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Solutions 08 01

The document discusses the concepts of ideal and non-ideal solutions, focusing on Raoult's law and deviations in vapor pressure. It explains positive and negative deviations, their characteristics, and introduces colligative properties, which depend on the number of solute particles. Additionally, it covers the effects of solutes on boiling point and vapor pressure, providing examples and calculations related to these concepts.

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0% found this document useful (0 votes)
31 views32 pages

Solutions 08 01

The document discusses the concepts of ideal and non-ideal solutions, focusing on Raoult's law and deviations in vapor pressure. It explains positive and negative deviations, their characteristics, and introduces colligative properties, which depend on the number of solute particles. Additionally, it covers the effects of solutes on boiling point and vapor pressure, providing examples and calculations related to these concepts.

Uploaded by

tanvichauhan4342
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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LECTURE 8

12
Example

(i) Benzene and toluene

(ii) CCl4 and SiCl4

Vapour Pressure
(iii) n-hexane and n-heptane

(iv) C2H5Br and C2H5Cl

XA=0 XA=1
(v) PhCl and PhBr XB=1 Mole Fraction XB=0

(vi) n-butylchloride and n-


butylbromide
Those solutions which do not obey Raoult's law over the entire range of
concentration, then they are called non-ideal solutions.

PA  PoA XA

PB  PoB XB

Ps  PA0 XA + PB0 XB
For non ideal solutions : A–A interactions or B-B interactions  A-B interactions.

A------A  B------B  A------B

Non ideal solutions show either positive or negative deviations from Raoult's
law.
A positive deviation solution possesses the following characteristics :

Vapour pressure Intermolecular interactions

PA > PoA XA ; PB > PoB XB A −−−A + −− −


>A−−
Ptotal > (PoA XA + PoB XB ሻ 2

Change in volume (Vmix) Change in enthalpy (Hmix)


Positive (+) Positive (+)
A + B = A-----B
10 ml 10ml 20.2ml Attraction force decrease
.
A positive deviation solution possesses the following characteristics :

Change in Gibb’s energy (Gmix) Change in entropy (Smix)

Negative(-) Positive (+)

Mixing of solutions is Entropy increases as a


spontaneous process result of mixing.

Boiling point

(B.P.)Th > (B.P.)Exp


'A' and 'B' escape easily showing
higher vapour pressure than the
expected value.
Vapour Pressure

XA=0 XA=1
XB=1 Mole Fraction XB=0
Example

(i) Ethanol and cyclohexane


(ii) Ethanol and Water
(iii) Ethanol and Acetone
(iv) Methanol and H2O
(v) CCl4 and Benzene
(vi) CCl4 and Toluene
(vii) CCl4 and CHCl3
(viii) CCl4 and Methanol
(ix) Benzene and Acetone
(x) CS2 and Acetone
A negative deviation solution possesses the following characteristics :

Vapour pressure Intermolecular interactions


A −−−A + −− −
PA < PoA XA ; PB < PoB XB <A−−
2
Ptotal < (PoA XA + PoB XB ሻ

Change in volume (Vmix) Change in enthalpy (Hmix)


Negative (–) Negative (Exothermic)
A + B = A----B
10 ml 10ml 19.8mL Heat is evolved as result of
mixing. (attraction force
increases)
A negative deviation solution possesses the following characteristics :

Change in Gibb’s energy (Gmix) Change in entropy (Smix)

Negative (–) Positive (+)


Mixing of solutions is Entropy increases as a
spontaneous process result of mixing.

Boiling point
(B.P.)Th < (B.P.)Exp
'A' and 'B' escape easily showing
less vapour pressure than the
expected value.
Vapour Pressure

XA=0 XA=1
XB=1
Mole Fraction XB=0
Example

(i) CHCl3 and CH3COCH3


(ii) CHCl3 and C6H6
(iii) CHCl3 and C2H5OC2H5
(iv) CHCl3 and HNO3

Vapour Pressure
(v) CHCl3 and CH3COOH
(vi) H2O and HCl
(vii) H2O and HNO3
(viii) CH3COOH and CH3OH
XA=0 XA=1
(ix) CH3COOH and C5H5N XB=1
Mole Fraction XB=0

(x) CH3COCH3 and Aniline


Difference between ideal and non ideal solutions

Properties Ideal Solutions Positive Deviation Negative deviation

A −−−A + −− − A −−−A + −− −
Intermolecular A-----A = B-----B =
interactions A-----B 2 2
>A−− <A−−

Vapour
Ps = PA0 XA+ PB0 XB Ps > PA0 XA+ PB0 XB Ps < PA0 XA+ PB0 XB
Pressure

Zero Positive (+) Negative(–)


A + B = A-----B A + B = A-----B A + B = A-----B
10ml 10ml 20ml 10ml 10ml 20.2ml 10ml 10ml 19.8ml
Vmix
Intermolecular Intermolecular Intermolecular
distances are same. distances increase. distances decreases.
Difference between ideal and non ideal solutions

Properties Ideal Solutions Positive Deviation Negative deviation

Zero Positive
Negative(Exothermic)
(Endothermic)
Hmix No heat is absorbed
Heat is evolved as
or evolved as a result Heat is absorbed as
result of mixing.
of mixing. result of mixing.

Gmix Negative (-) Negative (-) Negative (-)

Smix Positive (+) Positive (+) Positive (+)


Colligative
Properties
Those physical properties of a solution which depends upon the relative
number of particles of solute but do not depend on nature of solute
particles are called colligative properties.

Number of solute particles

Number of molecules ( in the solution of non electrolyte)

Number of ions (in the solution of electrolytes)

Number of moles of solute

Mole fraction of solute


1 Relative lowering in vapour pressure

2 Elevation in boiling point

3 Depression in freezing point

4 Osmotic Pressure of Solution


Relative Lowering in Vapour Pressure

When a non-volatile solute is dissolved in a pure solvent, the vapour


pressure of the solvent is lowered.
The vapour pressure of a solution is always lower than that of pure
solvent, because the escaping tendency of solvent molecules decreases
(due to lesser solvent molecules per unit surface area).
Relative Lowering in Vapour Pressure

If at a certain temperature P° is the vapour pressure of pure solvent, and


Ps is the vapour pressure of solution then :

According to Raoult’s law

PA0 −PS nB
Relative lowering in vapour pressure = =
PA0 NA + nB
Relative Lowering in Vapour Pressure

For a dilute solution nB << NA

PA0 −PS nB nB
≈ ΔP ∝
PA0 NA NA

Relative lowering depends upon relative number of solute particles.


Therefore, it is called colligative properties.
Example

Calculate weight of urea added to 100 g H2O to reduce its vapour pressure
by 25%.
Example

3 gm urea is dissolved in 45 gm H2O, then find out RLVP.


Example

Calculate lowering in vapour pressure of 1 molal aqueous solution of a


non-volatile solute at 100oC.
Boiling Point

The boiling point of a liquid is that temperature at which its vapour


pressure becomes equal to the atmospheric pressure.
1 atm
Atmospheric pressure
1 atm

Vapour pressure
1 atm

Boiling point of water = 100°C


The boiling point of a liquid is that temperature at which its vapour
pressure becomes equal to the atmospheric pressure.

1
Boiling Point 
Vapour Pressure of liquid at room temperature
Atmospheric pressure
1 atm 1 atm

101
Vapour pressure
1 atm

Boiling has stopped


Atmospheric pressure
1 atm 1 atm

Vapour pressure
101

1 atm

Elevation in boiling point takes place


When a non-volatile solute is dissolved in a pure solvent, its vapour pressure
is decreased. Hence, we have to make it equal to Pext for boiling, so
temperature is increased. Thus, boiling point increases.

1 atm

nt
l v e

Vapour Pressure
S o
re
Pu
P n
io
So lu t
PS
Tb

T
b Tb
Temperature

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