Chemistry Form 6 Sem 1 04

CHAPTER 4 :

CHEMISTRY FORM 6

CHEMISTRY FORM 6

CHEMISTRY FORM 6

CHEMISTRY FORM 6

PHYSICAL CHEMISTRY

PHYSICAL CHEMISTRY

PHYSICAL CHEMISTRY

PHYSICAL CHEMISTRY

CHAPTER 4 :

STATE OF MATTER

2.1 THE KINETIC THEORY OF MATTER

The Gaseous State

A gas is composed of atoms or molecules that are separated from each other by distances far greater than their own size.

The gas particles can be considered as ‘point’ particles – they possess mass but have negligible volume

There are no forces between the gas particles

Gas particles can vibrate, rotate and move anywhere within the container where the gas is placed.

A gas has no fixed shape or volume and can be easily compressed
A gas has no fixed shape or volume and can be easily compressed

Particles of gas are in constant random motion, moving in straight line unless they collide with the wall of container or with other gas particles
As the particles collide with the wall, they exert a pressure on container
The collision are perfectly elastic. There’s no loss of kinetic energy during

collision

The average kinetic energy of the particles is directly proportional to the absolute temperature (Kelvin scale) of the gas

2.2 The Gas Laws 2.2.1 Boyle’s Law

Stated by Robert Boyle (1662) during his investigation of

relationship between volume occupied by gas to the pressure of gas using the apparatus below.

The total pressure on the gas is the sum of the atmospheric

pressure and the pressure exerted due to the difference in the height (h) of the mercury in the column

A plot of the result yields the following graph

Based on the results, he formulated Boyle’s Law which state that

“The volume occupied by a fixed mass of gas at constant temperature is inversely proportional to its pressure

P

α 1/V

P

x V = constant

P

Ideal gas obeys Boyle’s Law under all conditions of temperature and

pressure.

Ideal laws does not exist as real gases obeys Boyle’s Law closely only at low

pressures and high temperatures. Real gases do not obey Boyle’s Law closely at high pressures and low temperatures.

Pressure = force per unit area, SI unit of pressure = Newton (N) per

square metre

1 Nm-2 is called a pascal (Pa) 1 kPa = 1000 Pa or 1000 N m-2
Atmospheric pressure is expressed in units of milimetres of mercury (mm

Hg). Standard atmospheric pressure is 760 mm Hg or 101325 Pa or 101325 N m-2_.

The SI Unit of volume is cubic metre, m3.

1 m3 = 103 dm3 = 106 cm3 ; 1dm3 = 1 litre

(a) 1/P against V 1/P V (b) V against 1/P V 1/P (c) PV against P PV P (d) V against P V P

2.2.2 CHARLES’ LAW

JAC Charles investigated the relationship between the volume

occupied by a gas, at constant pressure with temperature which is describe by the apparatus setting below.

He found that the relationship is a linear, as shown in the graph

V

α T

V

_{= constant x T}

V

/T

= V

/T

Temperature must

When the temperature of a sample of gas is increased, the

kinetic energy of the gas particles will increase, thus increasing, the rate of collision between the gas particles and the wall of container. At the same time, the collision are more energetic and this lead to an increase in the pressure exerted by gas. Hence, in order to maintain the original pressure, the volume occupied by the gas must now increase.

Avogadro’s Law

V

α number of moles (n)

V

_{= constant x n}

V

/n

= V

/n

2.2.3 Avogadro’s Law

~ state that equal volumes of all gases at the same temperature

and pressure contain equal number of atoms/molecules.

The volume occupied by 1 mole of any gas (molar volume)

depends on the temperature and pressure of the gas. The molar volume of gas at standard temperature & pressure is 22.4 dm3_.
The condition of s.t.p are

:-Temperature : 0 o_{C (273 K)}

Pressure : 101 kPa (1 atm)

Under room temperature and pressure, molar volume of gas is

24.4 dm3 _{/ mol.}

The condition of r.t.p are

:-Temperature : 25 o_{C (298 K)}

Ideal Gas Equation

Charles’ law: V α T (at constant n and P)

Avogadro’s law: V α n (at constant P and T)

Boyle’s law: V α (at constant n and T)

1

P

nT

V

α

nT

P

V

_{= constant x = R}

nT

P

nT

P

R

_{is the gas constant}

2.2.4 Ideal Gas Equation

For a fixed mass of gas
P V / T = constant

Pressure, P Volume, V Gas constant, R Temperature, T

Pa or N m-2 _m3 _{8.31 J K}-1 _mol-1 _K

atm dm3 _{0.0821 dm}3 _{atm K}-1 _mol-1 _K

2 2 1 1

T

V

P

T

V

P

₌

T

T

m _{is the mass of the gas in g}

M _{is the molar mass of the gas}

Molar Mass (M _{) of a Gaseous Substance}

dRT P

1. Calculate the volume of the following gases at s.t.p. (at 0o_{C and 1 atm = 101 kPa = 760 mm Hg)}

a) 200 cm3 _{hydrogen at 30°C and 2.00 x 10}4 _Pa

b) 4.25 dm3 _{hydrogen chloride at 50°C and 650 mm Hg}

P₁V₁ / T₁ = P₂V₂ / T₂ (2.00 x 104_{)(200) / (273+30) = (101 x 10}3_)(V 2) / (0 + 273) V₂ = 35.7 cm3 P₁V₁ / T₁ = P₂V₂ / T₂ (650 mmHg)(4.25) / (273+50) = (760 mmHg)(V₂) / (0 + 273) V = 3.07 dm3

c) 600 cm3 _{oxygen at 308 K and 2.20 atm}

d) 1.70 dm3 _{neon at 95°C and 1.45 atm}

V₂ = 3.07 dm3 P₁V₁ / T₁ = P₂V₂ / T₂ (2.20)(600) / (308) = (1.00)(V₂) / (0 + 273) V₂ = 1170 cm3 P₁V₁ / T₁ = P₂V₂ / T₂ (1.45)(1.70) / (273+95) = (1.00)(V₂) / (0 + 273) V₂ = 1.83 dm3

2.

An aerosol can containing helium gas at 30°C and 1.8 x 10

_{Pa is}

heated to 60°C. What is the pressure of helium in the can now?

3.

Determine the density of SO

gas at 25°C and 101 kPa.

P₁ / T₁ = P₂ / T₂ (1.8 x 103_{) / (273+30) = P / (60 + 273)} P₂ = 2.0 x 103 _Pa PV = nRT @ P = mRT / M_RV => P = dRT / M_R d = (101 x 103_{) (32 + 2(16) / 8.31 (25 +273)} d = 2610 g / m3 _{@ 2.61 x 10}-3 _{g / cm}3

4.

4. A 20.0 m3 _{steel tank was constructed to store liquefied natural gas (LNG) which}

contained mainly methane at 160°C and 101 kPa

(a) How many grams of methane can be stored in the container if the density of the liquid is 416 kg m-3_?

(b) Calculate the volume of a storage tank capable of holding the same mass LNG as a gas at 28°C and 101 kPa.

Since density = 416 kg m-3 _;

Mass = 20.0 x 416 = 8320 kg @ 8.32 x 106 _g

Since the temperature of gas decrease Since the temperature of gas decrease

V₁ / T₁ = V₂ / T₂

20.0 / 160 + 273 = V₂ / 28 + 273 V2 = 13.9 m3

5. An organic compound consists of 24.24% carbon, 4.04% hydrogen

and 71.72% chlorine by mass. If a 1.803 g sample occupies 546 cm

_at

84.2°C and 745 mm Hg (Given 760 mm Hg = 101 kPa)

(a) calculate its molar mass

(b) determine its molecular formula

Pressure in kPa = 745 mm Hg x 101 kPa / 760 mm Hg Pressure in kPa = 99.0 kPa

PV = nRT (99.0 x 103_{)(546 x 10}-6_{) = (1.803 / M}

R) 8.31 (84.2 + 273)

M_R = 99.0

(b) determine its molecular formula

Element C H Cl Mass 24.24 4.04 71.72 Mol 24.24 12 = 2.02 4.04 1 = 4.04 71.72 35.5 = 2.02 Ratio 2.02 / 2.02 = 1 4.04 / 2.02 = 2 2.02 / 2.02 = 1 Empirical formula = CH₂Cl (CH₂Cl)n = 99.0 [(12(1) + 2(1) + 35.5(1)]n = 99 n = 2 Molecular formula = (CH₂Cl)2 = C₂H₄Cl₂

2.3 Dalton’s Law Partial Pressure

Partial Pressure – In a mixture of gases which do not interact

with one another, each gas in the mixture will exert its own pressure independent of the other gases.

Dalton’s Law of Partial PressureDalton’s Law of Partial PressureDalton’s Law of Partial PressureDalton’s Law of Partial Pressure –

In a mixture of gases which do not interact with one another, the total pressure of the mixture is the sum of the partial pressure of the

constituent gases

P_T = P_a + P_b + P_c + …….. P_T = P_a + P_b + P_c + ……..

The partial pressure of a gas is also given by the following expression:

P_a = mole fraction of A in mixture (X_a) X Total pressure Mole fraction of A (X_a) = moles of A

Dalton’s Law of Partial Pressures

V

_{and T}

are

P

P

P

=

P

+

P

Consider a case in which two gases, A and B, are in a container of volume V.

P

=

n

RT

V

P

=

n

_RT

V

A is the number of moles of A

n

B is the number of moles of B

n

n

P

=

P

+

P

X

=

n

n

+

n

X

=

n

n

+

n

P

=

X

P

P

=

X

P

P

_{= X}

P

A sample of natural gas contains 8.24 moles of CH₄, 0.421

moles of C₂H₆, and 0.116 moles of C₃H₈. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C₃H₈)?

P

_{= X}

P

0.116

P

= 1.37 atm

X

=

0.116

8.24 + 0.421 + 0.116

= 0.0132

P

Example 2

A mixture of gases at 200 kPa contains the following composition of gases by volume. (20% CO₂ ; 50% CO ; 30% O₂)

a) What is the partial pressure of each gas in the mixture?

b) If CO₂ is removed from the vessel, what would be the partial pressure of O₂ and CO

For CO₂

P_CO2 = x_CO2 . P_tot

P_CO2 = (20) / (20 + 50 + 30) x 200 kPa P_CO2 = 40 kPa For CO For CO P_CO = x_CO . Ptot P_CO = (50) / (20 + 50 + 30) x 200 kPa P_CO = 100 kPa For O₂ P_O2 = x_CO . Ptot P_O2 = (30) / (20 + 50 + 30) x 200 kPa P_O2 = 60 kPa

Partial pressure of O2 and CO will remain as 60 kPa and 100 kPa

respectively, since removing CO₂ will not affect the partial pressure of each gas in the vessel

Example 3 : 5.00 dm3 _{of H}

2 at 200 kPa, 12.0 dm3 of N2 at 300 kPa and 1.50 dm3 of Cl2 at 120 kPa were forced into a vessel of capacity 10.0 dm3 _{at constant temperature.}

Calculate the pressure in the vessel as a result of the mixture of these gases.

For hydrogen P₁V₁ = P₂V₂ (200 kPa)(5.00) = P₂(10.00) P₂ = 100 kPa For nitrogen P₁V₁ = P₂V₂ (300 kPa)(12.00) = P₂(10.00) P₂ = 360 kPa For chlorine Total pressure = P_H2 + P_N2 + P_Cl2

= 100 kPa + 360 kPa + 18 kPa = 478 kPa

For chlorine P₁V₁ = P₂V₂

(120 kPa)(1.50) = P₂(10.00) P₂ = 18 kPa

Nitrogen monoxide and oxygen gas are mixed in the vessel as in the diagram below

Calculate its partial pressure of each gas and total pressure if the stopper is lifted up from the vessel.

For NO For O₂ For NO P₁V₁ = P₂V₂ (0.5 atm)(4) = P₂(4 + 2)) P₂ = 0.333 atm For O₂ P₁V₁ = P₂V₂ (1.0 atm)(2) = P₂(4 + 2) P₂ = 0.333 atm Total pressure = P_NO + P_O2 = 0.333 + 0.333 = 0.666 atm

2.5

THE DISTRIBUTION OF MOLECULAR SPEEDS IN A GAS

Speed of various gas molecules differ widely and are constantly

changing. This is due to their frequent collision and the resultant

changes in energy.

At any time, the speed instantly increase and spread widely. This spread

of molecular speeds is called Maxwell-Boltzmann distribution.

There are 2 factors which can only affect the distribution of the graph

1.Temperature.

1.Temperature.

At constant pressure, the Maxwell Boltzmann distribution graph can be

varies with temperature

The graph below shows the distribution of the graph at 3 different

temperature

THE DISTRIBUTION OF MOLECULAR SPEEDS IN A GAS

Speed of various gas molecules differ widely and are constantly

changing. This is due to their frequent collision and the resultant changes in energy.

At any time, the speed instantly increase and spread widely. This

spread of molecular speeds is called Maxwell-Boltzmann distribution.

Important points to note about the distribution curves for molecular

speed in a gas.

Area directly proportional to the total number of molecules in the gasArea directly proportional to the total number of molecules in the gas
At any temp., a small proportion have high/low speed – mostly

average

T ↑, distribution curve shift to right – peak become lower – number

of gas molecules with high speed increase ; average speed decrease.

When heat is supplied, the energy is used to increase the average

kinetic energy and to increase the motion of the gas molecules.

2.

Molecular mass

At constant temperature and pressure, the molecular gas distribution

can also be influenced by the mass of gas itself.

Distribution graph below shows the distribution of different gases

molecules with different molar mass

Important points to note about the distribution curves for

molecular speed in a gas.

Area directly proportional to the total number of molecules in the gas
At constant temperature, a small proportion of molecules have high/low

speed – but mostly average

Smaller the molecular mass of the gaseous molecules, distribution curve

shift to right & at the same time, peak become lower. This indicated lesser molecules travel at average speed while more molecules travel at high speed.

This is due to lighter molecules can move faster than a heavier molecule.
This is due to lighter molecules can move faster than a heavier molecule.

More light molecules are able to travel at high speed, causing lesser molecules to travel at average speed (represent by lowered peak)

4.5

Real gases – Deviations from ideality

Real gases area gases that do not obey the ideal gas equation, PV = nRT

Real gases behave ideally at _____________ & ________________

At high pressure and low temperature, gases would ______________

The deviation of real gases from ideal gas can be explained using

PV/RT against P. The graph below shows how 3 real gases behave

At very low pressure, the gas molecules are far apart and attraction

forces between the molecules _____________________________

low pressure high temperature liquefied

can be ignored

forces between the molecules _____________________________

However, when pressure increased, the attraction forces become

__________ since the molecules are now __________. Thus, when

molecules are at _________ deviation, these attractions make the gas

more compressible. Volume of the gas _________ more than

expected. Thus, PV …… nRT and the gas exhibit _______________

deviation from ideality.

can be ignored

significant closer negative

lower

Graph of PV / RT against P

PV

RT

1

H

CH

H

NH

P

1

At higher pressure, the molecules are pushes very close together, and

now repulsive forces operate between them. These repulsive forces

make the gas harder to compress. As a result, a higher pressure has to

be used to compress the gas. Thus, PV …. nRT. Moreover, as high

pressure, volume of the gas become smaller and hence the actual

volume of the gas molecules in comparison with the volume of the

containing vessel cannot be ignored. Thus, the gas exhibit _________

deviation from ideality.

The same curve could also be given at different temperature.

>

positive

The same curve could also be given at different temperature.

Example, for carbon dioxide gas at 100 K, 200 K and 300 K

Graph of PV / RT against P

PV

RT

1

P

1

At low temperatures, when a molecule is about to collide with the wall of

container, the intermolecular forces of attraction between molecules will the force exerted by the impact. As a result, the pressure exerted by the gas is reduced. This will cause the value of PV …… RT (negative deviation)

At very high temperatures, the kinetic energy of the gas molecules is very

high and there are no chances for molecules to interact with each other causing no intermolecular forces occur.

Negative deviation caused by polar bond

Light gases such as ………. and ……… have small molecular

<

hydrogen helium

Light gases such as ………. and ……… have small molecular

mass and are non-polar. Hence, they possessed very weak intermolecular forces, and behave almost to ideal (especially at room condition)

Gas such as ammonia, carbon monoxide and nitrogen dioxide are polar

molecules so they have strong intermolecular forces. These gases will show deviation from ideal behavior

3.1 Properties of Liquid

Average content of kinetic energy.

Molecules arrangements are loose and a little apart.

Intermolecular forces among particles are intermediate.

Particles in liquid state can move around freely, rotate and vibrate freely.
Has fixed mass and volume.

Shape follow its container, and must be filled from beneath of the container.

Molar latent heat of fusion, ∆H_fusion :- amount of energies required to change 1 mole of solid to liquid under standard condition.

Molar latent heat of vaporization, ∆H_vaporization:- amount of energies required to change 1 mole of liquid to gas under standard condition

3.1.1 : Vapour Pressure

Particles in liquid are in constant motion. Due to the presence of intermolecular

forces, the particles can only move throughout the ‘body’ of liquid and not free from attraction by other liquid particles.

When some of these particles at surface of liquid have enough energy to overcome

intermolecular forces, it ‘escape’ as vapour. This process is called evaporation.

In open container, evaporation on will continue until no liquid remain.

In a close container, vapour will trap in the container and collide with the walls of

container to form vapour pressure. container to form vapour pressure.

Vapour pressure will increase with time. When the rate of vaporation = rate of

condensation it will attain a maximum value of pressure ~ called as saturated vapour pressure.

Saturated vapour pressure – vapour pressure exerted in a close container, where

Saturated vapour pressure is directly proportional to the temperature. Higher the

temperature more liquid can escape as vapour and the collision of vapour with container will be more vigorous.

P

3.1.2 : Boiling

Boiling point of liquid is the temperature where saturated vapour pressure is equal

to the external pressure.

VAPOUR PRESSURE/ ATM

ETHER ETHANOL WATER 1

35 78 100 temperature

For example, the boiling point of water is 100 0C at 1 atm, while under the same

pressure, boiling point of ether and ethanol are 35 and 78o_{C respectively.}

Bubbling at the surface of water is the sign of that the saturated vapour pressure is

Constant heat is supply during boiling process, so that molecules in water

receive enough energy to overcome the attraction forces among the liquid molecules.

Volatility of liquid refers to the liquid with high saturated vapour

pressure at any given temperature.

Higher the vapour pressure, higher the volatility. Among the 3 liquids,

volatility increased from

Intermolecular forces holding the particles in the liquid state are weak, so the

tendency of liquid to ‘escape’ as vapour, is high. More volatile a liquid, lower tendency of liquid to ‘escape’ as vapour, is high. More volatile a liquid, lower the boiling point, faster for it to achieve saturated vapour pressure at low temperature.

Furthermore, when the pressure is lowered, the attraction forces between the

particles become weaker. Hence, liquid are easier to vapourise and decrease its boiling point

3.2 LATTICE STRUCTURE & CRYSTAL SYSTEM

2 types of solids – a) crystalline and b) amorphous

In crystalline structure, atoms, molecules or ions are closely packed and have an

ordered 3-D arrangement.

Regular arrangement of atoms, molecules & ion is called as crystal lattice (space

lattice).

Lattice - pattern of point repeat

Unit cell ~ small repeating unit that made up crystal.

Crystal system (primitive unit cells) is a method of classifying crystalline substance

on the basis of their unit cell.

lattice point

Unit Cell

point

Shared by 8 unit cells

Shared by 2 unit cells

From the lattice formed, the number of particles involved can be calculated. In

each position, it only contributes a fraction of its volume or mass to its unit cell

Shared by 8 unit cells

Shared by 2 unit cells

1 atom/unit cell (8 x 1/8 = 1) 2 atoms/unit cell (8 x 1/8 + 1 = 2) 4 atoms/unit cell (8 x 1/8 + 6 x 1/2 = 4)

EXAMPLE 1 :

CALCULATE THE NUMBER OF PARTICLES IN THE UNIT CELL OF THE a) FACE-CENTRED CUBIC LATTICE b) BODY CENTRED CUBIC LATTICE

EXAMPLE 2 :

DIAGRAM SHOWS A UNIT CELL OF AN OXIDE OF ELEMENT X. WHAT IS THE FORMULA OF THE X OXIDE?

a) FCC = 8 corner + 6 faces = 8 x 1/8 + 6 x ½ = 1 + 3 = 4

b) BCC = 8 corner + 1 body center = 8 x 1/8 + 1 2 X = 1 body + 2 face = 1 (1) + 2 (1/2) = 2 O = 8 corner = 8 (1/8) = 1 Formula = X₂O O = 12 middle edge = 12 (1/4) = 3 Rh = 8 corner = 8 (1/8) = 1 b) Formula = RhO₃

* Other than the regular 3 unit cell, there are other structure that can be found, such as a hexagonal close-pack (h.c.p.) and cubic close pack (c.c.p.).

3.3 Allotropes

When an element exists in two or more physical forms in the same state, they

exhibit allotropy

Usually, allotropes have the different types of crystalline structure, due to different

arrangement of atoms

Examples of elements which exhibit allotropies are sulphur and carbon

Allotropes can be changed under certain temperature and pressure. For example,

sulphur allotropes change its crystalline structure at 95.6o_{C at 1 atm.}

Tm = 95.6o_C

Rhombic Sulphur Monoclinic Sulphur

This temperature is known as transition temperature, Tm

Both sulphur allotropes have 8 atoms, S8, which look like a crown shape

ALLOTROP ES ALPHA-SULPHUR (RHOMBIC) BETA-SULPHUR (MONOCLINIC) DIAGRAM SHAPE COLOUR DENSITY MELTING POINT orthorhombic monoclinic Light yellow Deep yellow

Greater Lower

ALLOTRO

PES DIAMOND GRAPHITE

DIAGRAM

- Each carbon is surrounded by 4 - Each carbon is surrounded by 3

EXPLANA TION

- Each carbon is surrounded by 4 other carbon atoms in tetrahedral shape, building a strong giant covalent network among C–C. Because of this, diamond is extremely hard and strong.

- Since all 4 valence electron in carbon are strongly bonded, so there is no delocalized electron around the structure, making diamond an insulator

- Each carbon is surrounded by 3 other carbon atoms in a planar, building a sheet like honeycomb hexagon structure. Each layer of graphite is hold by mere weak Van Der Waals forces.

- Since there are only 3 electrons used for bonding, so there’s 1 unbonded ‘delocalised’ electron which is locate among/within the layer

ALLOTRO

PES DIAMOND GRAPHITE

DIAGRAM TYPE OF TYPE OF HYBRIDIS ATION ELECTRIC AL CONDUC TIVITY DENSITY APPLICAT ION / USES sp3 sp2 insulator _conductor higher lower

Other allotrope of carbon is fullerene, which is with the formula C₆₀. It is also known

as buckminsterfullerene, where it consists of 12 pentagonal faces and 20 hexagonal faces (similar to a soccer ball shape).

Each carbon atom is _____________ hybridized.

It is used to make superconductor, lubricant, micro-ball bearing in nano machine and

Maximum Density 40_C

Density of Water Water is a Unique Substance

Ice is less dense than water

Uses of Solid Carbon Dioxide (Dry Ice)

• Solid carbon dioxide is used as a refrigerant for foodstuffs such as ice cream since it does not melt on warming.

• it is also used in stage performance to give a smoky effect. This is due to the effect of condensation of water vapour

around the atmosphere by the cold air due to sublimation of carbon dioxide.

• It is also used as fire extinguisher to put off fire. Fire

extinguisher is stored in liquid phase under high pressure. extinguisher is stored in liquid phase under high pressure.

Once the gauge is released, the pressure is reduced and CO₂ gas is given off.

• It is also used in ‘cloud seeding’ to encourage the formation of ice crystals in clouds. As the dry ice sublimes, it absorbs heat in the clouds, thereby lowering the Temperature and causing the water vapour to condense and form water. Cloud seeding is carried out to induce rainfall especially after long periods without rain

4.10 Colligative properties - effect of a non volatile solute on the vapour pressure of a solvent

• Colligative properties are those properties of solutions that depend on the number of dissolved particles in solution, but not on the identities of the solutes.

• For example, the freezing point of salt water is lower than that of pure water, due to the presence of the salt dissolved in the water.

• To a good approximation, it does not matter whether the salt • To a good approximation, it does not matter whether the salt dissolved in water is sodium chloride or potassium nitrate; if the molar amounts of solute are the same and the number of ions are the same, the freezing points will be the same

• The four commonly studied colligative properties are freezing point depression, boiling point elevation, vapor pressure

lowering, and osmotic pressure. In this Chapter, we shall study the first two while at the end of the Chapter we shall discuss the next two.

4.10.1 The effect of a solute on the Vapour Pressure of a solvent

The addition of a non-volatile solute to a pure liquid caused the

vapour pressure of liquid Y to decrease. This occurs because the

presence of solute molecules decrease the number of molecules of

solvent that can reach the surface of the liquid. As a result, the number

of solvent that can 'escape' as vapour decrease

Freezing point depression Boiling point elevation

↓ ∆T_f is defined as the freezing point of the pure solvent (To

f) minus the

freezing point of the solution(T_f)

↓ the freezing point is lower

compared to the pure solvent.

↓ Freezing involves a transition from the disordered state to the ordered state. For this to happen, energy must be removed from the system.

↑ ∆T_b is defined as the boiling point of the solution (T_b) minus the boiling point of the pure solvent (T°b)

↑ the boiling point is higher compared to the pure solvent

↑ Boiling involved a transition from liquid to gaseous state. During the process occur, energy is absorbed to

must be removed from the system. Because a solution has greater disorder than the solvent, more energy needs to be removed from it to create order than in the case of a pure solvent. Therefore, the solution has a lower freezing point than its solvent. Note that when a solution freezes, the solid that separates is the pure solvent component

process occur, energy is absorbed to the system. Due to the presence of solute in solvent (which cause interference of other intermolecular forces) , more energy is required to vaporise the solvent, hence increase the boiling point of the solvent.