1 Define a solution.

A solution is a homogeneous mixture of two or more pure substances whose composition can be
varied within certain limits.
A solution consisting of two components (solute-present in relatively small quantity & solvent
present in relatively large quantity) is called a binary solution.
2 Define the various modes of expressing the concentration of a solution. Which of these modes are
independent of temperature and why?
Concentration is the amount of solute dissolved in a particular amount of the solvent or
solution. (i) i) Mass percentage (w/w): It is defined as the mass of solute per 100g of solution.

ii) Mole
fraction(Χ): It is the ratio of number of moles of one component to the total number of (ii) moles
present in the solution.
Mole fraction of component A = No. of moles of component A
No. of moles of A + No. of moles of B

Mole fraction of component B = No. of moles of component B

No. of moles of A + No. of moles of B

i.e.,
iii) Molarity (M): It is the number of moles of solute dissolved in one litre of the solution.
Molarity, M = n/V = w2 x 1000
M2 x V(mL)
iv) Molality (m): It is the number of moles of the solute dissolved in one kilogram of the solvent.
Molality = number of moles of solute/ Weight of solvent in kg m = w2 x 1000
M2 x w1(kg)
where w2 is the mass of the solute and w1 is the mass of the solvent. M2 is the molar mass of solute.
v)Parts per million (ppm): It is the parts of a component per million (106) parts of the solution. It
is used when a solute is present in trace quantities.
Another unit which is commonly used in medicine and pharmacy is mass by volume
Percentage (w/V). It is the mass of solute dissolved in 100 mL of the solution.

➢ Since volume depends on temperature and undergoes a change with change in temperature, the
molarity will also change with change in temperature. On the other hand, mass does not change with
change in temperature, as a result other concentration terms, mass percentage, ppm, molality and
mole fraction, are independent of temperature.
3 Define solubility.
It is the maximum amount of substance that can be dissolved in 100 g of the solvent at a specified
temperature.
The solubility of a solid in a liquid depends upon:
(a) the nature of the solute (b) nature of solvent and (c) the temperature of the system In a nearly
saturated solution, if the dissolution process is endothermic (∆solH > 0), the solubility should increase
 with rise in temperature and if it is exothermic (∆solH < 0) the solubility should decrease.
4 Explain the solubility rule “like dissolves like” in terms of intermolecular forces that exist in
solutions.
A substance (solute) dissolves in a solvent if the intermolecular interactions are similar in both the
components; for example, polar solutes dissolve in polar solvents and non-polar solutes in non-polar
solvents thus we can say “like dissolves like”.
5 State Henry’s law.
The effect of pressure on the solubility of a gas in a liquid is governed by Henry’s Law. “The partial
pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (Χ) in the
solution” and is expressed as: p = KHx, where KH is Henry’s Law constant.
Characteristics of KH
• Different gases have different KH values at the same temperature. This suggests that KH is a
function of the nature of the gas.
• Higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid.
• The solubility of gases increases with decrease of temperature.
Applications of Henry’s law
• To increase the solubility of CO2 in soft drinks, the soda water bottles are sealed under high
pressure.
• To avoid the toxic effects of high concentration of nitrogen in the blood of the scuba divers (the
painful condition is known as bend), the tanks are filled with air diluted with Helium. • At high
altitude, low blood oxygen causes climbers to become weak and unable to think clearly, a condition
known as anoxia.
6 Explain the following phenomena with the help of Henry’s law.
(i) Painful condition known as bends.
Deep sea divers depend upon compressed air for breathing at high pressure under water. The
compressed air contains N2 in addition to O2. At great depths the pressure is far higher than the
surface atmospheric pressure and more N2 get dissolved in the blood. When the diver comes towards
the surface, the pressure decreases, so N2 comes out of the body forming bubbles in the blood stream.
These bubbles restrict blood flow, affect the transmission of nerve impulses. This causes a painful
and life-threatening condition called as bends.
(ii) Feeling of weakness and discomfort in breathing at high altitude (Anoxia) At high altitude,
partial pressure of oxygen islessthan that of ground level. Thisleadsto low concentrations of oxygen
in blood and tissue of people living at high altitudes or climbers. The low blood oxygen causes
climbers to become weak and unable to think clearly, symptoms of a condition known as anoxia.

7 What is the effect of temperature on solubility of gases in liquid?

As dissolution is an exothermic process, according to Le Chatelier’s principle solubility should
decrease with rise in temperature.
8 Why are aquatic species more comfortable in cold water in comparison to warm water? 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 comfortable in cold water. 9 What is vapour pressure?
It is the pressure exerted by the vapours above the liquid surface, in equilibrium with the liquid at a
given temperature.
10 State Raoult’s law for the solution containing volatile components.
Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each volatile
 component in the solution is directly proportional to its mole fraction.
Consider a binary solution of two volatile liquids 1 and 2. If p1 and p2 are the partial vapour pressures
of the two liquids and x1 and x2 are their mole fractions in solution.
Then, p1 = p10 x1 and p2 = p20 x2
Where p10 and p20are the vapour pressures of the pure components 1 and 2 respectively.
Limitations:
(i) Applicable only to dilute solution. (ii) Applicable to solutions of only non- electrolyte. 11
Explain the fact that Raoult’s Law is a special case of Henry’s Law. [What is the similarity between
Raoult’s Law and Henry’s Law?]
In the solution of a gas in a liquid, one of the components is so volatile that it exists as a gas. Its
solubility is given by Henry’s law, p = KHX.
If we compare the equations for Raoult’s law and Henry’s law, it can be seen that the partial pressure
of the volatile component or gas is directly proportional to its mole fraction in solution. Thus,
Raoult’s law becomes a special case of Henry’s law in which KH becomes equal to p10.
12 How is vapour pressure of solvent affected when a non-volatile solute is dissolved in it? The
addition of a non-volatile solute to a volatile solvent decreases the escaping tendency of the
solvent molecules from the surface of solutions as some of the surface area is occupied by non
volatile solutes. It results in decrease of the vapour pressure of solution.
13 Explain why on addition of 1 mol of NaCl to 1 litre of water, the boiling point of water increases,
while addition of 1 mol of methyl alcohol to one litre of water decreases its b.p? NaCl is a
non-volatile solute, therefore, addition of NaCl to water lowers the vapour pressure of water. As a
result, boiling point of water increases.
Methyl alcohol on the other hand is more volatile than water, therefore its addition increases, the total
vapour pressure over the solution and a decrease in boiling point of water results.
14 State Raoult’s law for the solution containing non-volatile solute.
It states that the relative lowering of vapour pressure is equal to mole fraction of solute when solvent
alone is volatile and is
expressed as:

15 Explain ideal solutions with the help of a diagram.

Ideal Solutions: A solution which obeys Raoult’s law at all concentrations and temperatures.
Liquids having similar nature and structure are likely to form ideal solutions.
e.g., Mixtures of methanol and ethanol, n- hexane and n-heptane, benzene and toluene.

The plot of vapour pressure and mole fraction of an ideal solution at constant temp. where
component 1 is that of a solvent and 2 is of a
solute.

Characteristics of an ideal solution:

➢ Obeys Raoult’s law.
➢ F1–2 = F1–1 = F2–2
➢ ∆solV = 0, i.e., there is no change in
volume
 when an ideal solution is formed.
➢ ∆solH= 0; i.e., heat is neither evolved nor
absorbed during the formation of an ideal
solution.
➢ Does not form azeotrope
16 Explain non-ideal solutions.
Non-ideal Solutions: Those solutions which show deviation from Raoult’s law.
Characteristics of a non-ideal solution
0 0
➢ Raoult’s law is not obeyed. i.e., P1 ≠ P1 X1 and P2 ≠ P2 X2
➢ Δmix H ≠ 0
➢ Δmix V ≠ 0
➢ Forms azeotrope
Types of deviations from Raoult’s law:
a) Non-ideal solutions showing positive deviation: The solution shows positive deviation from
Raoult’s law if its vapour pressure is higher than that predicted by Raoult’s Law.
b) Non-ideal solution showing negative deviation: The solution shows negative deviation if its
vapour pressure is lower than that predicted by Raoult’s Law.
17 Differentiate between a solution showing positive deviation and a solution showing negative
deviation.
Positive deviation Negative deviation.

p1 > p10x1 and p2 > p20x2 a) p1 < p10x1 and p2 < p20x2

The solvent – solute interactions are weaker The solvent – solute interactions are stronger
b) than solvent – solvent or solute – solute
than solvent – solvent or solute – solute interactions i.e., F1–2 > F1–1 and F2–2
interactions. i.e., F1–2< F1–1 and F2–2

Δmix H > 0, Δmix V > 0, c) Δmix H < 0, Δmix V < 0

e.g., Ethyl alcohol and water, Acetone and e.g., Nitric acid and water, Chloroform and
CS2, Acetone and benzene, ethanol and acetone, Acetic acid and pyridine
acetone

a)
18 What type of deviation is shown by a mixture of ethanol and acetone? Why is it showing such
kind of deviation?
Positive deviation from Raoult’s law because some of the Hydrogen bonds between pure ethanol
molecules breaks by the adding acetone into ethanol. Due to weakening of the interactions, ethanol
gets easily vaporized, increasing the vapour pressure of the solution.
19 What type of deviation from Raoult’s law is observed by mixing chloroform and acetone? Why is
an increase in temperature observed on mixing chloroform and acetone?
It shows negative deviation from Raoult’s law. It is because CHC13 molecules form H-bond with
acetone, resulting in the release of energy. As a result, there is an increase in temperature. The increase in
the force of interaction between chloroform and acetone, decreases the escaping tendency of each
component, which results in decrease of vapour pressure of the solution. 20 Define azeotropes. What
are the two types of azeotropes?
Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boil at a
constant temperature (known as constant boiling mixtures).
• These are formed by non-ideal solutions.
• It is not possible to separate the components by fractional distillation.
(i) Minimum boiling azeotropes
• Formed by non-ideal solutions which show positive deviation from Raoult’s law. • They are the
binary mixtures whose boiling points are lower than either of the two components. ➢ e.g., C2H5OH
(95%) + H2O (5% by mass).
(ii) Maximum boiling azeotropes
• Formed by non-ideal solutions which show negative deviation from ideal behaviour. •
The binary mixtures whose boiling points are higher than either of the two components. • e.g.,
68% nitric acid and 32% water by mass, with a boiling point of 393.5 K.
21 What are colligative properties? Give examples.
The properties of solution which depend only on the number of solute particles in the solution, but
independent of their nature are called colligative properties.
a) Relative lowering of vapour pressure
The ratio of the lowering of the vapour pressure of the solvent divided by the vapour pressure of the
pure solvent is called the relative lowering of vapour pressure.
According to Raoult’s Law, the relative lowering of vapour pressure of a solution is equal to the mole
fraction of the solute.

Relative lowering of vapour pressure =

Where, p1º = vapour pressure of the pure solvent & p1 = vapour pressure of the solvent in the solution
and the lowering of vapour pressure, ∆p = (p10-p1)
b) Elevation of boiling point
The difference in the boiling point of solution (Tb) and that of pure solvent (Tb0) is called elevation of
boiling point (∆Tb). ∆Tb = Tb - Tb0
For a dilute solution, the elevation in boiling point is found to be proportional to the molality of the
solution. Thus,
∆Tb∝ m or ∆Tb= Kb.m
where ‘m’ is the molality and Kb is the molal elevation constant or ebullioscopic constant.
• Unit of Kb is K kg mol-1
 • c) The depression in freezing point (∆Tf)
The difference in the freezing point of pure solvent (Tf0) and that of the solution (Tf) is called
depression in freezing point (∆Tf).
Depression in freezing point, ∆Tf = Tf0 - Tf
For a dilute solution, the depression in freezing point is found to be proportional to the molality of
the solution. Thus,
∆Tf ∝ m or ∆Tf= Kf. m
where ‘m’ is the molality and Kb is the molal depression constant or cryoscopic constant.
• Unit of Kf is K kg mol-1
• d) Osmotic pressure (π)
• The spontaneous flow of solvent molecules from a dilute solution to a concentrated solution when the
two are separated by a perfect semipermeable membrane is called osmosis.
• Osmotic pressure (π) is the excess pressure which must be applied to the solution to prevent osmosis.
Mathematically, osmotic pressure is proportional to molarity (C) of the solution at a particular
temperature, T. Thus,
π = CRT = w2 x R x T
M2 x V
where π is the osmotic pressure of the solution, w2 is the mass of solute, V is the volume of the
solution in litres, R is the gas constant, and T is the temperature in Kelvin.
22 Define i) Molal elevation constant, Kb ii) Molal depression constant, Kf i) Molal elevation constant
(ebullioscopic constant) is defined as the elevation in boiling point when one mole of a `solute is
dissolved in 1000 grams of the solvent.
ii) Molal depression constant (cryoscopic constant) is defined as the depression in freezing point
when one mole of a non-volatile solute is dissolved in 1000g of the solvent.
23 How does sprinkling of salt help in clearing the snow-covered roads in hilly areas? Explain the
phenomenon involved in the process.
When salt is sprinkled on snow-covered roads, it lowers the freezing point of water. This is due to the
phenomenon of depression in freezing point of water when a non-volatile solute is dissolved in it . As
a result, the snow starts melting from the surface, and the ice layer formed on the road is broken
down. This helps in clearing the roads . Common salt acts as a de-icing agent in this process .
24 Which one of the following will have higher osmotic pressure in 1M KCl or 1M urea solution? 1M
KCl will have higher osmotic pressure because it dissociates into K+ and Cl- ions while urea does
not dissociate into ions in the solution.
25 Define reverse osmosis. Write one of its applications. Name one SPM that can be used in the process
of reverse osmosis [RO].
If pressure larger than osmotic pressure is applied to the solution side, the pure solvent flows out of
the solution through the SPM. As the direction of osmosis is reversed by this way, the process is
called reverse osmosis.
➢ Reverse osmosis is used in the desalination of sea
water to get pure water.

➢ SPM- Cellulose acetate, cellophane, parchment

paper
26 What are hypertonic, hypotonic and isotonic solutions?
➢ Hypertonic solution: A solution having higher osmotic pressure than that of the other solution from
which it is separated by a semipermeable membrane.
➢ Hypotonic solution: A solution having lower osmotic pressure than that of the other solution. ➢
Isotonic solution: Two solutions having same osmotic pressure at a given temperature are called isotonic
solutions. These solutions have same molar concentration. When such solutions are separated by a
semipermeable membrane, no osmosis occurs between them.
For isotonic solutions, π1 = π2. Also, C1 = C2.
➢ 0.91% solution of pure NaCl is isotonic with human RBC’s. All intravenous injections must be
isotonic with body fluids.
➢ If we place our blood cells in a solution containing more than 0.9% (hypertonic) sodium chloride
solution, water will flow out of the cells and they would shrink.
➢ On the other hand, if they are placed in a solution containing less than 0.9% (hypotonic) NaCl,
water will flow into the cells and they would swell.
27 What do you expect to happen when Red blood cells are placed in 0.5%NaCl solution? RBC are
isotonic with 0.9% NaCl solution, so they will swell and even may burst when placed in 0.5% NaCl
solution.
28 What is the advantage of using osmotic pressure method over other colligative properties? a.
Osmotic pressure measurement is around the room temperature
b. Molarity of the solution is used instead of molality, which can be determined easily.
c. The magnitude of osmotic pressure is large even for every dilute solution.
d. It is useful for determination of molar masses of biomolecules as they are generally not stable at
higher temperatures and for polymers which have poor solubility.
29 Name the colligative property which is used to find the molecular mass of macromolecules.
Osmotic pressure method
30. Define Abnormal molar mass and van’t Hoff factor (i).
Abnormal molar mass: When the molar mass of a substance determined by using any of the
colligative properties deviate from the theoretically expected molar mass, the substance is said to
have an abnormal molar mass. It is due to either association or dissociation of molecules. Van’t Hoff
factor (i)
➢ It is used to express the extent of association or dissociation of solute particles in solution. ➢ It may
be defined as the ratio of observed colligative property to the calculated colligative property. ➢ i.e., i =
Normal molar mass
Observed molar mass
Or, i = Observed colligative property
Calculated colligative property
➢ In case of association, i < 1
➢ But in case of dissociation, i> 1
➢ When there is neither association nor dissociation, i = 1
Note:
➢ When the solute molecules undergo association, the number of particles becomes less and
molecular mass determined with the help of colligative property will be more.
 ➢ Dissociation leads to increase in number of particles, therefore, increase in colligative property,
therefore, decrease in molecular weight because colligative property is inversely proportional to
molecular weight.

31 Why is the value of Van’t Hoff factor for ethanoic acid in benzene close to 0.5? Molecules of ethanoic
acid dimerise in benzene due to hydrogen bonding (association). Thus, the number of particles
reduced to nearly half of initial value due to dimerization. Therefore, value of van’t Hoff factor is
close to 0.5.
32 Incase of association or dissociation van’t Hoff factor ‘i’ should be included, which modifies the
equations for colligative properties as follows:
Elevation of boiling point, ∆Tb= i Kb m
Depression of freezing point, ∆Tf= i Kf m
Osmotic pressure, π = i CRT
33
Salt No. of ions Van’t Hoff factor ‘i’ for
on dissociation complete dissociation of
solute

NaCl 2 2

MgSO4 2 2

K2SO4 3 3

AlCl3 4 4

Solve these.
1. Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g
octane to reduce its vapour pressure to 80%.
2. A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point
of 5% glucose in water if freezing point of pure water is 273.15 K.
3. 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.
4. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate
the solubility of methane in benzene at 298 K under 760 mm Hg.
5. Determine the amount of CaCl2 dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm
at 27° C.
6. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water
at 25° C, assuming that it is completely dissociated.
7. 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water
observed is 1.00 C. Calculate the Van’t Hoff factor and dissociation constant of fluoroacetic acid. 8. 0.3g of
acetic acid is dissolved in 30g (M= 60 g mol-1) of benzene shows a depression in freezing point of 0.45oC.
Calculate the percentage association of acid if it forms a dimer in the solution. [Given Kf for benzene =
5.12 K kg mol-1).
9. The freezing point of a solution containing 5 g of benzoic acid (M = 122 g mol-1) in 35 g of benzene is
 depressed by 2.94 K. What is the percentage association of benzoic acid if it forms a dimer in solution?
(Kf for benzene = 4.9 K kg mol-1)
10. 18 g of glucose, C6H12O6 is dissolved in 1kg of water in a saucepan. At what temperature will this
solution boil? (Kb for water = 0.52 K Kg mol-1)
11. Calculate the freezing point of a solution when 3g of CaCl2 (M= 111gmol-1) was dissolved
in 100g of water, assuming CaCl2 undergoes complete ionisation.
12. 30g urea dissolved in 846g of water. Calculate the vapour pressure of water for this solution if vapour
pressure of pure water at 298 K is 23 8 mm Hg.

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