CHEMISTRY (XI, XII & MEDICAL) by VIJAY KUMAR (M.Sc., B.Ed.

Solutions are homogeneous mixtures containing two or more than two components. Generally, the
component that is present in the largest quantity is known as solvent. Solvent determines the physical
state in which solution exists. One or more components present in the solution other than solvent are
called solutes.

Solutions containing only two components are called binary solutions. Here each component may be
solid, liquid or in gaseous state.

EXPRESSING CONCENTRATION OF SOLUTIONS

Composition of a solution can be described by expressing its concentration. Concentration is defined as
the amount of solute present in the given of solution. The concentration of a solution can be expressed by
several ways.
1. Mass percentage (w/w): The mass percentage of a component of a solution is defined as the mass
of a component present in 100 parts by mass of the solution.

Mass of the component in the solution

Mass % of a component = ------------------------------------------------ × 100
Total mass of the solution

For e.g. 10% glucose in water by mass means that 10 g of glucose is dissolved in 90 g of water
resulting in a 100 g solution. Concentration described by mass percentage is commonly used in
industrial chemical applications.

2. Volume percentage (v/v): The volume percentage is defined as the volume of a component
present in 100 parts by volume of the solution.

Volume of the component

Volume % of a component = --------------------------------- ×100
Total volume of solution

For example, 10% ethanol solution in water means that 10 mL of ethanol is dissolved in 90 mL of
water such that the total volume of the solution is 100 mL. Solutions containing liquids are
commonly expressed in this unit.

3. Mass by volume percentage (w/v): It is the mass of solute dissolved in 100 mL of the solution. It
is commonly used in medicine and pharmacy.

Mass of the component in the solution

Mass/volume % of a component = ------------------------------------------------ × 100
Total volume of the solution

4. Parts per million (ppm): When a solute is present in trace quantities (i.e. very small amounts), its
concentration is expressed in parts per million (ppm). It is defined as the number of parts of a
particular component in million parts of the solution.

Number of parts of the component

Parts per million (ppm) = ------------------------------------------------------------------------- × 10 6
Total number of parts of all the components of the solution

Concentration in parts per million can be expressed as mass to mass, volume to volume and mass
to volume. The concentration of pollutants in water or atmosphere is expressed in terms of μg mL–1
or ppm.

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 CHEMISTRY (XI, XII & MEDICAL) by VIJAY KUMAR (M.Sc., B.Ed.)

5. Mole fraction (ꭓ): It is the ratio of the number of moles of a particular component to the total
number of moles of the solution.
For a solution containing n2 moles of the solute dissolved in n 1 moles of the solvent, mole fraction
of solute in the solution -
n2 n1
χ2 = ----------- & χ1 = ---------- (N/B – χ1 + χ2 = 1)
n1 + n2 n1 + n2

Mole fraction is useful in describing the calculations involving gas mixtures.

6. Molarity (M): It is defined as the number of moles of solute dissolved per litre of solution.

Number of moles of solute

Molarity (M) = ------------------------------------
Volumes of solution in litre

Number of moles of solute

Molarity (M) = ----------------------------------- × 1000
Volume of solution in mL

(N/B – Often in a chemistry laboratory, a solution of a desired concentration is prepared by

diluting a solution of known higher concentration called 'stock solution'. For diluting solutions-
M 1 × V1 = M2 × V2
(stock solution) ( solution to be prepared)
where M1 & M2 and V1 & V2 are molarities and volumes of corresponding solutions.)
For example, 0.25 M solution of NaOH means that 0.25 mol of NaOH is dissolved in one litre of
solution.

7. Molality (m): It is defined as the number of moles of the solute present per kilogram (kg) of the
solvent.

Number of moles of solute

Molality (m) = ------------------------------------
Mass of solvent in kg

For example, 1 molal (m) solution of KCl means that 1 mol (74.5 g) of KCl is dissolved in 1 kg of
water.

8. Normality (N): It is defined as the number of gram equivalents of solute present per litre of the
solution.
Number of gram equivalents of solute
i.e., Normality (N) = -----------------------------------------------
Volume of solution in litre

Mass of solute × 1000

Or, Normality (N) = ---------------------------------------------------------
Eq. mass of solute × Volume of solution in mL

(N/B- Among the different methods for expressing the concentration of solution, mass %, ppm,
mole fraction and molality are independent of temperature; whereas molarity, normality and
volume % are depend on temperature. This is because volume depends on temperature and the
mass does not.)

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 CHEMISTRY (XI, XII & MEDICAL) by VIJAY KUMAR (M.Sc., B.Ed.)

SOLUBILITY
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent. It
depends upon the nature of solute and solvent, temperature and pressure.

1. Solubility of a Solid in a Liquid

It is observed that polar solutes dissolve in polar solvents and non polar solutes in non polar
solvents. In general, a solute dissolves in a solvent if the intermolecular interactions are similar in
the two or the general principle related to solubility is that “like dissolves like”.

Saturated and Unsaturated solutions

A solution in which no more solute can be dissolved at the same temperature and pressure is called
a saturated solution. Here there is a dynamic equilibrium between the dissolved solute and the
undissolved solute. An unsaturated solution is the one in which more solute can be dissolved at the
same temperature.

Effect of temperature
The solubility of a solid in a liquid mainly depends on temperature. Since the dissolution of a solid
in a liquid is an equilibrium process, it should follow Le Chateliers Principle. In general, if in a
nearly saturated solution, the dissolution process is endothermic (ΔsolH > 0), the solubility should
increase with rise in temperature and if it is exothermic (ΔsolH > 0) the solubility should decrease.

Effect of pressure
Since solids and liquids are highly incompressible, pressure does not have any significant effect
on solubility of solids in liquids.

2. Solubility of a Gas in a Liquid

Solubility of gases in liquids is greatly affected by pressure and temperature. The solubility of a
gas increases with increase of pressure.

A quantitative relation between pressure and solubility of a gas in a liquid was first given by
Henry, which is known as Henry’s law. “The law states that at a constant temperature, the
solubility of a gas in a liquid is directly proportional to the pressure of the gas.” Dalton stated that
the solubility of a gas in a liquid solution is a function of partial pressure of the gas. If we use the
mole fraction of a gas in the solution as a measure of its solubility, then we can state that the mole
fraction of gas in the solution is proportional to the partial pressure of the gas over the solution.

The most commonly used form of Henry’s law states that “the partial pressure of the gas in
vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution ” and is
expressed as:
p = KH.ꭓ

Here KH is the Henry’s law constant. Different gases have different KH values at the same
temperature. i.e. KH is a function of the nature of the gas. Higher the value of K H at a given
pressure, the lower is the solubility of the gas in the liquid. A graph of partial pressure of the gas
against mole fraction of the gas in solution is a straight line with slope is equal to KH.

As the temperature increases solubility of a gas in a liquid decreases. It is due to this reason that
aquatic species are more comfortable in cold waters rather than in warm waters.

Applications of Henry’s law

• To increase the solubility of CO 2 in soft drinks and soda water, the bottle is sealed under high
pressure.

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• Scuba divers carry oxygen cylinders to breath under deep sea. At high pressure under water, the
solubility of atmospheric gases in blood increases. 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 the blood. This blocks capillaries and creates a medical condition known as
bends. To avoid bends the cylinders used by scuba divers are filled with air diluted with helium
(11.7% helium, 56.2% nitrogen and 32.1% oxygen).

• At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to
low concentrations of oxygen in the blood and tissues of people living at high altitudes or
climbers. Low blood oxygen causes climbers to become weak and unable to think clearly. This
condition is known as anoxia.

Effect of Temperature
Solubility of gases in liquids decreases with rise in temperature. When dissolved, the gas
molecules are present in liquid phase and the process of dissolution can be considered similar to
condensation and heat is evolved during this process. The dissolution process involves a dynamic
equilibrium and thus follows Le- Chatelier’s Principle. As dissolution is an exothermic process,
the solubility should decrease with increase of temperature.

VAPOUR PRESSURE OF LIQUID SOLUTIONS

Liquid solutions are formed when solvent is a liquid. The solute can be a gas, a liquid or a solid.
Generally, the liquid solvent is volatile. The solute may or may not be volatile. Based on the volatility of
solute, the vapour pressure of the solution is greater or less than that of the solvent.

VAPOUR PRESSURE OF LIQUID-LIQUID SOLUTIONS – RAOULT’S LAW

A quantitative relationship between the vapour pressure and mole fraction of solute in a solution was first
given by a French chemist F.M Raoult and it is known as Raoult’s Law. It states that for a solution of
volatile liquids, the partial vapour pressure of each component in the solution is directly proportional
to its mole fraction.

Consider a binary solution of two volatile liquids 1 and 2. Let p 1 and p2 be the partial vapour pressures of
the two components 1 and 2 respectively and ptotal be the total vapour pressure. Let x 1 and x2 be the mole
fractions of the two components 1 and 2 respectively.
Then according to Raoult’s law, for component 1
p1 α x1
or, p1 = p10.x1
similarly for component 2,
p2 α x2
or, p2 = p20.x2

where p10 and p20 are the vapour pressures of the pure components 1 & 2 respectively.

According to Dalton’s law of partial pressures, the total pressure

(pT) will be the sum of the partial pressures of the components of the
solution.

So, pT = p1 + p2

Substituting the values of p1 and p2, we get

pT = p20 + (p10 – p20).x1

The composition of vapour phase in equilibrium with the solution is

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 CHEMISTRY (XI, XII & MEDICAL) by VIJAY KUMAR (M.Sc., B.Ed.)

determined by the partial pressure of the components. If y 1 and y2 are the mole fractions of the
components 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressures:

y1 = p1/pT and y2 = p2/pT

Raoult’s Law as 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, solubility of the gas is given by Henry’s law which states
that p = KH.x. If we compare the equations for Raoult’s law and Henry’s law, we can see that the partial
pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the
proportionality constant KH differs from pi0. Thus, Raoult’s law becomes a special case of Henry’s law in
which KH becomes equal to pi0.

VAPOUR PRESSURE OF SOLUTIONS OF SOLIDS IN LIQUIDS

The vapour pressure of a liquid is the pressure exerted by the vapour in equilibrium with its own liquid..
If a non-volatile solute is added to a pure solvent, the vapour pressure of the resulting solution is always
lower than that of the pure solvent. This is because in a pure solvent, there are only solvent molecules,
which can vapourise. But when a non-volatile solute is added to the solvent, a fraction of the surface is
occupied by solute molecules. So the number of solute molecules passing to the vapour phase decreases
and hence the vapour pressure also decreases. The decrease in the vapour pressure of solvent depends on
the quantity of non-volatile solute present in the solution and not on its nature.

For such a solution the Raoult’s law can be stated as, for any solution the partial vapour pressure of
each volatile component in the solution is directly proportional to its mole fraction.
Consider a binary solution containing a solvent 1 and solute 2. Since the solute is non-volatile, only the
solvent molecules are present in vapour phase and contribute to vapour pressure. Let p 1 be the vapour
pressure of the solvent, x1 be its mole fraction, p10 be its vapour pressure in the pure state. Then according
to Raoult’s law,
p1 α x1
or, p1 = p10.x1 = pT

IDEAL AND NON-IDEAL SOLUTIONS

Liquid – liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law.

1. Ideal solutions: These are solutions which obey Raoult’s law over the entire range of concentration.
For such
solutions, the enthalpy of mixing (ΔHmix) of the pure components is zero and also volume of mixing
(ΔVmix) is zero. i.e. for an ideal solution,
p1 = p10.x1 and p2 = p20.x2

ΔHmix = 0 and ΔVmix = 0

Ideal behaviour can be explained by considering two components A and B. in pure components, the inter
molecular attractive interactions will be of types A-A and B-B. In solution, in addition to these two
interactions, A-B type of interaction will also be present. If the A-A and B-B interactions are nearly equal
to the A-B interaction, the solution behaves ideally. i.e. solute-solute interactions and solvent-solvent
interactions are nearly equal to solute-solvent interaction.

A perfectly ideal solution is rare. But some solutions are nearly ideal in behaviour. e.g. solutions of n-
hexane and n-heptane, bromoethane and chloroethane, benzene and toluene etc.

2. Non-ideal solutions: These are solutions which do not obey Raoult’s law over the entire range of
concentration. The vapour pressure of such a solution is either higher or lower than that predicted

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 CHEMISTRY (XI, XII & MEDICAL) by VIJAY KUMAR (M.Sc., B.Ed.)

by Raoult’s law. If it is higher, the solution exhibits positive deviation and if it is lower, it exhibits
negative deviation from Raoult’s law. The plots of vapour pressure against mole fractions for such
solutions are as follows:

Positive deviation from Rault's law Negative deviation from Rault's law
• In the case of positive deviation from • In case of negative deviation from Raoult’s
Raoult’s law, A-B interactions are weaker law, the A-B interactions are stronger than
than A-A and B-B interactions. i.e., in this A-A and B-B interactions. i.e. solute-
case solute-solvent interactions are weaker solvent interactions are stronger than
than solute-solute and solvent-solvent solute-solute interaction and solvent-
interactions. So more molecules are escaped solvent interaction. So number of
to vapour phase and hence the vapour molecules escaped to vapour phase
pressure of the solution increases. decreases and hence the vapour pressure of
• Examples - solutions of ethanol and acetone, the solution decreases.
acetone and CS2, acetone and CCl4 etc. • Examples - solution of phenol and aniline,
• In pure ethanol, molecules are hydrogen chloroform and acetone etc.
bonded. On adding acetone, its molecules get • In the mixture of phenol and aniline case
in between the host molecules and break the intermolecular hydrogen bonding
some of the hydrogen bonds between them. between phenolic proton and lone pair on
Due to weakening of interactions, the nitrogen atom of aniline is stronger than
solution shows positive deviation from the respective intermolecular hydrogen
Raoult’s law. In a solution of carbon bonding between similar molecules.
disulphide and acetone, the dipolar Similarly, a mixture of chloroform and
interactions between solute-solvent acetone forms a solution with negative
molecules are weaker than the respective deviation from Raoult’s law. This is
interactions among the solute-solute and because chloroform molecule is able to
solvent-solvent molecules. This solution also form hydrogen bond with acetone
shows positive deviation. molecules.
• Graph • Graph

AZEOTROPES
They are binary mixtures having the same composition in liquid and vapour phase and boil at a constant
temperature. For such solutions, it is not possible to separate the components by fractional distillation.
There are two types of azeotropes:
1. Minimum Boiling Azeotrope : The solutions which show a large positive deviation from Raoult’s
law form minimum boiling azeotrope at a particular composition. e.g. 95% ethanol solution by
volume.
2. Maximum Boiling Azeotrope : The solutions which show large negative deviation from Raoult’s
law form maximum boiling azeotrope at a particular composition. e.g. a mixture of 68% Nitric
acid and 32% water by mass forms a maximum boiling azeotrope at 393.5 K.

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COLLIGATIVE PROPERTIES AND DETERMINATION OF MOLARMASS

Colligative properties are those properties which depend only on the number of solute particles and not on
their nature. The important colligative properties are
1. Relative lowering of Vapour pressure
2. Elevation of Boiling point
3. Depression of Freezing point
4. Osmotic Pressure.

1. Relative lowering of Vapour Pressure

When a non-volatile solute is added to a pure solvent, the vapour pressure (V.P) of the resulting solution
is lower than that of the pure solvent. The difference between the vapour pressure of pure solvent and that
of the
solution is called lowering of vapour pressure (ΔP).

Consider a binary solution containing a non-volatile solute 2 dissolved in a solvent 1. Let P 10 be the
vapour pressure of pure solvent 1 and P1 be the vapour pressure of solution. Then according to Raoult’s
law,
p1 = p10.x1
The lowering of vapour pressure of the solvent, Δp= p10 – p1= p10 – p10.x1
Or, Δp = p10.(1- x1)
But, x1 + x2 = 1.
Therefore, 1- x1 = x2
So, Δp = p10.x2

Δp
Or, ------ = x2, the mole fraction of the solute.
p10
Where Δp/p10 is called relative lowering of vapour pressure. It is defined as the ratio of the lowering of
vapour pressure to the vapour pressure pure solvent.
n2
But, x2 = ----------
n1 + n2

where n1 and n2 are the number of moles of solvent and solute respectively.

For dilute solutions, n2 << n1 and hence n2 in the denominator can be neglected.
So, x2 = n2/n1

Δp n2 w2/M2
---- = ---- = ---------
p10 n1 w1/M1

Δp w2 × M1
---- = ------------
p10 w1 × M 2

where w1 and w2 are the masses and M1 and M2 are the molar masses of solvent and solute respectively.

2. Elevation of Boiling Point (ΔTb)

Boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric
pressure. When a non-volatile solute is added to a pure solvent, the boiling point of the resulting solution
is always greater than that of the pure solvent. The difference between the boiling point of solution (ΔT b)
and that of the pure solvent (Tb0) is called elevation of boiling point (Tb).

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i.e. ΔTb = Boiling point of solution – Boiling point of pure solvent

Or, ΔTb = Tb – Tb0
If we plot a graph between the vapour pressure and temperature, we get the following graphs for the pure
solvent and the solution.
For dilute solutions, the elevation of boiling point is directly proportional to molality (m).
i.e. ΔTb α m
or, ΔTb = Kb.m
Where Kb is a constant called Boiling Point Elevation Constant or Molal Elevation Constant or
Ebullioscopic Constant. It is defined as the elevation of boiling point for 1 molal solution. The unit of K b
is Kkg/mol. For water, Kb = 0.52 Kkg/mol.

w2 × 1000
But, molality, m = -------------
M2 × w1

Kb × w2 × 1000
Therefore, ΔTb = ---------------------
M 2 × w1

where w1 = mass of solvent, w2 = mass of solute, M2 = molar mass of solute. By using this equation, we
can calculate the molar mass of unknown solute.

3. Depression of Freezing point (ΔTf)

According to Raoult’s law, when a non-volatile solute is added to a pure solvent, its vapour pressure
decreases. Now it would become equal to that of solid solvent at lower temperature. Thus the freezing
point of the solvent decreases. The difference between the freezing point (f.p) of pure solvent (T f0)and that
of the solution (Tf) is called depression of freezing point (ΔTf).
i.e. ΔTf = Tf0 - Tf
The vapour pressure – temperature graph representing the freezing point of pure solvent and solution is as
follows:

For dilute solutions, it is found that the depression of freezing point (ΔT f) is directly proportional to
molality (m) of the solution.
Thus, ΔTf α m
Or, ΔTf = Kf.m
where Kf is a constant called Freezing Point Depression Constant or Molal Depression Constant or
Cryoscopic Constant. It is defined as the depression of freezing point for 1 molal solution. The unit of K f
is Kkg/mol. For water, Kf = 1.86 Kkg/mol.

w2 × 1000
But, molality, m = -------------
M2 × w1

Kf × w2 × 1000
Therefore, ΔTf = --------------------
M 2 × w1

where w1 = mass of solvent, w2 = mass of solute, M2 = molar mass of solute. By using this equation, we
can calculate the molar mass of unknown solute.

4. Osmosis and Osmotic Pressure

Osmosis is the process of flow of solvent molecules from pure solvent to solution through a
semipermeable membrane. Or, it is the flow of solvent molecules from lower concentration side to a

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 CHEMISTRY (XI, XII & MEDICAL) by VIJAY KUMAR (M.Sc., B.Ed.)

higher concentration side through a semi-permeable membrane (SPM). A membrane that allows the
passage of solvent molecules only is called a semi-permeable membrane. E.g. egg membrane, all animal
and plant membrane. Cellulose acetate is an example for artificial SPM.

Osmotic pressure is defined as the excess pressure that must be applied on solution side to stop
osmosis. Or, it is the pressure that just stops the flow of solvent molecules. It is denoted by π. It is a
colligative property, since it depends on the number of solute molecules and not on their nature.
For dilute solutions, osmotic pressure is proportional to the molarity (C) and temperature (T).
i.e. π = C.R.T

Here R is the universal gas constant. (R= 0.0821 Latm/K/mol or R = 0.083Lbar/K/mol). But C = n 2/V, the
concentration of the solution.
n2.R.T w2.R.T
Therefore, π = -------- = ------------
V M2.V

Where V is the volume of the solution, w2 is the mass of solute and M2 is the molar mass of solute. Thus
by knowing all other values, we can calculate the molar mass of the unknown solute by this equation.

Advantages of osmotic pressure measurement over other colligative property measurement

1. Osmotic pressure measurement can be done at room temperature, while other colligative property
measurement can be done at either higher or lower temperature than room temperature.
2. Here molarity of the solution is used instead of molality, which can be determined easily.
3. The magnitude of osmotic pressure is large even for very dilute solutions.
4. This method can be used for the determination of molar masses of Biomolecules (which are
generally not stable at higher temperatures) and for polymers (which have poor solubility)

Example for osmosis

• Raw mango placed in concentrated salt solution loses water and shrink.
• Wilted flowers revive when placed in fresh water
• Blood cells collapse when suspended in saline water.
• The preservation of meat by salting and fruits by adding sugar protect against bacterial action.
Through the process of osmosis, a bacterium on salted meat or candid fruit loses water, shrinks
and dies.

Isotonic, hypertonic and hypotonic solutions

• Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.
When such solutions are separated by a semi-permeable membrane, no osmosis occurs. For e.g.
our blood cells are isotonic with 0.9% (mass/volume) sodium chloride solution, called normal
saline solution. So it is safe to inject intravenously.
• A solution having higher osmotic pressure than another is called hypertonic solution. While a
solution having lower osmotic pressure than another is called hypotonic solution.
• If we place our blood cells in a solution containing more than 0.9% (mass/volume) sodium

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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% (mass/volume) NaCl, water will flow into the
cells and they would swell.

Reverse osmosis and water purification

The direction of osmosis can be reversed if a pressure larger
than the osmotic pressure is applied to the solution side.
Now the pure solvent flows out of the solution through the
semi permeable membrane. This phenomenon is called
reverse osmosis and is used in desalination of sea water.
When pressure more than osmotic pressure is applied, pure
water is squeezed out of the sea water through the
membrane. Commonly used SPM is cellulose acetate. A
schematic representation of reverse osmosis is as follows:

ABNORMAL MOLARMASS
The molar mass obtained by colligative property measurement is incorrect, if there is association or
dissociation of particles. Such molar masses are called abnormal molar masses. For e.g. acetic acid
dimerises in benzene due to hydrogen bonding. So the number of molecules in solution decreases and
hence the colligative property decreases and molecular mass increases. In order to correct the abnormal
molar masses, van’t Hoff introduced a factor called van’t Hoff factor (i).
van't Hoff factor (i) is defined as:

Normal Molar Mass Total number of moles of particles after association/dissociation

i = ----------------------------- = --------------------------------------------------------------------------------
Abnormal Molar Mass Total number of moles of particles before association/dissociation

• In the case of dissociation, the value of i > 1 . i = 1 + α(n – 1) where α is degree of dissociation.

• In the case of association, the value of i < 1. i = 1 ― α + α/n where α is degree of association.

Thus for NaCl, i = 2, for K2SO4, i = 3, for CaCl2, i = 3 and for acetic acid which undergo dimerization in
benzene, i = 0.5.

Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows:

1. Relative lowering of vapour pressure,

Δp w2 × M1
--- = i. -----------
p10 w1 × M 2

2. Elevation of Boiling point, ΔTb = i.Kb.m

3. Depression of freezing point, ΔTf = i.Kf.m

4. Osmotic Pressure , π = i.C.R.T

life is chemistry....
dilute your sorrow
evaporate your worries
filter your mistakes
boil your ego
“you will get the crystal of happiness”

SOLUTIONS 10