Repeaters 2025-Jee Main - Chemistry(Study Material)

02
CHAPTER -
SOLUTIONS

SYNOPSIS
Solution: A homogeneous mixture of two or more non-reacting substances. Binary solution contains only
two components
Solvent is the component that is present in relatively higher amount and the other component is solute. If
the two components are in different phases (physical states), the component which is in the same physical
state as that of the solution is considered as the solvent (even though the amount is small).
Considering the three states of matter, 32 = 9 types of binary solutions are possible.
Solvent Solute Examples
I. Gaseous solutions Gas Gas Mixtures of O2 and N2
Gas Liquid CHCl3 in N2 gas
Gas Solid Camphor in N2 gas; I2 vapour in air
II. Liquid solutions Liquid Gas O2 dissolved in water
Liquid liquid Ethanol in water
Liquid Solid Sugar in water
III. Solid solutions Solid Gas N2 in Ti, H2 in Pd
Solid Liquid Hg in gold, Amalgam of Hg with Na
Solid Solid Alloys like brass, bronze etc
(Substitutional solid solution)
Expressing concentrations of solutions
(i) Mass percentage (w/w) : Mass % of a component =
Mass of the component in the solution 100
Total mass of the solution
Volume of the component 100
(ii) Volume percentage (v/v): vol% of a component =
Total volume of the solution
Solutions containing liquids are commonly expressed in this unit
(iii) Mass by volume percentage (w/v) : This is commonly used in pharmacy
Mass of the component 100mL
Mass by volume % of a component =
Vol : of the solution in mL
(i.e mass of the solute dissolved in 100ml of the solution)

Number of parts of the component 106

(iv) Parts per million (ppm) =
Total no : of parts of all the components of the solution
ppm can be expressed as mass to mass, volume to volume or mass to volume. ppm unit is used when the
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amount of the solute is in traces. The concentration of pollutant in water or atmosphere is often expressed
in terms of the  g mL–1 or ppm
(v) Mole fraction (Symbol x)
Number of moles of the component
Mole fraction of a component =
Total number of moles of all the components
In binary solutions x1 denotes the mole fraction of solvent and x2 denotes the mole fraction of solute.
If there are several components in the solution, mole fraction of the ith component

ni n
=  i
n1  n 2  n 3  .....n i
 ni
x1 + x2 = 1 in a binary solution
x1 + x2 + x3 = 1, in a ternary solution
(vi) Molarity (M)

Moles of solute
Molarity =
Volume of the solution in litre
i.e molarity (M) is the number of moles of solute per litre of the solution
(One litre = one cubic decimetre i.e 1L = 1dm3)
(vii) Molality (m) It is the no: of moles of the solute per kg of the solvent

number of moles of solute

i.e molality =
mass of solvent in kg
(viii) Formality (F) For ionic substances like NaCl, KCl etc, where the crystal does not contain discrete
molecules, but only ions, formality can be used.

No. of formula weights of the ionic solute

Formality (F) =
Vol. of the solution in litre
(ix) Normality (N) It is the number of gram equivalents of the solute per litre of the solution.

No : of gram equivalent of solute

N
Vol.of the solution in litre

Mass of the component

(x) Mass fraction = Total mass of all the components

Molality, mole fraction and mass fraction are independent of temperature as they do not involve volumes.
But molarity, normality and formality involves volume and they may vary slightly with temperature. For
very accurate work, therefore, molality, mole fraction or mass fraction are preferred.
Important relationS
1. V1 N1 = V2 N2 (used in volumetric titrations and during dilution of solutions)
2. V1N1 + V2N2 = V3 N3 (Two solutions at different normalities of the same solute are mixed together)
3. V1M1 = V2M2  used during dilutions

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 Repeaters 2025-Jee Main - Chemistry(Study Material)

M1V1 M 2 V2
4. 
n1 n 2 (when the two substances are reacting not in equimolar amounts.
For eg: n1A + n2B  Products.
But when n1 = n2, M1V1 = M2V2 can be used for volumetric calculations.
Solubility: Solubility of a substance is the maximum amount that can be dissolved in a specified amount
of solvent at a specified temperature. It depends on the nature of the solute and solvent as well as
temperature and pressure.
Generally polar solutes dissolve in polar solvents while non polar solutes dissolve in nonpolar solvents
– “like dissolves like”.
If the dissolution of a solid in a solvent is endothermic i.e H sol  0 , solubility should increase with
temperature. Solutes like KCl, NH4Cl, KNO3 etc. dissolves in water by the absorption of heat. But if the
dissolution of a solute in a solvent is exothermic i.e  Hsoln < 0 (heat liberated), solubility should decrease
with rise in temperature.
Eg: Co2(SO4)3, Na2CO3.H2O, Li2 CO3 dissolves in water with liberation of heat. Pressure has no significant
effect on the solubilities of solids and liquids. (Reason: - solids and liquids are incompressible)
Solubility of gases in liquids
Solubility of gases depends on the nature of the gas and the solvent. NH3, HCl etc are highly soluble in
water, while N2, O2, H2 etc are less soluble at the same temperature and pressure.
Henry’s law: The mass of a gas dissolved in a given volume of the solvent at a particular temperature is
directly proportional to the pressure of the gas at equilibrium with the liquid.
m = KH. P
(KH = Henry’s law constant, unit of KH in this expression will be(mass) (pressure)–1)
m = mass of the gas; P = Pressure
Instead of mass, we can use mols litre–1 also, then the unit of KH should be mol litre–1 atm–1.
Henry’s law is also expressed as x = KH. P where x is the mole fraction of the gas in the solution and P is
the partial pressure of gas in equilibrium with the liquid (Here unit of KH is atm–1 or bar–1)
The most commonly used form of Henry’s law: States that the partial pressure of the gas in the
vapour phase (P) is proportional to the mole fraction (x) of the gas in the solution.
P = KH . x (Here the unit of KH will be same as that of pressure)
This form of equation is used when different gases are dissolving in the same solvent. Higher the value of
KH at a given pressure, the lower is the solubility of the gas in the liquid. For many gases KH value
increases with increasing temperature, indicating that the solubility of gases decrease with temperature.
NB:- Henry’s law is not valid, when the gas is highly soluble in a solvent and it enters into chemical
combination and dissociation for eg: NH3, HCl etc in H2O.
NH3 + H2O  NH4 OH  NH4+ + OH–
HCl + H2O  H3O+ + Cl–
Henry’s law and Scuba divers : During deep sea diving, some nitrogen from the oxygen air mixture will
dissolve in the blood (due to high pressure). But when the diver comes to the surface, due to the decrease
of pressure, dissolved N2 forms bubbles in blood and this causes severe pain to the diver and this
condition is called ‘bends’ or caisson disease. To decrease the solubility of N2 in blood vessels, the
oxygen - nitrogen mixture for breathing is diluted with Helium (11.7% He, 56.2% N2 and 32.1% O2)
Vapour pressure of liquid solutions: For a binary solution of two volatile liquids, let P1 and P2 be the partial
vapour pressures and x1 and x2 their mole fractions in the liquid mixture, then Raoult’s law states that P1 =
x1 P10 and P2 = x2 P20. P10 and P20 are the vapour pressures of the pure components 1 and 2 at the same
temperature at which P1 and P2 are measured. According to Dalton’s law of partial pressures, total vapour
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pressure above the solution is P1 + P2 i.e Ptotal = P1 + P2. Substituting the values of P1 and P2
Ptotal = x1P10 + x2P20 = (1 –x2) P10 + x2P20

or Ptotal = P10 + (P20 – P10) x2   A 

(OR) Ptotal = x1P10 + (1 – x1) P20
= P20 + (P10 – P20) x1  (B)
From equations A and B it is clear that total vap: pressure over the solution is linearly dependent on the
mole fraction of any one component. When a solution obeys Raoult’s law over the entire range of
concentration, it is called an ideal solution, otherwise it is non ideal. Examples of ideal solution:
1. n hexane + n heptane
2. Ethyl chloride + ethyl bromide
3. Chlorobenzene + bromobenzene
4. Benzene + Toluene
For ideal solutions Vmix  0 and H mix  0 i.e total vol: does not change on mixing. Similarly no heat
change on mixing.
In pure liquids A and B there are only A – A interactions and B – B interactions. But on mixing them there
will be additional A– B interactions. If the A – B interactions are nearly the same as A – A and B – B
interactions, the solution will be ideal.
Non ideal solution : Non ideal solutions may exhibit +ve or –ve deviations from Raoult’s law. If the total
vapour pressure is higher than that predicted by Raoult’s law, it is +ve deviation. This is due to weaker
A – B interactions compared to more strong A – A and B – B interactions. For such non-ideal systems
exhibiting +ve deviations from Raoult’s law, Vmix  0 and  Hmix is +ve or greater than zero
Example of systems exhibiting +ve deviations
1. Acetone + Ethanol
2. Acetone + CS2
3. Water + ethanol
4. Water + methanol
5. CCl4 + Toluene
6. CCl4 + CHCl3
7. Acetone + Benzene
8. Cyclohexane + Ethanol
9. CCl4 + methanol
10. Water + Propanol
If the total vapour pressure is less than that predicted by Raoult’s law, we have –ve deviations. Here A –B
interactions are stranger than A – A or B – B attractions. Here  Vmix < 0 and  Hmix < 0 (H
is –ve).
Examples for –ve deviations
1. Acetone + CHCl3 2. Acetone + Aniline
3. Phenol + Aniline 4. Methanol + Acetic acid
5. H2O + HCl 6. H2O + HNO3
7. H2O + H2SO4 8. CHCl3 + diethyl ether
9. CHCl3 + Benzene 10. Acetic acid + Pyridine

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 Repeaters 2025-Jee Main - Chemistry(Study Material)

Azeotropes : Several non ideal solutions form azeotropes or constant boiling mixture. These are solutions
which cannot be separated completely into pure components by fractional distillation. Example. Ethyl
alcohol water mixture shows +ve deviation and it forms an Azeotropic mixture with minimum boiling point.
95.5% ethanol and 4.5% water by weight is a constant boiling mixture. i.e it will distill without any change
in composition. i.e the vapour will also have the same composition.
Boiling point of H2O is 373K
B.P of Ethanol is 351.3K
B.P of Azeotrope is 351.10K
Non ideal systems with –ve deviations will form Azeotropes of maximum boiling point. A mixture containing
68% HNO3 and 32% H2O by mass has the boiling point 393.5K
Vapour pressure curves for ideal and non ideal solutions
dotted line for ideal solutions

Raoult’s law is a special case of Henry’s law

According to Raoult’s law, the vapour pressure of a volatile component in a given solution is given by Pi =
xi Pi0. In the solution of a gas in a liquid, one of the components is too volatile that it exists as a gas and for
gases, dissolved in a liquid solvent we have Henry’s law P = KH x [This is the most commonly used form of
Henry’s law].
If we compare those two equations, only the proportionality constant is differing. Thus, Raoult’s law
becomes a special case of Henry’s law in which KH becomes equal to Pi0.
Composition of the vapour phase in ideal solutions

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If y1 and y2 are the mole fractions of the components 1 and 2 respectively in the vapour phase, then

P1 P2
P1 = y1 Ptotal;  y1 = P and y 2  P
total total

x1P10 x 2 P20
we know P1 = x1 P and P2 = x2P
0 0  y1  y
and 2 
1 2 ; x1P10  x 2 P2 0 x1P10  x 2 P2 0
Vapour pressure of solutions of solids in liquids
When a solute like, urea, glucose, cane sugar or sodium chloride is dissolved in water, vapour pressure
of the resulting solution is less than that of the pure solvent (water) at the same temperature. Similarly I2 or
sulphur dissolved in CS2(liquid solvent) lowers the vapour pressure of the solvent. The decrease in the
vapour pressure of water by adding 1 mole of sucrose to 1 kg of water is nearly similar to that produced by
adding 1 mole of urea to the same quantity of water at the same temperature.
Raoult’s law in its general form
For any solution, the partial vapour pressure of each volatile component in the solution is directly
proportional to its mole fraction. In a binary solution containing a non volatile solute, only the solvent
molecules are present in the vapour phase. Let P1 be the vapour pressure of the solvent and x1 its mole
fraction in the solution. Let P10 be its vapour pressure in the pure state. According to Raoult’s law.
P1  x1 or P1 = x1P10 (P10 is the proportionality constant)
But x1 = 1 – x2
 P1 = (1 – x2) P10 ie P1 = P10 – x2P10

P10  P1
 P10 –P1 = x2 P10 or  x2
P10
P10 is the vapour pressure of the pure solvent
P1 is the vapour pressure of the solution at the same temperature. Since the solute is non volatile, P1 can
be replaced by Ps

P10  Ps
  x 2 ( P = Vap. pressure of solution)
P10 s

i.e Relative lowering of vapour pressure in a solution containing a non volatile solute is equal to the mole
fraction of the solute. This is the form of the Raoult’s law generally used for dealing with solutions of
solids in liquids.
Lowering of vapour pressure = P10 – Ps

P10  Ps
 Relative lowering  P10

n2
x2 = mole fraction of solute = n2 = no. of moles of solute, n1 = no. of moles of solvent
n1  n 2

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 Repeaters 2025-Jee Main - Chemistry(Study Material)

P10  Ps n2
 
P10
n1  n 2
But in a dilute solution n1 >> n2  n2 in the denominator can be neglected so that

P10  Ps n 2 w2 W1
 but n 2  w2 = mass of solute ; M2 = Molecular mass of solute and n1 
P10 n1 M2 M1 , W 1 =
mass of solvent; M1 = Molecular mass of solvent
This expression can be used to find out the molecular mass of unknown solute using Raoult’s law ( i.e
from lowering of vapour pressure)
Colligative properties and determination of molar mass or molecular mass of solute
When a non volatile solute is added to a volatile solvent, the escaping tendency of solvent molecules to
the vapour phase is decreased and it leads to a lowering of vapour pressure. There are many properties
of solutions which are connected with this decrease of vapour pressure. These are (1) Relative lowering
of vapour pressure of solvent (2) Depression of freezing point of the solvent (3) Elevation of boiling point
of the solvent and (4) Osmotic pressure of the solution. Since these properties are all bound together
through their common origin, they are called colligative properties. Common origin is the lower value of
chemical potential of the solvent in the solution, compared to its chem: potential in the pure solvent. All the
four colligative properties depend on the number of solute particles irrespective of their nature relative to
the total no: of particles present in the solution.
Elevation in boiling point (  Tb) : Boiling point is the temperature at which vapour pressure becomes equal
to the atmospheric pressure (1 atm or 1.013 bar or 760 torr or 760mm Hg). Since vapour pressure of a
solution is less than that of pure solvent, the boiling point of solution is greater than that of pure solvent.

 Tb is proportional to the molality (m) of the solution

i.e Tb  m or Tb  K b  m

Kb is called molal elevation constant (Ebullioscopic constant)

when m = 1, Tb  K b

Therefore Kb is defined as the elevation in boiling point of a 1 molal solution.

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The unit of Kb = K kg mol–1 or K/m (Kelvin per molality)

Kb is characteristic of the solvent. It has nothing to do with solute
Solvent Kb value
Water 0.52
Benzene 2.67
Methanol 0.80
CHCl3 3.88
Ethanol 1.20
CCl4 5.03

Tb  K b  molality of solution

 W2 
W2 1000  
 Tb  K b   But molality =
 M 2   no.of moles of solute
M2 W1  W1  mass of solvant in Kg
 
 1000 

K b  W2  100
 M2  Here W 2 = mass in gms of solute
Tb  W1
W 1 = mass in gms of solvent
Kb is related to heat of vaporization of the solvent

RTb 2
Kb  where Tb = boiling point of pure solvent in Kelvin
1000  v
R = gas constant

RTb 2 .M1
  v = latent heat of vapourisation per gm of solvent
1000H vap

H vap
But  v 
M1

H vap = molar heat of vapourisation of solvent

M1 = Molecular mass of solvent (or molar mass)

Depression in freezing point  Tf 

The freezing point of a solution is less than that of pure solvent Tf  Tf 0  Tf

 Tf  molality or  Tf = kf  m (Kf = cryosopic constant)

Molecular mass of solute can be calculated using the expression,

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 Repeaters 2025-Jee Main - Chemistry(Study Material)

K f  W2  1000
M2 
Tf  W1
Molal depression constant or cryoscopic constant for some solvents.
Solvent Kf(K kg mol–1)
Water 1.86
C6H6(Benzene) 5.12
Phenol 7.27
Acetic acid 3.90
Naphthalene 6.90
Cyclohexane 20.0
Camphor 39.70
Cause of depression in F.P
The lowering of vapour pressure of a solution causes a lowering of the freezing point compared to that of
the pure solvent. The freezing point of a substance is the temperature at which the vapour pressure of the
substance in its liquid phase is equal to that of the vapour pressure in the solid phase. A solution will
freeze when its vapour pressure equals the vapour pressure of the pure solid solvent (see diagram)

Kf is also characteristic of the solvent. It is related to the latent heat of fusion of the solid solvent.
RTf 2
Kf  where
1000 f
Tf = F.P of the solvent (In Kelvin)
R = gas constant
 f = heat of fusion per gram of solvent
(If R is in Joules  f also should be in Joules)

H f
but  f 
M1 where M1 = Molecular mass of solvent and H f is the molar heat of fusion

RTf 2 , M1
 Kf 
H fus 1000

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Experimental methods for Tf measurement are Beckmann method and Rast method.
For measuring Tb , Landsberger method and Cottrell’s method are used.
Osmosis and Osmotic pressure
The migration of solvent particles from pure solvent or dilute solution towards the concentrated solution
when they are separated by a semipermeable membrane is called osmosis.
Osmotic pressure (  ) is the excess hydrostatic pressure which builds up as a result of osmosis (OR) The
excess pressure that must be applied to a solution to prevent osmosis.
Example of Semipermeable membranes
(1) Natural :- (1) Skin of an egg (2) Pig’s bladder
(3) Membrane of plant cells (4) RBC membrane.
(2) Artificial:- (1) Cu2[Fe(CN)6] (2) Ca3(PO4)2
Reverse osmosis:– If a pressure greater than osmotic pressure is applied to the solution side, the solvent
particles migrate from solution to solvent side, through the membrane. Desalination of sea water is by this
technique.
Isotonic solutions:- Solutions having the same osmotic pressure. They are also called Isopiestic solutions.
A 0.9% aq: solution: of NaCl is isotonic with blood: Average osmotic pressure of blood is about 7.6 atm.
Hypertonic and hypotonic solutions:- When comparing the osmotic pressure of two solutions, the one
with lower value of O.P is hypotonic and the other with higher O.P is called hypertonic solutions.
Vant Hoff’s eqn: for O.P :  = CRT, where C = molarity of the solution.
n
C n = no: of moles of solute; V = Vol: of the solution in litres
V
n
  = CRT or   RT
V
Here  is in atmospheres, when R is in litre atmosphere and V is in litres

nRT WB RT
 can be written as   
V MB V

WB RT
or M B  . Hence mol: wt. of solute (or molecular mass) can be determined.
V
Osmotic pressure method is suitable to find out the molecular mass of macromolecules (or polymers)
Abnormal colligative properties -: For electrolytes like NaCl, KBr etc which dissociates in aqueous
solution, the values of the colligative properties will be higher than the theoretical values. Acetic acid,
benzoic acid etc, which associates in benzene, give a lower value of the colligative properties. Hence
molecular masses of such solutes determined by colligative property measurements will be abnormal
values.

KCl  K   aq   Cl  aq  
 Dissociation increases the number of Particles .
CaCl 2 aq   Ca 2   aq   2Cl   aq  

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 Repeaters 2025-Jee Main - Chemistry(Study Material)

O H ------ O
C CH3
2CH3COOH CH3 C
O H O

dimer
Association decreases the no: of particles
1
molecular mass 
colligative property
Molecular mass of KCl obtained by colligative property method is 40.5 (actual mol. mass = 74.5). The
value of mol. mass obtained shows that KCl is not completely dissociated. Had there been 100%
1
dissociation, the molecular mass obtained would have ben  74.5   37.25 . The mol: mass of acetic
2
acid using benzene solvent is 120. (Actual mol. mass = 60). This shows that acetic acid exists as a dimer
in benzene.
Van’t Hoff factor (i) :- To compensate for the variation in colligative properties, a parameter ‘i’ called Van’t
Hoff factor is used in the equation for colligative properties.

P 0  Ps
 ix 2 Tb  i K b  m Tf  i K f  m  = i CRT
P0
In the case of association i < 1 and for dissociation i > 1 when there is dissociation, i = 1 +  (n–1)
Where  is the degree of dissociation and ‘n’ is the no. of particles (ions) obtainable theoretically from
one molecule. For KCl, n = 2, for CaCl2 n = 3 etc

1 
When there is association, i = 1 +    1
n 
Here  is the degree of association and n is the no: of molecules associated together. For acetic acid,
benzoic acid etc in benzene n = 2 since dimer is formed.

i 1  i  1 n  i 1 
Degree of association    ; Degree of dissociation    
 1  1  n   n 1 
  1
n 

Normal molecular mass M Observed value of colligative property

i  ;i 
Observed molecular mass M 0 Theoretical value of colligative property

No.of particles after association or dissociation

i
No.of particles before association or dissociation

 M theoretic  M obs   M obs  M theor  n

Degree of dissociation   ; Degree of association    
M obs  n  1  M obs  n 1

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PART I - (JEEMAIN )
SECTION - I- Straight objective type questions
1. From the following, select the temperature independent concentration term

1) Molarity 2) Normality 3) Molality 4) Volume percent

2. 5 L of a solution contains 25 mg of CaCO3. What is its concentration in ppm?
1) 25 2) 1 3) 5 4) 2500
3. If 100 cm3 of 0.3 M NaCl and 150 cm3 of 0.1 M CaCl2 solutions are mixed, the molarity of Cl– in the final
mixture will be:
1) 0.24 2) 0.18 3) 0.5 4) 0.6
4. Which of the following gases has the highest Henry’s constant (KH/kbar) for dissolution in water at 293
K?
1) Ar 2) CO2 3) HCHO 4) CH4
5. Which of the following statements are CORRECT with respect to Henry’s law constant (KH)?
I) KH is a function of nature of the gas
II) KH of N2 is 76.48 kbar at 303 K and 88.84 kbar at 293 K
III) Highly soluble gases have low KH (in kbar units) at a given temperature and pressure
IV) KH has a unit of pressure or (pressure)–1
1) I, IV only 2) I, II, III only 3) II, III, IV only 4) I, III, IV only
6. The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above the liquid
solution. The molar solubility of oxygen in water is ........... × 10–5 mol dm–3. [Given : Henry’s law constant
KH = 8.0 × 104 kPa for O2, Density of water with dissolved oxygen = 1.0 kgdm–3]
1) 1389 2) 1289 3) 1487 4) 1287
7. Vapour pressure of liquids depends on
1) Surface area and volume of liquid 2) Temperature and volume of liquid
3) Temperature only 4) Temperature and nature of liquid
8. Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapour pressure of
pure A and pure B are 7×103 Pa and 12×103 Pa, respectively. The vapour pressure of a solution of A
and B containing 40 mol percent of A at this temperature is:
1) 1.9 × 104 Pa 2) 104 Pa 3) 9.5 × 103 Pa 4) 103 Pa
9. At a given temperature, total vapour pressure (in torr) of a mixture of volatile components A and B is
given by PTotal  120  75X B (XB is the mole fraction of B in the liquid phase). The vapour pressure of
pure A and B (in torr) respectively are
1) 120, 75 2) 120, 195 3) 120, 45 4) 75, 45

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 Repeaters 2025-Jee Main - Chemistry(Study Material)

10. Which of the following correctly represents the changes in thermodynamic properties during the formation
of an ideal binary solution?

1)  H mix  0,  G mix  0, T  S mix  0

2) H mix  0, Gmix  ve, T Smix  ve

3) H mix  0, Gmix  ve, T Smix  ve

4) H mix  ve, Gmix  0, T Smix  ve

11. Assertion : Both enthalpy of mixing and volume of mixing are negative for a binary mixture of aniline
and phenol.
Reason : The intermolecular forces of attraction between aniline and phenol is stronger than aniline-
aniline or phenol-phenol interactions.
In the light of the above statements choose the correct option:
1) Both assertion and reason are true and the reason is the correct expalantion of the assertion
2) Both assertion and reason are true but reason is not the correct expalantion of the assertion
3) Assertion is true but reason is false
4) Assertion is false but reason is true
12. Statement-I : Raoult’s law and Dalton’s law can be used to predict the vapour pressure of a liquid-liquid
solution
Statement-II : Raoult’s law and Dalton’s law can be used to predict the composition of vapour phase of
a liquid-liquid solution.
In the light of the above statements choose the correct option
1) Both statement-I and statement-II are true
2) Statement-I is true, but statement-II is false
3) Both statement-I and statement-II are false
4) Statement-I is false but statement-II is true
13. Match the following
Column-I Column-II
A) n-hexane + n-heptane P) Ideal solution
B) Nitric acid + water Q) Maximum boiling azeotrope
C) Chlorobenzene + bromobenzene R) Will not follow Raoult’s law
D) Ethanol + water S) Minimum boiling azeotrope
1) A-P, B-QR, C-P, D-RS

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2) A-PR, B-QR, C-PR, D-RS

3) A-P, B-RS, C-P, D-QR
4) A-PR, B-RS, C-PR, D-QR
14. The colligative properties of a solution depend on
1) Molar mass of the solute
2) Concentration of non-volatile solute particles
3) Volume of solution
4) Nature of solute
15. Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point
for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is
1) Kb = 0.5 Kf 2) Kb = 2 Kf 3) Kb = 1.5 Kf 4) Kb = Kf
16. In the depression of freezing point experiment, it is found that
i) The vapour pressure of the solution is less than that of pure solvent
ii) The vapour pressure of the solution is more than that of pure solvent
iii) Only solute molecules solidify at the freezing point
iv) Only solvent molecules solidify at the freezing point
1) i, ii 2) ii, iii 3) i, iv 4) i, ii, iii
17. Which of the following solutions will show the maximum vapour pressure at 300 K?

1) 1M NaCl 2) 1M CaCl2 3) 1M AlCl3 4) 1M C12 H 22O11

18. The molecular weight of benzoic acid in benzene as determined by depression in freezing point method
corresponds to
1) dissociation of benzoic acid
2) dimerization of benzoic acid
3) trimerization of benzoic acid
4) normal molar mass

19. The freezing point of equimolal aqueous solution will be the lowest for

1) C6H5COOH 2) Ca  NO3  2 3) Al  NO3 3 4) C6 H12O6

20. Which of the following aqueous solutions are isotonic at a given temperature? [Assume 100%
dissociation/association in the case of dissociating/associating solutes]

i) 0.15 M urea ii) 0.05 M CaCl2 iii) 0.1M MgSO4 iv) 0.15 M glucose

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 Repeaters 2025-Jee Main - Chemistry(Study Material)

1) i & iv 2) ii & iii 3) i, ii & iv 4) ii, iii & iv

SECTION - II

Numerical Type Questions

21. How many of the following is/are examples of solid solutions?

1) glucose dissolved in water 2) Camphor dissolved in nitrogen

3) Copper dissolved in gold 4) Amalgam of mercury with sodium

5) Solution of hydrogen in palladium

22. Vapour pressure of water at 293 K is 17.535 mmHg. The vapour pressure of the solution at 293 K
when 25 g of glucose is dissolved in 450 g of water is .......... mmHg (molar mass of glucose = 180
gmol–1)

23. If K2SO4 is completely dissociated in water then its van’t Hoff factor will be ................

24. The osmotic pressure of 0.01 m aqueous solution of sodium chloride is x × 10–1 times that of 0.01 m
urea solution. The value of x is ............. (assume NaCl is 90% dissociated).

25. A solution of 12.2 g of benzoic acid (molar mass = 122 gmol–1) in 100 g of benzene has a elevation in
boiling point of 1.25°C. The degree of dimerisation of benzoic acid in this solution is ............ (given Kb
of benzene = 2.5 K kg mol–1).

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