CHAPTER

01 SOLUTIONS

1. Define the term “Solution”, write briefly about types of solutions based on the
physical states of solute and solvent. Give two examples for each type.
2. Define : (a) molarity (b) molality (c) molefraction (d) mass percentage (e) PPM write
formulae, units & effect of temperature on them.
3. Write the effect of temperature and pressure on
a) the solubility of solid in liquid
b) the solubility of gas in liquid
4. State “Henry’s law”. Explain its applications (Anoxia & Bends). Write its
limitations.
5. State “Raoult’s law” for a solution of volatile liquids. Write formulae in solution
phase and in vapour phase.
6. Define ideal and non-ideal solutions. Write examples, intermolecular forces, Hmix
and Vmix.
7. Explain positive and negative deviations from Raoult’s law.
(a) Intermolecular forces (b) Hmix
(c) Vmix (d) Examples
8. Define azeotropes. Explain types of Azeotropes with examples.
9. Define colligative properties.
10. Define: (a) Elevation in Boiling point (b) Depression in freezing point
11. Define and write formula and units for
(a) Ebullioscopic constant (b) Cryoscopic constant
12. Define:
(a) Osmosis (b) Osmotic pressure
(c) Reverse osmosis (d) Isotonic solutions
(e) Abnormal molar mass (f) Van’t Hoff factor (i)
(g) Edema
13. What is the similarity between Raoult’s law and Henry’s law?
14. Out of two 0.1 molar solution of glucose and KCl, which one will have a higher
boiling point and why?
15. How is it that measurement of osmotic pressure is more widely used for
determining molar masses of macromolecules than the elevation in boiling point
(or) depression in freezing point of their solutions?

16. When and why is molality preferred over molarity in handling solutions in
chemistry?
17. What is an antifreeze? Give an example.
18. What would be the value of Van’t Hoff factor for a dilute solution of K2SO4 in
water?
19. Why is glycol and water mixture used in car radiators in cold countries?
20. Why are aquatic species more comfortable in cold water in comparison to warm
water?
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21. What type of deviation from Raoult’s law is shown by a solution of chloroform and
acetone, why?
22. Give reasons :
(a) Cooking is faster in pressure cooker than in cooking pan.
(b) Red blood cells (RBC) shrink when placed in saline water but swell in distilled
water.
23. (i) On mixing liquid “X” and liquid “Y”, volume of the resulting solution decreases.
What type of deviation from Raoult’s law is shown by the resulting solution? What
change in temperature would you observe after mixing liquids X & Y.
(OR)
(ii) When two liquids A & B are mixed the solution becomes warmer, which
deviation solution exhibits from Raoult’s law?
*****
Solutions – ANSWERS
1. A solution is a homogeneous mixture of two (or) more pure substances.
Types of solutions based on states of solute and solvent.
S.No. Type of solution Solute Solvent Examples
1. Gaseous solution Gas Gas Mixture oxygen and nitrogen gases
Liquid Gas Chloroform mixed with nitrogen gas
Solid Gas Camphor in N2 gas
2. Liquid solutions Gas Liquid Oxygen dissolved in water
Liquid Liquid Ethanol dissolved in water
Solid Liquid Glucose dissolved in water
3. Solid solutions Gas Solid Solution of hydrogen in palladium
Liquid Solid Amalgam of mercury with sodium
Solid Solid Copper dissolved in gold
2. (a) Molarity : (M)
It is defined as number of moles of solute dissolved in one litre of solution.
n
M
V  in L 
W 1000
M 
GM .W V  in mL 
 Units : mol L1
 Molarity changes with change in temperature as volume changes with change in
temperature.
 Molarity of solution decreases with increase in temperature as volume increases.
(b) Molality (m) :
It is defined as the number of moles of solute present in 1 kg of solvent.
n
m
m  in kg  of solvent
W 1000
m 
G.M .W mass of solvent (in g)

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Units :
mol kg 1
 Molality is independent of temperature.
(c) Molefraction : ()
It may be defined as the ratio of the number of moles of one component to the
total number of moles of all the components present in the solution.
nA nB
A  ; B 
nA  nB nA  nB
Mole fraction has no units and it is independent of temperature.
 A   B  1.
(d) Mass percentage (w/W) % :
It may be defined as mass of solute per 100 g of solution.
mass of solute
mass % of solute = 100
total mass of the solution
(e) Parts per Million (PPM) :
It is the parts of a component per million (106) parts of the solution.
Number of parts of the component
PPM = 106
Total number of parts of all components of the solution
3. (a) (i) For endothermic process, if temperature increases, the solubility of solid in
liquid increases.
(ii) for exothermic process, if temperature increases, the solubility of solid in
liquid decreases.
* Pressure has no effect on solubility of solid in liquid.
(b) (i) The solubility of a gas in liquid is an exothermic process. [H = -ve].
If temperature increases, the solubility of gas in liquid decreases.
1
Solubility of gas in liquid  .
temperature
(ii) If pressure increase, the solubility of gas in liquid also increases.
Solubility of gas in liquid  pressure.
4. Henry’s law :
At constant temperature, the solubility of a gas in liquid is directly proportional
to the pressure of the gas.
Solubility  Pgas
*The partial pressure (P) of a gas in vapour phase is proportional to the
molefraction of gas () in the solution.
Pgas  gas
Pgas = KH gas
Where KH = Henry’s law constant.

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*Applications :
 To increase the solubility of CO2 in soda water and soft drinks, the bottle is sealed
under high pressure.
 Anoxia : At high altitudes, low blood oxygen causes climbers to become weak and
make them unable to think clearly, which are symptoms of a condition known as
anoxia.
 Bends : Scuba divers must cope with high concentrations of dissolved gases while
breathing air at high pressure underwater. Increased pressure increases the
solubility of atmospheric gases in blood. When the divers come towards surface,
the pressure gradually decreases. This releases the dissolved gases and leads to
the formation of bubbles of nitrogen in blood. This blocks capillaries and creates
a medical condition known as “bends”.
* Limitations :
(i) The pressure of gas is not too high and temperature is not too low.
(ii) The gas should not undergo any chemical change.
(iii) The gas should not undergo association or dissociation in the solution.
5. Raoult’s law for solutions of volatile liquids :
It states that for a solution of volatile liquids the partial pressure of each
component of solution is directly proportional to its molefraction.
PA   A PB   B
PA  PAo  A PB  PBo  B
PT  PA  PB
PT  PAo  A  PBo  B
PT  PAo   PBo  PAo   B
In vapour phase : PA  YA PTotal PB  YB PTotal .
6. Ideal solution : A solution is called an ideal solution if it obey’s Raoult’s law over
a wide range of concentration at a specific temperature.
 PT  PAo  A  PBo  B
 fA-B = fA-A = fB-B
 Hmix = 0
 Vmix = 0
Eg :
Mixture of methanol and ethanol.
Mixture of n-hexane and n-heptane.
Mixture of benzene and toluene.
* Non-Ideal solution :
A solution which does not obey Raoult’s law for all concentrations is called a non-
ideal solution.
 PT  PAo  A  PBo  B
 Hmix  0
 Vmix  0

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7. (i) Positive deviation from Raoult’s law:
PT  PAo  A  PBo  B
 fA-B < fA-A and fB-B
 Hmix > 0 i.e. +ve
 Vmix > 0 i.e. +ve
Eg. (i) A mixture of ethyl alcohol and water.
(ii) A mixture of acetone and carbon disulphide.
(iii) A mixture of carbon tetrachloride and benzene.
(iv) A mixture of acetone and benzene.
(ii) Negative deviation from Raoult’s law :
 PT  PAo  A  PBo  B
 fAB > fA-A and fB-B
 Hmix < 0 i.e. -ve
 Vmix < 0 i.e. –ve
Eg. (i) A mixture of HNO3 and water (ii) A mixture of chloroform and acetone
(iii) A mixture of acetic acid and pyridine (iv) A mixture of HCl and water
8. Azeotropes (or) Azeotropic mixtures :
Azeotropes are binary mixtures having the same composition in liquid and vapour
phase and boil at constant temperature.
Types of azeotropes :
(i) Minimum boiling azeotrope :
The non-ideal solution which shows a large positive deviation from Raoult’s law
form minimum boiling azeotrope.
Eg. A mixture of 95.5% ethyl alcohol and 4.5% water by volume.
(ii) Maximum boiling azeotrope :
The solution that shows the large negative deviation from Raoult’s law form
maximum boiling azeotrope.
Eg. A mixture of 68% HNO3 and 32% H2O by mass.
9. The properties which depend on the number of solute particles (molecules, atoms
or ions) but not upon their nature are called colligative properties. The following
are colligative properties.
(i) Relative lowering of vapour pressure of the solvent.
(ii) Elevation of boiling point of the solvent.
(iii) Depression in freezing point of the solvent.
(iv) Osmotic pressure of the solution.
10. Elevation of boiling point :
When a non-volatile solute is added to a volatile solvent, the vapour pressure of
the solvent decreases. In order to make this solution boil, its vapour pressure
must be increased by raising that temperature above the boiling point of the pure
solvent. The difference in the boiling point of solution (Tb) and that of pure solvent
T  is called elevation of boiling point  T  .
b
o
b

Tb  Tb  Tbo  Tb  K b m

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* Depression in freezing point :
Whenever a non-volatile solute is added to the volatile solvent, its vapour pressure
decreases and it would become equal to that of solid solvent at a lower
temperature. The difference in the freezing point of pure solvent T fo  and that of

the solution (Tf) is known as depression in freezing point  T f  .

Tf  Tfo  Tf
Tf  K f  m
11. (a) Ebullioscopic constant : (Kb)
It is elevation in boiling point when one mole of a non-volatile solute is dissolved
in one kilogram (1000 g) of solvent.
i.e. Tb  K b  m
Tb  K b [if m=1]
(units of Kb  K kg mol1 )
Kb WB 1000
Tb  [B=solute, A=solvent]
M B WA
(b) Cryoscopic constant : (Kf)
It is depression in freezing point when one mole of non-volatile solute dissolved in
one kilogram (1000g) of solvent.
Tf  K f  m
Tf  K f [if m=1]
1
Units of Kf : K kg mol
K f  WB 1000
T f 
M B  WA
12. (a) Osmosis : When a solution is separated from its solvent by a semipermeable
membrane (SPM) there is a spontaneous flow of solvent molecules from solvent
compartment to solution compartment. This phenomenon is called osmosis.
(b) Osmotic pressure (): The excess pressure that must be applied to a solution
side to prevent osmosis is called osmotic pressure.
  CRT
W  R T
 B
M B Vin L
(c) Reverse osmosis : If a pressure larger than the osmotic pressure is applied to
the solution side, the pure solvent (or water) flows out of the solution through the
semipermeable membrane. In this way direction of osmosis is reversed and so
the process is called reverse osmosis.
 Reverse osmosis is used in desalination to get pure water from sea water.
(d) Isotonic solutions : When two solutions exert same osmotic pressure because
they have same molar concentration, these are called isotonic solutions.

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(e) Abnormal molar mass : When the mass of a substance determined by using
any of the colligative properties comes out to be different than the theoretically
expected molar mass, the substance said to show abnormal molar mass.
(f) Van’t Hoff factor : (i) Van’t Hoff factor (i) gives the extent of association (or)
dissociation of the solute particles in the solution. It may be defined as the ratio
of observed colligative property to calculated colligative property.
observed colligative property
i ;
calculated colligative property
number of moles of particles after dissociation (or) association
i
number of moles of particles before dissociation (or) association
(g) Edema : It is a common observation that people taking a lot of salt or food rich
in salt develop swelling called edema. This is due to retension of water in tissues
and intercellular spaces in their body due to osmosis.
13. The similarity between Raoult’s law and Henry’s law is that in both the Laws, the
partial vapour pressure of the volatile component (or) gas is directly proportional
to its molefraction in the solution.
14. 0.1M KCl solution will have higher boiling point as KCl dissociates in the solution.
15. The osmotic pressure method has the advantage over elevation in boiling point
(or) depression in freezing point for determining molar masses of macromolecules
because :
(i) Osmotic pressure is measured at the room temperature and the molarity of
solution is used instead of molality.
(ii) Compared to other colligative properties, its magnitude is larger even for very
dilute solutions.
16. Molality is preferred in studies that involves changes in temperature of some of
the colligative properties of solutions. This is because molality depends on masses
of solvent which do not change with temperature.
17. A substance such as ethylene glycol which is added to water to lower its freezing
point is called as antifreeze.
18. In dilute solution, K2 SO4   2K   SO42
number of moles of particles after dissociation
Van’t Hoff factor, i 
number of moles of particles before dissociation
3
i  3.
1
19. Ethylene glycol lower the freezing point of water. Due to this, coolant in radiators
will not freeze otherwise, radiator will burst due to freezing of coolant (water).
20. At a given pressure, the solubility of oxygen in water increases with decrease in
temperature. Presence of more oxygen at lower temperature makes the aquatic
species more comfortable in cold water.
21. A solution of chloroform and acetone shows negative deviation from Raoult’s law.
This is because chloroform molecule is able to form H-bond with acetone molecule
as shown below.

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22. (a) The use of pressure cooker reduces cooking time because the weight over the
lid does not allow the steam to go out. As a result, pressure inside the cooker
becomes high. Higher the pressure higher is the boiling point and faster is
cooking.
(b) As the concentration of saline solution is higher than the concentration inside
the cell. Thus water will move outside the cytoplasm and the cell will shrink while,
distilled water is hypotonic, water will enter the cell through simple diffusion and
lead to cell swelling.
23. (i) The solution show negative deviation from Raoult’s law.
Temperature will rise.
(ii) ‘A’ and ‘B’ on mixing produces warm solution. i.e. heat is released during their
process. The reaction is exothermic i.e. Hmix<0.
The solution is negatively deviated from Raoult’s law.
*****
SOLUTIONS
1) Both assertion and reason are correct, and reason is the correct explanation of the
assertion
2) Both assertion and reason are correct, but reason is not the correct explanation of
the assertion
3) Assertion is correct, but reason is incorrect
4) Both assertion and reason are incorrect
1. Assertion (A): Solutions are homogeneous mixtures.
Reason (R) : Solvent determines the physical-state in which solution exists.
2. Assertion (A): The solubility of a solid in a liquid is significantly affected by
temperature changes.
Reason (R) : If the dissolution process is endothermic, the solubility should
increase with rise in temperature and if it is exothermic, the solubility should
decrease as per Le-chatelier’s principle.
3. Assertion (A): To avoid bends, the tanks used by scuba divers are filled with air
diluted with helium.
Reason (R) : Helium is very less soluble in blood even at high pressure.
4. Assertion (A): A mixture of chloroform and acetone forms a solution with negative
deviation from Raoult’s law.
Reason (R) : Chloroform forms hydrogen bonds with acetone.
5. Assertion (A): Azeotropes are the binary mixtures having the same composition in
liquid & vapour phase.
Reason (R) : The components of an Azeotrope can be seperated by fractional
distillation.
6. Assertion (A): Abnormal molecular masses are evaluated when the substances
undergo dissociation (or) association in solution.
Reason (R) : In case of dissociation (or) association of solute in the solution
observed-colligative property becomes abnormal.
7. Assertion (A): Molarity of a solution decreases with an increase of temperature.
Reason (R) : As the temperature increases volume of the solution increases.
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8. Assertion (A): One molar aqueous solution (d = 1 gr/cc) has always higher
concentration than one molal.
Reason (R) : The molality of a solution does not depends on temperature where as
molarity depends.
9. Assertion (A): At the same temperature water has higher vapour pressure than
acetic acid.
Reason (R) : Hydrogen bonding in water is weaker than in acetic acid.
10. Assertion (A): In diffusion the flow of solvent molecules occur in one direction only.
Reason (R) : In osmosis the flow of solvent molecules occur in both directions.
11. Assertion (A): Van’t Hoff factor for electrolytes in water is always greater than
unity.
Reason (R) : Electrolytes undergoes dissociation in polar solvents like water.
12. Assertion (A): Benzene and toluene form an ideal solution.
Reason (R) : Successive homologues generally form ideal solutions.
13. Assertion (A): When CuSO4 is dissolved in water, the solution is warmed up.
Reason (R) : Hydration energy of CuSO4 is higher than its lattice-energy.
14. Assertion (A): If blood cells are placed in pure water, they swell and burst.
Reason (R) : Due to osmosis, the movement of water molecules in to the cell dilutes
the salt content.
15. Assertion (A): The molecular mass of acetic acid in Benzene is more than the actual
value of the solute.
Reason (R) : Molecules of the acetic acid dimerise in benzene due to hydrogen
bonding.
16. Assertion (A): Solutions of identical concentrations always have same osmotic
pressure.
Reason (R) : Number of particles present in solutions are same, if molar
concentrations are same.
17. Assertion (A): Ethylene glycol is added to water and used as antifreeze in the
radiators of auto-mobiles.
Reason (R) : Addition of ethylene glycol lowers the freezing point of water.
18. Assertion (A): If two liquids on mixing form a solution with liberation of heat, it is
a non-ideal solution with negative deviation.
Reason (R) : Solutions with negative deviations are accompanied by decrease in
volume.
19. Assertion (A): A pressure cooker reduces cooking time.
Reason (R) : The boiling point of water inside the cooker is increased.
20. Assertion (A): The vapour pressure of mixture of ethanol and water is greater than
pure water.
Reason (R) : Ethanol & water form an azeotropic mixture.
21. Assertion: Colligative properties depend on the nature of the solute, not its
chemical identity.
Reason: Colligative properties are related to the number of solute particles in the
solution.

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22. Assertion: Henry's law is not applicable to solutions with high solute
concentrations.
Reason: Henry's law assumes the solute concentration is low, and the solution
behaves ideally.
23. Assertion: The boiling point of a solution is always higher than the boiling point
of the pure solvent.
Reason: Boiling point elevation is a colligative property, and it depends on the
concentration of solute particles in the solution.
24. Assertion: Adding more non-volatile solute to a solution will increase its osmotic
pressure.
Reason: Osmotic pressure is directly proportional to the molality of the solute.
25. Assertion: Solutions with positive deviations from Raoult's law exhibit ideal
behavior.
Reason: In such solutions, the interactions between the solute and solvent are
weaker than between solute-solute and solvent-solvent.
26. Assertion : Azeotropic mixtures are formed only by non-ideal solutions and they
may have boiling points either greater than both the components or less than
both the components.
Reason : The composition of the vapour phase is same as that of the liquid phase
of an azeotropic mixture.
27. Assertion : When methyl alcohol is added to water, boiling point of water
increases.
Reason : When a NON volatile solute is added to a volatile solvent elevation in
boiling point is observed.
28. Assertion : Molarity of a solution in liquid state changes with temperature.
Reason : The volume of a solution changes with change in temperature.
29. Assertion : If a liquid solute more volatile than the solvent is added to the solvent,
the vapour pressure of the solution may increase i.e., ps > po.
Reason : In the presence of a more volatile liquid solute, only the solute will form
the vapours and solvent will not.
30. Assertion : When a solution is separated from the pure solvent by a semi-
permeable membrane, the solvent molecules pass through it from pure solvent
side to the solution side.
Reason : Diffusion of solvent occurs from a region of high concentration solution
to a region of low concentration solution.

Q.No Key Q.No Key Q.No Key

1 2 11 1 21 4
2 1 12 2 22 1
3 1 13 1 23 1
4 1 14 1 24 3
5 3 15 1 25 4
6 2 16 1 26 2
7 1 17 1 27 4
8 2 18 2 28 1
9 1 19 1 29 3
10 3 20 2 30 4
******

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NUMERICALS
1. Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing
20% of C2H6O2 by mass.
2. Calculate the molarity of a solution containing 5g of NaOH is dissolved in 450 ml
solution.
3. Calculate the molality of 2.5g of ethanoic acid (CH3COOH) in 75g of benzene.
4. If N2 gas is bubbled through water at 293K, how many millimoles of N2 gas would
dissolve in 1 litre of water? Assume that N2 exerts a partial pressure of 0.987 bar.
Given that Henry’s law constant for N2 at 293K is 76.48 K.bar.
5. The vapour pressure of pure benzene at a certain temperature is 0.85 bar. A non-
volatile, non-electrolyte solid weighing 0.5g is added to 39g of benzene (molar
1
mass 78 g mol ) vapour pressure of the solution, then, is 0.845 bar. What is the
molar mass of the solid substance?
6. 18g of glucose, (C6H12O6) is dissolved in 1 kg of water in a sauce pan. At what
1
temperature will water boil at 1.013 bar? (Kb for water is 0.52 K kg mol )
7. The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute
was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate
1
the molar mass of the solute (Kb for benzene is 2.53 k kg mol )
8. 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate
(a) the freezing point depression and
(b) the freezing point of solution
1
(Kb for water is 1.86 k kg mol )
9. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing
point of benzene by 0.40K. The freezing point depression constant of benzene is
5.12 k kg mol1 . Find the molar mass of the solute.
10. 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The
osmotic pressure of such a solution at 300 K is found to be 2.57 103 bar.
Calculate the molar mass of the protein.
11. Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4)
if 22 g of benzene is dissolved in 122 g of CCl4.
12. Calculate the mole fraction of benzene in a solution containing 30% by mass in
carbon tetrachloride.
13. Calculate the molarity of each of the following solutions.
(a) 30 g of Co(NO3)2.6H2O in 4.3L of solution.
(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
14. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal
aqueous solution.
15. Calculate :
(a) molality
(b) molarity and
(c) mole fraction of KI if the density of 20% (mass/mass) aqueous solution of KI is
1.202 g mL1 .
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16. H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis, if
the solubility of H2S in water at STP is 0.195m, calculate Henry’s law constant.
17. Henry’s law constant for CO2 in water is 1.67 108 Pa at 298K. Calculate the
quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure
at 298 K.
18. The vapour pressure of pure liquids A & B are 450 and 700 mm Hg respectively,
at 350 K. Find out the composition of the liquid mixture if total vapour pressure
is 600 mm Hg. Also find the composition of vapour phase.
19. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2)
is dissolved in 850 g of water. Calculate the vapour pressure of water for this
solution & its relative lowering.
20. Boiling point of water at 750 mm Hg is 99.63o C . How much sucrose is to be added
to 500 g of water such that it boils at 100o C ?
21. Calculate the mass of ascorbic acid (vitamin C, C6H8O6) to be dissolved in 75 g of
acetic acid to lower its melting point by 1.5o C . K f  3.9 K kg mol1
22. Calculate the osmotic pressure in Pascals exerted by a solution prepared by
dissolving 1.0g of polymer of molar mass 185,000 in 450 mL of water at 37o C .
23. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and
200g of water. Calculate the molality of the solution. If the density of the solution
1
is 1.072 g ml then what shall be the molarity of the solution?
24. An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1.004
bar at the normal boiling point of the solvent. What is the molar mass of the
solute?
1
25. Calculate the mass of a non-volatile solute (molar mass 40 g mol ) which should
be dissolved in 114 g octane to reduce its vapour pressure to 80%.
26. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K.
Calculate the freezing point of 5% glucose in water if freezing point of pure water
is 273.15 K.
27. Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when
6.5g of C9H8O4 is dissolved in 450 g of CH3CN.
28. 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing
point of water observed is 1.0o C . Calculate the Van’t Hoff factor and dissociation
constant of fluoroacetic acid.
29. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour
pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
30. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that
its osmotic pressure is 0.75 atm at 27o C .
31. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of
K2SO4 in 2 litres of water at 25o C , assuming that it is completely dissociated.
1
32. A 3% solution of glucose (molar mass  180 g mol ) is isotonic with 2.5% solution
of an unknown organic substance. Calculate the molecular weight of the
unknown organic substance.
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33. The vapour pressure of a solvent at 283 K is 100 mm Hg. Calculate the vapour
pressure of a dilute solution containing 1 mole of a strong electrolyte AB in 50
moles of the solvent at 283 K (assuming complete dissociation of solute AB).
Numericals – ANSWERS
1. 20% C2H6O2 by mass means 20 g ethylene glycol dissolved in 100 g of solution.
Mass of solute (ethylene glycol) = 20g
Mass of solvent (water) = 100 – 20 = 80g.
20
nC2 H6O2   0.322
62
80
nH 2O   4.44
18
nC2 H6O2 0.322
C2 H6O2    0.068
nC2 H6O2  nH 2O 0.322  4.44
WB 1000 5 1000
2. Molarity  M   
M B V  ml  40  450
M = 0.278 mol L1
WB 1000 2.5 1000
3. Molality  m     0.556 mol kg 1
M B WA  g  60  75
4. PN2  KH   N2
PN2 0.987 bar
 N    1.29 105
2
KH 76480 bar
1000
 nH 2O   55.55
18

n nH2O 
nN 2 nN
N   2  nH2O , nN2  nH2O
nN2  nH 2O nH 2O
2 N2

nN2   N2  nH2O  1.29 105  55.5

nN2  7.16 104 moles
= 0.716 milli moles
P  Ps WB  M A
o
5. A

PAo M B  WA
0.850 bar  0.845 bar 0.5 g  78 g mol1

0.850 bar M B  39 g
0.005 0.5g  78g mol1

0.850 M B  39 g
M B  170 g mol1 .
Kb WB 1000
6. Tb 
M B WA  g 
0.52 18 1000
  0.052 K
180 1000
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Tb  Tbo  Tb   373.15  0.052 K
Tb  373.202 K
7. Tb  354.11  353.23  0.88 K
Kb WB 1000
Tb 
M B WA  g 
Kb WB 1000 2.53 1.8 1000
MB  
WA  Tb 0.88  90
 58 g mol 1
K f  WB 1000 1.86  45 1000
8. (a) T f    2.2 K
M B  WA  g  62  600
(b) Tf  Tfo  Tf  273.15  2.2  270.95K
K f  WB 1000
9. T f 
M B  WA  g 
K f  WB 1000 5.12 11000
MB    256 g mol1
T f  WA  g  0.40  50
n
10.   CRT  RT
V
W RT
V  nRT  B
MB
WB RT
MB 
V
1.26  0.083  300
 3
g mol1
2.57 10  0.2
M B  61038.91 g mol1
mass of C6 H 6
11. Mass % of C6 H 6  100
total mass of solution
22
 100  15.27%
144
mass of CCl4
Mass % of CCl4  100
total mass of solution
122
 100  84.72% .
144
12. 30% by mass of benzene in CCl4 means 30g of C6H6 (benzene) dissolved in 70g of
CCl4.
30 70
nC6 H 6   0.38 mol ; nCCl4   0.45 mol
78 154
nC6 H6 0.38
 C6 H6    0.457
nC6 H6  nCCl4 0.38  0.45

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WB
13. (a) Molarity  M  
M B V  in L 
30
  0.024 mol L1
290.7  4.3
(b) M 1V1  M 2V2
30  0.5  M 2  500
30  0.5
M2   0.03M
500
WB 1000
14. Molality  m  
M B WA  g 
m  M B  WA
WB 
1000
0.25  60  2500

1000
WB  37.50 g
WB 1000
15. (a) Molality of KI 
M B WA  g 
20 1000
  1.5m
166  80
WB 1000  density
(b) Molarity of KI 
M B  weight of solution  g 
20 1000 1.202
  1.44M
166 100
20 80
(c) nKI   0.12 mol ; n H2O   4.4 mol
166 18
nKI 0.12
 KI    0.0265
nKI  nH2O 4.4  0.12
16. Molality = 0.195 m
i.e. 0.195 moles of H2S dissolved in 1000g of solvent (water)
1000
nH 2 S  0.195; nH 2O   55.55
18
nH 2 S 0.195
xH 2 S    0.0035
nH 2 S  nH 2O 0.195  55.55
 According to Henry’s law
PH2S  KH  H2S
PH 2 S 0.987
KH    282 bar  At STP, P = 0.987 bar  .
H S2
0.0035
17. According to Henry’s law PCO2  KH CO2 .
PCO2 2.5 101325Pa
CO  
2
KH 1.67 108 Pa
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 1.517 103
Number of moles of water present in 500 mL of soda water.
[500 ml = 500 g because density of water = 1 g/cm3]
500
nH 2O   27.78 moles
18
nCO2
xCO2   1.517 103
27.78
nCO2  42.14 103 moles
Mass of CO2  nCO2  GMW of CO2 = 42.14 103  44  1.854 g .
18. PT  PAo  A  PBo  B
PT  PAo  A  PBo 1   A    A  B  1
600  450   A   700 1   A 
 A  0.4 ;  B  1   A  0.6
In vapour phase
PA PAo  A 450  0.4
YA     0.3
PT PT 600
PB PBo X B 700  0.6 
YB     0.7 (OR)
PT PT 600
YB  1  YA  1  0.3  0.7 .
50 850
19. nurea   0.83 mol ; nH2O   47.2 mol
60 18
nurea 0.83
  urea    0.017
nurea  nH2O 0.83  47.2
PAo  PA
   urea
PAo
23.8  PA
 0.017
23.8
PA  23.4mm
20. Tb  Tb  Tbo  100  99.63  0.37o C
Kb WB 1000
Tb 
M B WA  g 
M B WA  Tb 342  500  0.37
WB    121.7 g
1000  Kb 0.52 1000
K f  WB 1000
21. Tf 
M B  WA  g 
1.5 176  75
WB   5.077 g
3.9 1000
22. V  nRT

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WB  R  T 1 0.0821 310
   3.057 104 atm
M B V 185000  0.45
 3.057 104 1.013 105 Pa
= 30.96 Pa.
WB 1000 222.6 1000
23. Molality  m     17.95m
M B WA  g  62  200
Mass of solution = mass of glycol + mass of water
= 222.6 + 200 = 422.6 g.
mass of solution
Volume of solution V  
density of solution
422.6
 mL = 394.21 mL
1.072
WB 1000 222.6 1000
Molarity    9.10M
M B V  in mL  62  394.21
(OR)
WB  density 1000 222.6 1000 1.072
Molarity = 
MB  mass of solution  g  62  422.6
= 9.10M.
PAo  Ps WB  M A
24. 
PAo
M B  WA  g 
WB  M A PAo
MB   o [ PA = vapour pressure of pure water = 1.013]
o

WA PA  Ps
2 18 1.013

98 1.013  1.004
M B  41.35 g mol1 .
25. Suppose the vapour pressure of pure octane = 100 bar and vapour pressure of
solution = 80 bar
PAo  Ps WB  M A

PAo M B  WA
PAo  Ps M B  WA
WB  
PAo MA
100  80 40 114
   8g
100 114
K  WB 1000
26. T f  f
M B  WA  g 
K f  5 1000
For sugar, 2.15 
342  95
2.15  K f  0.154 …..(1)

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K f  5 1000
For glucose, Tf 
180  95
Tf  K f  0.292 …..(2)
Compare equation (1) & (2)
2.15 K f  0.154

T f K f  0.292
2.15  0.292
T f   4.08K
0.154
Freezing point of glucose solution = 273.15 – 4.08
= 269.07 K.
mass of solute
27. Mass % of solute (asprin) = 100
mass of solution
6.5
 100
450  6.5
= 1.424%.
K  WB 1000 1.86 19.5 1000
28. M B  observed   f 
T f  WA  g  1 500
 72.54 g mol 1
M B  calculated  78
i   1.0753
M B  observed  72.54
Calculation of dissociation constant :
CH 2 FCOOH CH 2 FCOO   H 
C O O
C C  C C
Total number of moles  C 1   C  C 
 C 1 
Total number of moles after dissociation
i
Total number of moles before dissociation
C 1  

C
i  1  1.0753
 i 1  1.0753 1  0.0753
W 1000
Molality  m   B
M B  WA  g 
19.5 1000
  0.5m
78  500
C  .C  C 2
Ka  
C 1   1 

0.5  0.0753
2

  3.07 103 .
1  0.0753

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PAo  PS WB  M A
29. 
PAo M B  WA
17.535  PS 25 18

17.535 180  450
17.535  PS 1
 
17.535 180
PS  17.44 mm Hg .
i WB  RT
30.   iCRT 
M B V
 V  M B0.75  2.5 111
WB    3.42 g
i  R T 2.47  0.0821 300
31. Since K2SO4 dissociate completely
K2 SO4 2K   SO42
3
i 3
1
i  WB  R  T 3  25 103  0.0821 298
 
M B V 174  2
  5.27 103 atm .
32.  1  C1RT ;  2  C2 RT
Since isotonic solutions  1   2
C1RT = C2RT
n1 n2

V V
3 2.5

180 M .Wt
2.5 180
M .Wt   150 g/mol .
3
33. Since AB is completely dissociates
i=2
P o  PS nB
 .i
Po nA
P o  PS 1
o
 2
P 50
100  PS 1

100 25
100
100  PS  4
25
PS  100  4
PS  96 mm Hg .

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CASE STUDY BASED QUESTIONS
1. Read the passage carefully and answer the questions that follow:
The Van't Hoff factor, named after the Dutch chemist Jacobus Henricus Van't
Hoff, is a crucial parameter in colligative properties of solutions. In the context of
class 12 chemistry, it is particularly relevant to understand the behavior of solute
particles in a solution, especially when dealing with ionic compounds.
The Van't Hoff factor (i) is defined as the ratio of the moles of particles produced
in a solution to the moles of solute dissolved. For non-ionic solutes, such as
glucose, the Van't Hoff factor is typically 1, as one mole of solute yields one mole
of particles. However, for ionic compounds that dissociate in solution, the Van't
Hoff factor is greater than 1.
In this case, one mole of NaCl dissociates into two moles of ions. Therefore, the
Van't Hoff factor for NaCl is 2. This factor becomes crucial in colligative properties
calculations, such as osmotic pressure and freezing point depression.
a) When roads are salted in winter, sodium chloride is commonly used. Discuss
how the Van't Hoff factor contributes to the effectiveness of salt in preventing the
freezing of road ice.
b) In a 0.1 M solution of acetic acid (CH3COOH), the Van't Hoff factor is less than
1. Explain why this is the case.
c) The value of van't Hoff factor for 0.1M Ba(NO3)2 solution is 2.74. Calculate its
degree of dissociation.
OR
c) Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such
that its osmotic pressure is 0.75 atm at 27°C.
2. Read the passage carefully and answer the questions that follow:
Henna is investigating the melting point of different salt solutions. She makes a
salt solution using 10 mL of water with a known mass of NaCl salt. She puts the
salt solution into a freezer and leaves it to freeze. She takes the frozen salt solution
out of the freezer and measures the temperature when the frozen salt solution
melts. She repeats each experiment.

a. One temperature in the second set of results does not fit the pattern. Which
temperature is that? Justify your answer.
b. Why did Henna collect two sets of results?
c. In place of NaCl, if Henna had used glucose, what would have been the melting
point of the solution with 0.6 g glucose in it?
OR
What is the predicted melting point if 1.2 g of salt is added to 10 mL of water?
Justify your answer.

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20 Velammal Bodhi Academy