Coligative property

1. Submitted by
Name of Student : Anshuman Singh
Class : XII – C
Board Roll No. :
Submitted to
Mr . Sandeep Sharma

2. CERETIFICATE
This is to certify that Anshuman Singh bearing Roll No. __________ of class-
XII(science.) Rawal Convent School has successfully completed the project on the
topic “COLLIGATIVE PROPERTY” under the guidance of Mr.Sandeep Sharma
(P.G.T.Chemistry). This project can be approved as a part of C.B.S.E in the session
2022-2023
Name and Signature
Internal Examiner
Name and Signature
External Examiner

3. I wish to express my deep sense of gratitude to them whose
encouragement and cooperation has been a source of
inspiration. It’s an honor to thank our principal Ms. Preeti N
Singh for providing me the opportunity to make this project
and learn through it. I would like to thank Mr. Sandeep Sharma
(PGT Chemistry) for providing us the right direction and for
his cooperation and help in successful completion of our
project. I am indebted to both of them for their help.
Acknowledgement

4. WHAT IS COLLIGATIVE PROPERTY?
Colligative properties are properties of solutions that depend upon the ratio of the
number of solute particles to the number of solvent molecules in a solution, and
not on the type of chemical species present.
The number ratio can be related to the various
units for concentration of solutions. The
independence of the nature of solute particles is
exact for ideal solutions, and approximate for
dilute real solutions. Here we consider only those
properties which result because of the dissolution
of nonvolatile solute in a volatile liquid solvent.
They are independent of the nature of the solute
because they are due essentially to the dilution of
the solvent by the solute.

5. Lowering Vapour Pressure (∆P) of solutions: Roult’s law:-
When a non-volatile solute is
added to a solvent, the vapour
pressure of the solution
decreases. According to
Roult’s Law, the vapour
pressure of a solvent (P1) in a
solution containing a non-
volatile solute is given by
According to Raoult's Law,
Vapour pressure of the pure
solvent = P1
Vapour pressure of the solvent in
solution = P1
P1 = x1P1 °
ΔP1 = P1° - P1
= P1° - x1P1 °
= P1° (1 - x1)
In a binary solution,1 - x1 =
x2
ΔP1 = P1° x2
ΔP1/P1° = (P1° - P1)/P1° = x2

6. The lowering of vapour pressure relative to the vapour pressure of pure solvent is called
relative lowering of vapour pressure. ΔP1/P1° → Relative lowering of Vapour pressure
Thus, the relative lowering in vapour pressure depends only on the concentration of
solute particles and is independent of their identity. If the solution contains more than
one non-volatile solute, then the relative lowering in vapour pressure of a solvent is 8
equal to the sum of the mole fractions of all the non-volatile solutes
If n1 and n2 are respectively the number of moles of the solvent and solute in a binary
solution, then the relative lowering in the vapour pressure of the solvent,
(P1° - P1)/P1° = x1 + x2 + x3 + ... + xn
if n1 and n2 are the number of moles of the solvent and solute,
(P1° - P1)/P1° = n2/(n1+n2)
For dilute solutions n2 << n1
(P1° - P1)/P1° = n2/n1
n1 = W1/M1 , n2 = W2/M2
(P1° - P1)/P1° = (W2xM1)/(W1xM2)
W1 = Mass of solvent
W2 = Mass of solute
M1 = Molar mass of solvent
M2 = Molar mass of solut

7. Boiling point elevation (ΔTB ):-
The exact relation between the boiling point of the solution and the mole
fraction of the solvent is rather complicated, but for dilute solutions the
elevation of the boiling point is directly proportional to the molal concentration
of the solute:
OR ΔTb = Kb.Cm Kb = Ebullioscopy
constant, which is 0.512°C kg/mol for the
boiling point of water. The vapour
pressure of a liquid increases with an
increase in temperature. When vapour
pressure of the liquid becomes equal to
the atmospheric pressure (or) external
pressure, then liquid starts boiling. The
temperature at which the vapour
pressure of the liquid is equal to the
external pressure is known as its boiling
point.

8. The temperatures at which the vapour pressure of the solvent and the solution
become equal to the atmospheric pressure are Tb0 and Tb.
Tb-Tb0 =∆Tb
Thus, it can be seen that the boiling point of a solution is greater than the boiling
point of the pure solvent. The boiling point of a solvent changes as the concentration
of the solute in the solution changes, but it does not depend on the identity of the
solute particles. The elevation of the boiling point depends upon the concentration of
the solute in the solution and is directly proportional to molality (m) of the solute in
the solution.
ΔTb = Tb - Tb °
Tb > Tb °
ΔTb ∝ Concentration of solute
ΔTb ∝ m (Molarity)
ΔTb = Kb m
Kb = Boiling point elevation constant or molal elevation constant or ebullioscopic
constant 11 Molal elevation constant is defined as the elevation in the boiling point
when 1mole of a solute is dissolved in 1kilogram of a solvent. If w2 grams of a solute
with M2 molar mass is dissolved in w1gram of a solvent, then molality (m) of the
solution is,
m = (W2x1000)/(W1xM2)
ΔTb = Kb (W2x1000)/(W1xM2)

9. The freezing point of a substance is defined as the temperature at which its solid
phase is in dynamic equilibrium with its liquid phase. At the freezing point, the
vapour pressure of the substance in its liquid phase is the same as the vapour
pressure of the substance in its solid
Freezing Point Depression (ΔTf ):
When a non-volatile solute is added to a solvent,
the freezing point of the solution gets lowered.
According to Roult’s law, the vapour pressure of a
solution containing a non-volatile solute is lower
than that of the pure solvent. Thus freezing point
of a solvent decreases when a non-volatile solute
is added to it. The depression in freezing point
depends upon the concentration of the solution.
For dilute solutions, depression in the freezing
point is directly proportional to molality (m).
Thus, ∆Tf =Kf m Where Kf =freezing point
depression constant (or) molal depression
constant (or) cryoscopic constant. Molal
depression constant Kf can be defined as the
depression in freezing point

10. when 1mole of solute dissolved in 1kg of solvent. The unit for Kf is kelvin kilogram
/mole. As Kf depends upon the nature of the solvent, its value is different for different
solvents.
The values of Kf can be calculated from this expression
Kf = (R x M1 x Tf 2 )/(1000 x ΔfusH)
R = Gas constant
M1= Molar mass of the solvent
Tf = Freezing point of the pure solvent
ΔfusH = Enthalpy for the fusion of the solvent
If w2 grams of a solute with molar mass M2 is dissolved in w1 grams of a solvent,
then molality m of the solution is given by W2 multiplied by 1,000 divided by w1
multiplied by M2 Substituting this value of molality in the freezing point
depression equation, we get depression in freezing point Molarity ,
m = (W2 x 1000)/(W1xM2)
ΔTf = Kf m
ΔTf = (Kf x W2 x 1000)/(W1xM2)
M2 = = (Kf x W2 x 1000)/(W1xΔTf)
Thus, the molar mass of a non-ionic solute can be calculated by using the
depression in freezing point.

11. Osmotic Pressure
osmotic pressure π of a solution is defined as the excess pressure that must be
applied to a solution to prevent osmosis from taking place. Osmotic pressure does not
depend on the identity of the solute, but on its concentration
Osmotic pressure for dilute solutions is
proportional to molarity of the solution at a
given temperature(T). π ∝ C (at given T)
π = C R T
R = Gas constant
C = n2/V
π = n2RT / V
If W2 grams of solute of molar mass M2 is
present in the solution,
n2 = W2/M2
π = W2RT / M2V
M2 = W2RT / πV
If the solutions have the same concentrations (C1 = C2), then π1= π2

12. BIOGRAPHY
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3. www.icbse.com
4. www.wikipedia.com
5. www.yahoo.co