

p x p x1

y p x

p (1 x )1 x x 1

p x

o oB Bo o

A A A

p p1p x p

o oB B

o oA A A A

p p11

y p x p


7/31/2019 41. Solution 1

31/38


An unknown compound is immisciblewith water. It is steam distilled at 98.0oCand P = 737 Torr.poH20 = 707 torr at98.0oC. This distillate was 75% by weight of water.Calculate the molecular weight of the unknown

7/31/2019 41. Solution 1

32/38

Solution

Using Daltons law of partial pressureP

total= 737 torr PoH

2O = 707 torr

Pounknown = 737 707 = 30 torr.

If water = 100 g the unknown = 75.0 g

22

o

unknown unknown

o

H OH O

P n 7518

n M 100P

30 75 18

707 m 100

m 318.15 gmol

7/31/2019 41. Solution 1

33/38

Non-ideal solution

Solute-solvent interaction are different than solute-solute andsolvent solvent in non ideal solutions.

These do not obey Raoults Law.

For non ideal solutions H 0 and V 0

7/31/2019 41. Solution 1

34/38

Non ideal solution

For solution showing negativedeviation from Raoult's law.

For solution showing positivedeviation from Raoult's law.

H 0 and V 0

s idealP >P

H 0 and V 0

s idealP P

7/31/2019 41. Solution 1

35/38

Azeotropic mixtures

Solution showing positive deviation fromRaoults form minimum boiling azeotrope

Interaction between AB < interactionbetween AA or BB

Liquid mixtures which distil without any

change in composition are called Azeotropesor Azeotropic mixtures.

7/31/2019 41. Solution 1

36/38

7/31/2019 41. Solution 1

37/38

Azeotropic mixtures

Solution showing negative deviation fromRaoults law form maximum boiling azeotropes

Interaction between A B > interaction between A A or B B

7/31/2019 41. Solution 1

38/38

Thank you

41. solution 1

Documents