Test - 4 TOPIC – Solution & Colligative properties Time : 45 min.

Marks : 23 Dhole Sir

Q1. State & Explain Van’t Hoff -Boyl’s law - 2M
Q2. State & Explain Van’t Hoff -Charles law - 2M
Q3 State & Explain Van’t Hoff -Avogadro’s law - 2M
Q4. Derive the relationship between relative Osmotic pressure & molar mass of solute 2M
Q5. Define 1) Osmotic pressure 2) Semipermiable membrane 2M
Q6. Define Osmosis ? Explain the Abbe Nollet Experiment 3M
Q7. Derive van’t Hoff general solution equation 2M
Q8. Derive the relationship between degree of dissociation & Van’t Hoff factors 2M
Q9. What is abnormal colligative property ? Explain the reasion 2M
Q10. 2.4 x 10-3 Kg of a substance dissolved in 0.120 dm3 of water gave an osmotic
pressure of 1.331 atmosphere at 293 K . calculate the molecular weight of the
substance ? 3M
Q11 Define Van,t Hoff factors ? 1M

Test - 4 TOPIC – Solution & Colligative properties Time : 45 min.

Marks : 23 Dhole Sir
Q1. State & Explain Van’t Hoff -Boyl’s law - 2M
Q2. State & Explain Van’t Hoff -Charles law - 2M
Q3 State & Explain Van’t Hoff -Avogadro’s law - 2M
Q4. Derive the relationship between relative Osmotic pressure & molar mass of solute 2M
Q5. Define 1) Osmotic pressure 2) Semipermiable membrane 2M
Q6. Define Osmosis ? Explain the Abbe Nollet Experiment 3M
Q7. Derive van’t Hoff general solution equation 2M
Q8. Derive the relationship between degree of dissociation & Van’t Hoff factors 2M
Q9. What is abnormal colligative property ? Explain the reasion 2M
Q10. 2.4 x 10-3 Kg of a substance dissolved in 0.120 dm3 of water gave an osmotic
pressure of 1.331 atmosphere at 293 K . calculate the molecular weight of the
substance ? 3M
Q11 Define Van,t Hoff factors ? 1M

Test - 4 TOPIC – Solution & Colligative properties Time : 45 min.

Marks : 23 Dhole Sir
Q1. State & Explain Van’t Hoff -Boyl’s law - 2M
Q2. State & Explain Van’t Hoff -Charles law - 2M
Q3 State & Explain Van’t Hoff -Avogadro’s law - 2M
Q4. Derive the relationship between relative Osmotic pressure & molar mass of solute 2M
Q5. Define 1) Osmotic pressure 2) Semipermiable membrane 2M
Q6. Define Osmosis ? Explain the Abbe Nollet Experiment 3M
Q7. Derive van’t Hoff general solution equation 2M
Q8. Derive the relationship between degree of dissociation & Van’t Hoff factors 2M
Q9. What is abnormal colligative property ? Explain the reasion 2M
Q10. 2.4 x 10-3 Kg of a substance dissolved in 0.120 dm3 of water gave an osmotic
pressure of 1.331 atmosphere at 293 K . calculate the molecular weight of the
substance ? 3M
Q11 Define Van,t Hoff factors ? 1M

Test - 4 TOPIC – Solution & Colligative properties Time : 45 min.

Marks : 23 Dhole Sir
Q1. State & Explain Van’t Hoff -Boyl’s law - 2M
Q2. State & Explain Van’t Hoff -Charles law - 2M
Q3 State & Explain Van’t Hoff -Avogadro’s law - 2M
Q4. Derive the relationship between relative Osmotic pressure & molar mass of solute 2M
Q5. Define 1) Osmotic pressure 2) Semipermiable membrane 2M
Q6. Define Osmosis ? Explain the Abbe Nollet Experiment 3M
Q7. Derive van’t Hoff general solution equation 2M
Q8. Derive the relationship between degree of dissociation & Van’t Hoff factors 2M
Q9. What is abnormal colligative property ? Explain the reasion 2M
Q10. 2.4 x 10-3 Kg of a substance dissolved in 0.120 dm3 of water gave an osmotic
pressure of 1.331 atmosphere at 293 K . calculate the molecular weight of the
substance ? 3M
Q11 Define Van,t Hoff factors ? 1M
 Test -4
Q1. State & Explain Van’t Hoff -Boyl’s law - 2M
Ans :- Statement :
A constant temperature, the osmotic pressure of dilute solution is directly proportional to its molar
concentration or inversely proportional to the volume (V) of the solution containing 1 mole of a solute. 1-Marks
Expression :
1) If ‘C’ be the molar concentration and p be the osmotice pressure, then according to Boyle’s van’t
Hoff law.
 C ............(1) 1/2-Marks
2) Let, 1 mole of solute dissolve in V dm3 of solvent then
C = n/V = 1/V (n = 1 mole) ............ (2)
3) Thus from equation (1) and (2)
 1/V
 V = Constant 1/2-Marks

Q2. State & Explain Van’t Hoff -Charles law - 2M

Ans :-
solutions :
Osmotice pressure of a dilute solution is directly proportional to its absolute temperature at constant
concentration (or volume) 1-Marks
Expression :
Let p be the osmotic pressure and T be the absolute temperature, then
 T
1-Marks
 /T = constant

Q3 State & Explain Van’t Hoff -Avogadro’s law - 2M

Ans :- Statment : -
At a given temperature Equal volume of two solutions having the same osmotic pressure contain
equal number of solute partical
OR
It is state as equal volumes of isotonic solutions contain an equal number of solute particales
at given temperature 1-Marks

for a given solution

V = nRT

For Solution 1 V1 = n1RT1 .....................................1

For Solution 2 V2 = n2RT2 ....................................2 1/2-Marks

Where n1 & n2 are the number of solutes in V1 & V2 liters of two solutions respectively

If  =  and T1 = T2 and V1 = V2

Than from equation 1 & 2

n1 = n2 1/2-Marks

Since , number of moles are equal , number of molecules are equal .

Q4. Derive the relationship between relative Osmotic pressure & molar mass of solute 2M
Ans :-
According to Van’t Hoff’s equation

V = n2RT 1/2-Marks

n2 R T
 =
V

If W2 is the mass of solute dissolved in V liters of solution & M2 be its molar mass 1/2-Marks

Than number of moles of solute is given by

W2
n2 = 1/2-Marks
M2

W2 R T
 =
M2V

W2 R T
M2 =
V 1/2-Marks
Thus by using above equation molar mass of solute can be calculate

Q5. Define 1) Osmotic pressure 2) Semipermiable membrane 2M

1) Osmotic pressure
Ans :- The minimum external pressure which should be applied on the solution so as to prevent
the migration of solvent molecules into the solution through semipermiable membrane is
called osmotic pressure
1-Marks

2) Semipermiable membrane
Ans :-
A memberane which allows only solvent molecules and not the solute molecules to pass thourgh it,
is called semipermeable membrane.
Example :- 1) Copper ferrocyanide Cu2[ Fe(CN)6) ] ( Artificial Prepared )
2) Cellulose 1-Marks

Q6. Define Osmosis ? Explain the Abbe Nollet Experiment 3M

Define Osmosis ?
Ans :-
The spontaneous & Unidirectional flow of solvent molecules from the solution of region of
lower concentration to the region of higher concentration through a semipermeable membrane
is called osmosis. 1-Marks

Semipermiable membrane

Flow of solvent

solvent Solution
 Explain the Abbe Nollet Experiment
1) Consider an aqueous solution of sugar placed in an inverted thistle funnel having a semipemeable
memberance like animal membrance or parchment paper attached to its bottom.
2) Now, the thistle funnel is kept in a jar containing water.
3)The nature of memberane is such that it allows only the molecules of the solvent and not the
solute molecules to pass through it.
1-Marks
4) Water molecules pass into the solution throughsemipermeable memberane and the level of
solution rises in the thistle funnel gradually. This process is called osmosis.
5) Osmosis can also take place when two solutions of different concentrations are separated by
a semipermeable memberane. In such case the flow of sovent molecules takes place from a
solution of lower concentration to the solution of higher concentration.

level after osmosis 

Original level 

 Sucrose solution 1-Marks

 Water
Semipermiable
membrane

Fig . Abbe Nollet’s Experiments

Q7. Derive van’t Hoff general solution equation 2M

Ans :-
a) Osmotice pressure of a dilute solution is inversely proportional to the volume containing 1 gm
mole of the solute. Thus  1/V .... (1) 1/2-Marks
b) Now, osmotic pressure of a dilute solution is directly proportional to absolute temperature
Thus,  T .... (2) 1/2-Marks
c) Combining equations (1) and (2)
 T/V
OR V = KT 1/2-Marks
Where, K is solution constant.
d) Thus, van’t Hoff solution equation is similar to gas equation.
i.e. PV = RT for 1 mole of solute.
e) However value of K is same as R.
Hence, V = RT (For 1 mole of solute)
and V = n RT (For n mole of solute)

 n RT
=
V

 = C RT
1/2-Marks
Where  = Osmotic pressure of dilute solution.
V = Volume containing 1 gm mol of solute.
n = Number of moles of solute R =Gas constant. T = Absolute temperature.
Q8. Derive the relationship between degree of dissociation & Van’t Hoff factors 2M

Vant Hoff factors ( i) & degree of dissociation () : -

By Using Van’t Hoff’s factors ‘i’ the degree of dissociation of electrolyte in the solution can be determine
Mathamatical expression :-
1. Consider an electrolyte Ax By dissolved in the solvent & it undergo dissociation into x ions of
A m+ & y ions of B n-
1/2-Marks
A x By  x A m+ + y B n-
2. Let m- molality of electrolyte i:e ‘m’ moles of solute in 1kg of solvent.
3. On dissolution solute dissociates .  - be the degree of dissociation
4. Then at equlibrium out of original ‘m’ mole , m moles dissociated & m - m = m ( 1-) moles
of solute remain undissociated
5. The dissociation of ma moles of solute produce x m moles of Am+ ions & ym moles of
B n- ions
6. Hence total number of moles .,mt in the solution at equlibrium will be , m ( 1- a) moles of Ax By &
x ( m) moles of m+ ions & y ( m) moles of B n- ions
Hence
mt = = m ( 1- ) + x (m) + y (m)
= m [ (1- x + y ]
= m [ 1 + ( x + y -1) ] 1/2-Marks
It is conveniant to represent the total number of ions produced by dissociation of one mole of solute
Ax By i:e x+ y = n’
Hence x+ y = n’

There fore mt = m [ 1 + ( n’ -1) ] 1/2-Marks

Observed number of moles

Now Van’t Hoff’s factors i =
Theoretical number of moles
mt
i = = m [ 1 + ( n’ -1) ]
m m
Hence i = [ 1 + ( n’ -1) ]

i- 1 1/2-Marks
The degree of dissociation , 
n’- 1
M ( Theoretical )
Futher i = = [ 1 + ( n’ -1) ]
M ( Observed )

M ( Theoretical ) - M ( Observed )
Hence The degree of dissociation 
M ( Observed ) ( n’ -1)

By determining experimentally observed M , by measuring any of the colligative properties

 can be calculate

Q9. What is abnormal colligative property ? Explain the reasion

Ans :- The difference in the experimental value ( Observed values ) & its normal value ( therotically )
due association or dissociation of the solute particles in the solution is called abnormal
osmotic pressure . 1-Marks
1) Dissociation of solute molecule :-
* When a solute like an electrolyte is dissolved in a polar solvents like water ,
* It undergoes dissociationwhich results in the increase in the number of particales in the solution
* Hence , the observed value of colligative property becomes , higher than the therotical value .
For example :- When one mole of KCl dissolved in the solution then due to dissociation
KCl  K + + Cl -
The number of partical increases , hence the colligative properties like
osmotic pressure , elevation in boling point etc increases
1/2-Marks
2) Association of solute molecule :-
* When a solute like an non-electrolyte is dissolved in a non -polar solvents like benzene ,
* It undergoes association forming a molecules of higher molecular mass
* Hence , the number of the particles in the solution decreases
* Hence the observed value of colligative property becomes ,Lower than the therotical value .
For example :- When one mole of acetic acid dissolved in the benzene then it associated
2CH3 COOH  ( CH3 COOH) 2
The number of partical decreases , hence the colligative properties like 1/2-Marks
osmotic pressure , elevation in boling point etc decreases

Q10. 2.4 x 10-3 Kg of a substance dissolved in 0.120 dm3 of water gave an osmotic
pressure of 1.331 atmosphere at 293 K . calculate the molecular weight of the
substance ?

Ans :- Formula used

W2 R T 1/2-Marks
M2 =
V

2.4 x 10-3 x 0.082 x 293 1/2-Marks

=
1.331 x 0.120

M2 = 361 Kg Mol-1 1 -Marks

Q11 Define Van,t Hoff factors ? 1M

Van’t Hoff Factor :- ( i )

Ans :- The ratio of observed colligative property produced by a given concentration of electrolyte
solution to the property observed for the same concentration of non electrolyte solution
Therefore ,

Observed colligative property of electrolyte solution

i =
Observed colligative property of non- electrolyte solution of same concentration

1-Marks
 Test -3

Q1. Define i) Freezing point ii) Boiling point - 2M

Ans:-
i) Freezing point:-
The temperature at which the vapour pressure of its liquids is equal to the vapour
pressure of the corresponding solid is called freezing point 1-Marks

ii) Boiling point

The temperature at which the vapour pressure of liquid is equal to the atmospheric
pressure is called boliling point of liquid 1-Marks

Q2. Define 1) Molal Elevation constant

2) Molal Depression constant 2M
1) Molal Elevation constant
Ans:- The elevation of boiling point when the molarity of the solution is unity
Or
The elevation of boiling point of solution produced by one mole of solute
is dissolved in 1 kg ( 1000g ) of solvent
 Tb = Kb m . 1-Marks

 Tb = Kb If m =1 molal

2) Molal Depression constant

The depression in freezing when the molarity of the solution is unity

Or
The depression in freezing solution produced by one gram mole of solute
is dissolved in 1 kg ( 1000g ) of solvent

  Tf = Kf m .

 Tf = Kf If m =1 molal 1-Marks

Q. Derive the relation between elevation of boiling point & molar mass of the solute
Ans :-
1. The elevation boiling point of dilute solution is diractly propertional to the 1/2-Marks
molal concentration of solution .
2. If  Tb is the elevation in the boiling point of a solution of molal concentration m then ,
 Tb  m

 Tb = Kb m .........................................................1 1/2-Marks
Where
Kb = Boiling point elevation constant or Molal elevation constant ( Ebulliscopic
constant )
If m =1 molal
 Tb = Kb

3. The molality of solution is given by

mass of solute in Kg
m= 1/2-Marks
molar mass of solute in kg x mass of solvent in Kg
W2
m=
- M2. W1
4. Put the value in equation no -1
W2
 Tb = Kb
M2. W1

Kb W2
There fore M2 = 1/2-Marks
 Tb W1
Q4. Define 1) Elevation of boling point ( Tb)
2) Depression in freezing point ( Tf) 2M

1) Elevation of boling point ( Tb)

Ans :- The difference between boiling point of solution & that of the pure solvent at any
constant pressure is known as elevation of boiling point . 1-Marks

It is denoted by  Tb

Elevation of boiling point = B.pt of the solution - B.pt of the pure solvent

 Tb = T - To

2) Depression in freezing point ( Tf)

Ans :- The difference between the freezing points of pure solvent ( To) and the freezing
points of pure solution ( T) is called depression in freezing point 1-Marks

It is denoted by  Tf
Depression in freezing point = freezing point of the pure solvent - freezing point of the solution then

 Tf = To - T

Q5. Derive the relation between Depression of freezing point & molar mass of the
solute 2M

Ans :- Let ‘W 2’ be the weight of solute dissilved in ‘W 1’ g of solvent ,

Let ‘M2’ be the molecular weight of solute 1/2-Marks

The molality of solution is given by

Moles of solute x 1000

molality =
Molar mass of solute X weight of solvent in grams
W2 x 1000
molality = 1/2-Marks
M2 x W 1
The relation between depression in freezing point & molality of solution is ,

Tf Kf x Molality 1/2-Marks

Where Kf is called molar Molal Depression constant or Cryoscopic Consatnt

Tf Kf x W2 x 1000

M2 x W 1

Tf  Kf x W2 x 1000

M2 x W 1
This relation can be used to calculate molecualr weight of solute from depression of
freezing point
Kf x W2 x 1000
M2  1/2-Marks
Tb x W1
Where
M2 = Molar mass of solute
W2 = Mass of solute
W1 = Mass of solvent
Kf = Molal Depression constant
Tb = Depression of freezing point
Q6. 1.35 g of a substance when dissolved in 55g of acetic acid produced a
depression of 0.618 0C in a freezing point . Calculate the molar mass of the
dissolved substance ( Kf = 3.865 K kg mol-1) 2M

Ans :- Formula used

Kf x W2 x 1000
M2  1/2-Marks
Tb x W1

3.865 x 1.35 x 1000

M2 
55 x 0.618 1/2-Marks

M2  153. 5 g mol -1

Ans :- Molar mass of the substance = 153. 5 g mol -1 1-Marks

 Model Answer bank
Test -2

Q1. State & Explain Henry’s law

Ans:- Statement :-
“ The solubility of gas in liquid at constanttemperature is propertional to the
pressure of the gas above the solution “ 1-Marks

S  P

S = KP

where S = Solubility of gas ( mol/dm3)

P = Pressure of gas in atmosphere
K = Henry’s constant 1-Marks

Thus solubility of gases increases with increase of pressure

Q2. Define 1) Mole Fraction 2) Molarity

1) Mole Fraction:-
The ratio of the number of moles of that component present in the solution to the total
number of moles of all the components of the solution is called mole fraction

It is represented by X 1-Marks

Expression :-

Number of moles of that component

Mole fraction of the components X =
total number of moles of all the components of the solution

2) Molarity :-
Defination :- The number of moles of solute per dm3 of solution is called molarity

It is represented by ‘M’

Unit :- Unit of molarity is mol. dm-3 ( moles per dm3 )

Expression :-

Number of mole of solute in Kg

Molality (m) = 1-Marks
mass of solvent in Kg

Q3.Define Colligative properties . Give examples .?

Ans :- 1-Marks
Defination :-
A properties of the solutions that depends only on the the number of solute
partical in the solution & not on the nature of the solute particals is called
colligative properties

Ex:- 1) Relative Lowering of vapour pressure

2) Elevation of Boiling point
3) Depression of freezing point
4) Osmotic pressure
1-Marks
Q4. Derive the relationship between relative lowering of vapour pressure & molar mass
of solute
Ans :-
1. Let W2 g of solute of molar mass M2 be dissolved in W1 g of solvent of molar mass M1 .

W1
Hence number of mole of solvent n1 =
M1

W2
Number of mole of solute n2 = 1/2 Marks
M2

n2
The mole fraction of solute, x2 is given by x2 = n1 +n2
W2/M2
= 1/2 Marks
W1/M1+ W2/M2

p p01 - p W2/M2
Also from relative lowering vapour pressure equation = = = x2 =
p 0
1 p0 1 W1/M1+ W2/M2

But for dilute solution n1 >> n2 Hence n2 be neglected in comparison with n1 1/2 Marks
Thus ,
p p01 - p W 2 x M1
= = = x2 =
p01 p0 1 M2 x W 1

p01 - p W 2 x M1
=
p0 1 M2 x W 1
1/2 Marks
Where p01 = Vapour pressure of pure solvent
p = Vapour pressure of solution
W1 = Weight of solvent
W2 = Weight of solute
M1 = Molar mass of solvent
M2 = Molar mass of solute

Q5. Define 1) Vapour pressure 2) Relative lowering vapour pressure

1) Vapour pressure:-
Ans ;-
Defination :-
The pressure exerted by the vapours when the vapour & the liquid phases are in
equilibrium at a particular temperature , is called vapour pressure of the liquid
1-Marks

2) Relative lowering vapour pressure

Ans :- The ratio of lowering of vapour pressure of pure solvent from solution to the
vapour pressure of pure solvent is called relative lowering of vapour pressure
of pure solvent
P
Relative Lowering vapour pressure =
P1o
P1 o - P
=
P1o

Where :- P1o = Vapour pressure of pure solvent

P = Vapour pressure of Solution 1-Marks
Q6. State and Explain Raoult’s law of Vapour pressure contaning volatile solute ?

Statment: -
The partial vapour pressure of any volatile component of a solution is the product
of vapour pressure of that pure component & the molefraction of the component of
a solution is called Raoults law . 1 Marks

Explanation :-
1. Consider the solution contaning two volatile components A1 & A2 with mole
fractions x1 & x2 respectively
2. let p01 & p02 be the vapour pressures of the pure components A1 & A2 respectively
3. According to Raoults law 1/2 Marks
The partial pressures p1 & p2 of the two components in the given solution
are given by
p1 = p01 x1 & p2 = p02 x2 1/2 Marks

Thus total vapour pressure of solution (pT ) contaning two volatile component is

The sum of the partial vapour pressure of the two components

Hence
pT = p1 + p2 1/2 Marks

pT = p01 x1 + p02 x2 .............................1

For binary solution x1 + x2 =1 1/2 Marks

x2 = 1 - x1
Hence pT = p01 x1 + p02 (1 - x1 )
pT = p01 x1 + p02 - p02x1

pT = p02 + ( p01 - p02 ) x1 1/2 Marks

Q7. The vapour pressure of pure benzene at certain temperature is 640mmHg .

A non- volatile solute of mass 2.175 x 10 -3 Kg is added to 39.0x 10-3 kg of benzene
The vapour pressure of the solution is 600mm Hg . What is the molar mass of
solute ? ( Given atomic masses C = 12 , H =1)
Ans : -
Molar mass of benzene ( C6H 6 ) = 78 x 10 -3 kg 1/2 Marks
Formula used
p01 - p W 2 x M1
=
p0 1 M2 x W 1

W2 x M1 x p0 1
M2 =
W1 x p01 - p 1/2 Marks

2.175 x 10 -3
X 78 x 10 -3
M2 = 1/2 Marks
39.0x 10-3 x 40

M2 = 69.6 x 10 -3 Kg mol-1

Molar mass of solute is M2 = 69.6 g mol-1 1/2 Marks

Q1. Write two differences between an ideal solution and a non-ideal solution.

Q2. What is meant by positive deviations from Raoult’s law? Give an example.
What is the sign of mixH for positive deviation?
Answer:
Those solutions in which force of attraction between A—B is less
than A—A and B—B, shows positive deviation from Raoult’s law,
e.g. ethanol and water show positive deviation from Raoult’s law.

mixH =+ve

Q3. What is meant by negative deviation from Raoult’s law? Give an example.
What is the sign of AmixH for negative deviation?
Answer: A solution is said to deviate negatively from Raoult’s law if the forces
of attraction between A—B are more than A—A and B—B,
e.g. CHCl3and CH3COCH3.
mixH =-ve

Q4. Define azeotropes. What type of azeotrope is formed by positive deviation

from Raoult’s law? Give an example.
Answer:
Azeotropes are the constant boiling mixtures which distill out unchanged
in their composition.Minimum boiling azeotropes are formed by the
solutions showing positive deviation,
e.g. cyclohexane and ethanol.

Q5. Define azeotropes. What type of azeotrope is formed by negative

deviation from Raoult’s law. Give an example.
Answer: Azeotropes are the constant boiling mixtures which distill out
unchanged in their composition. Maximum boiling azeotropes
are formed by the solution showing negative deviation,
e.g. H2O and HCl.

Q6. Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’?

Answer: Azeotropes are the constant boiling mixtures which distill
out unchanged in their composition.

Q7. What type of intermolecular attractive interaction exists in the pair of

methanol
and acetone?
Answer: H-bonding exists between methanol and acetone. It also has dipole-
dipole attraction.

Q8. . Define an ideal solution and write one of its characteristics.

Answer: Ideal solution is a solution which follows Raoult’s law.
Characteristics:
Q8 State Henry’s law. What is the effect of temperature on the solubility of a
gas in a liquid?
Answer: Henry’s Law: The mole fraction of a gas in a solution
(solubility of gas in liquid) is directly proportional to partial pressure
of gas over solution.

where KH is Henry’s law constant, %gas is mole fraction of gas and ^gas is
the partial pressure of gas.
The solubility of a gas decreases with increase in temperature.

Q9. . What type of deviation is shown by a mixture of ethanol and acetone?

What type of azeotrope is formed by mixing ethanol and acetone?
Answer: Positive deviation is shown by a mixture of ethanol and acetone.
Minimum boiling azeotropes are formed.

Q10. . (i) Why is an increase in temperature observed on mixing chloroform and

acetone?
(ii) Why does sodium chloride solution freeze at a lower temperature than water?
Answer: (i) It is because force of attraction between chloroform and acetone is more
than the force of attraction between CHC13—CHC13 or Acetone-Acetone.
Therefore, AH = -ve, i.e. exothermic process, therefore, there is increase
in temperature.

(ii) It is because when NaCl is added, vapour pressure of solution becomes less
and at a lower temperature, vapour pressure of solid and solution will
become equal, i.e. freezing point is lowered.
72. State Henry’s law and mention its two important applications.
Answer: Henry’s Law: It states that the partial vapour pressure of gas in vapour phase
is directly proportional to the mole fraction of the gas in the solution.
Applications of Henry’s Law:
(i) To minimise the painful effects of deep sea divers, oxygen diluted with less soluble
helium gas is used as breathing gas instead of air because nitrogen is more soluble in
blood and cause pain or bends.
(ii) To increase the solubility of COs in soft drinks and soda water, the bottle is sealed
under high pressure.
The Chapter 2 - Solutions of NCERT Class 12th Chemistry contains the
following subtopics:
1. Types of Solutions
2. Expressing Concentration of Solutions
3. Solubility
4. Vapour Pressure of Liquid Solutions
5. Ideal and Non-ideal Solutions
6. Colligative Properties and Determination of Molar Mass
7. Abnormal Molar Masses

Q1. Define the term solution. How many types of solutions are formed ?
Write briefly about each type with an example.
Ans :- Solution :- A Solution is a homogeneous mixture of two or more chemically
non-reacting substances.

Types of solution: There are nine types of solution.

Types of Solution Examples
1. Gaseous solutions:
(a) Gas in gas: Air, mixture of O2 and N2, etc.
(b) Liquid in gas: Water vapour
(c) Solid in gas: Camphor vapours in N2 gas, smoke etc.
2. Liquid solutions
(a) Gas in liquid: CO2 dissolved in water (aerated water), and O2 dissolved in water, etc.
(b) Liquid in liquid: Ethanol dissolved in water, etc.
(c) Solid in liquid: Sugar dissolved in water, saline water, etc.
3. Solid Solutions
(a) Gas in solid: Solution of hydrogen in palladium
(b) Liquid in solid:Amalgams,eg.Na-Hg.
(c) Solid in solid: Gold ornaments (Cu/Ag with Au).

Q2. Give an example of a solid solution in which the solute is a gas.

Ans :- Solution of hydrogen in palladium and dissolved gases in minerals

Question 11:
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Solution 11:
When gases are dissolved in water, it is accompanied by a release of heat energy, i.e.,
process is
exothermic. When the temperature is increased, according to Lechatlier’s Principle,
the
equilibrium shifts in backward direction, and thus gases becomes less soluble in
liquids.

Question 12:
State Henry’s law and mention some important applications.
Solution 12:
The effect of pressure on the solubility of a gas in a liquid is governed by Henry’s Law.
It states
that the solubility of a gas in a liquid at a given temperature is directly proportional to
the partial
pressure of the gas. Mathematically, P = KHX where P is the partial pressure of the
gas; and X is
the mole fraction of the gas in the solution and KH is Henry’s Law constant.
Applications of Henry’s law:
(i) In the production of carbonated beverages (as solubility of CO2 increases at high
pressure)
(ii) In the deep sea diving.
(iii) For climbers or people living at high altitudes, where low blood O2 causes climbers
to
become weak and make them unable to think clearly