Based on Maharashtra state Board this is upto Relative lowering of vapour pressure

1. SOLUTIONS [5/7]
Solution : A homogeneous mixture of two or more non-reacting substance is called
solution. (Solute(s) + Solvent = Solution)
Solute: Component of the solution which is present in smaller quantity is called solute.
Solvent: Component of the solution which is present in larger quantity is called solvent.
Note:
1. In a syrup (liquid solution) containing 60% sugar (solid) 40% water (liquid),
2. In such solution water is termed as the solvent
3. A solution of alcohol and water, larger quantity by mass = solvent.
4. If both liquids have same masses, it is difficult to differentiate.
True Solution :
A homogeneous mixtures of two or more substances, whose composition can be varied
within certain limits. The dissolved substances may be molecules or ions with size of the
order 10-8 cm.
Molecules : sugar , urea, etc. Ions : NaCl, NaOH, H2SO4, AgNO3, etc.
True solution can’t be separated into compounds by filtration, settling or centrifugal
action.
Solution containing 2 components are termed as binary solution.
Similarly 3 components – ternary solution, 4 components – quaternary solution.
Physical Chemistry NSMATHCHEM Solutions

2. Saturated solution:
The solution containing maximum amount of solute dissolved in given amount of solvent.
Dissolution ⇌ Crystallisation
Unsaturated solution:
A solution which contains less amount of solute than required for formation of saturated
solution. Here no equilibrium between dissolution and crystallization.
Dissolution > Crystallisation
Supersaturated solution:
It contains excess of solute than required for formation of saturated solution. In this
case, added excess solute does not dissolve.
Dissolution < Crystallisation
Aqueous solution:
The solution, which is prepared in Water.
Eg : NaOH in water.
Non aqueous solution:
The solution which is prepared in other Solvent (except water).
Eg : NaOH in alcohol.
SOLUTIONS DEPENDING ON SOLVENT
CAPACITY OF SOLUTION TO DISSOLVE SOLUTE
Physical Chemistry NSMATHCHEM Solutions

3. Types Of Solutions Depending Upon Physical States Of The Solute And The Solvent
Solute Solvent Examples
Gas Gas Air, mixtures of N2 & O2
Liquid Gas Rain, Chloroform &Nitrogen mixtures, humidity etc
Solid Gas Camphor , Naphthalene OR iodine in air
Gas Liquid Oxygen dissolved in water, CO2 in areated drinks
Liquid Liquid Ethanol dissolved in water, mixtures of all mixable liquids
Solid Liquid Aqueous solution of sugar, salt, urea etc
Gas Solid Absorption of gas on metal coal, H2 gas in Pd metal.
Liquid Solid Hg with Na. hydrated salt
Solid Solid Alloy Cu dissolved in gold bronzes etc.
Physical Chemistry NSMATHCHEM Solutions

4. Let, WA = Solute & WB = Solvent
∴ mass percentage of solute (A) =
𝑊 𝐴
𝑊 𝐴+𝑊 𝐵
x 100
Eg.
5% aqueous solution of sugar means 5gm sugar
is present in 100 gm of solution.
Ways of expressing concentration of a solution
Concentration of solution:
The amount of solute present in the given quantity of the solution.
Percentage by mass (W/W):
It is the mass of solute is grams present in the 100 grams of the solution.
∴ Percentage by mass =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100
M𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒+𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
x 100
Physical Chemistry NSMATHCHEM Solutions

5. NOTE:
Total volume of solutions may not be equal to the sum of volumes of
solute and solvent as some of the solute particles may occupy space.
Volume is temperature dependant quantity , hence percentage by
volume changes with change in temperature.
Percentage by volume (V/V) :
It is the ratios of number of parts by volume of the solute to 100 parts
by volume of the solution.
∴ % by volume =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100
Let VA & VB be the volume of solute & solution in volume
∴ Percentage by volume (A) =
𝑉 𝐴
𝑉 𝐵
x 100
Physical Chemistry NSMATHCHEM Solutions

6. Mole fraction (X):
The ratio of number of moles of a components to the total number of moles of all
component present in the solution is called mole fraction.
Explanation:
If solution containing n1 & n2 moles of solute & solvent respectively.
Then, mole fraction of solute (X1) & solvent (X2) is given by
𝐗 𝟏 =
𝐧 𝟏
𝐧 𝟏 + 𝐧 𝟐
Sum of mole fraction of all components of solution is always unity
∴ X1 + X2 =
n1
n1+n2
+
n2
n1+n2
=
n1+n2
n1+n2
= 1
𝐗 𝟐 =
𝐧 𝟐
𝐧 𝟏 + 𝐧 𝟐
OR
Physical Chemistry NSMATHCHEM Solutions

7. Molality (m)
The number of moles of solute present in 1 kg of the solvent is called molality
Molality =
number of moles of solute
mass of solvent in Kg
Molality =
number of moles of solute
mass of solvent in g
x 1000
Unit of molality = mol/Kg
Molarity (M)
The number of moles of solute present in 1 litre of the solution is called molarity.
Molarity =
Number of moles of solute
volume of solution in litre
Molarity =
Number of moles of solute
volume of solution in ml
× 1000
Unit of molarity = mol/dm3 OR mol/lit
Physical Chemistry NSMATHCHEM Solutions

8. Solubility : The maximum amount of solute which dissolves completely in a given amount of
solvent at a constant temperature is called solubility of the solute in a given solvent.
It is expressed in grams of solute per 1Lof the solvent or mol/dm3.
Solubility : Temperature dependent.
Effect of Temperature on Solubility :
The solubilities of gases, liquid and solid solutes are affected by change of temperature.
Solubility of solid in liquid solvent and Temperature :
Regularly solubility of solid in liquid increases with increase in temperature.
The solubility of solid solute in liquid solvent may be exothermic or endothermic process.
Depending on the nature of process, temperature has effect on solubility.
Effect of pressure on solubility :
The solids are incompressible, hence change in pressure has no effect on the solubility of solids in
liquids.
Solubility of gases in liquids :
The gases are soluble in liquids ,Water. The gases like O2, N2 have low solubility in water as they are
non-polar. CO2 and NH3 are more soluble in water as they react with water to form H2CO3(Carbonic
Acid) and ammonium hydroxide.
As gases are highly compressible then the increase in external pressure of gas increases its
solubility.
Physical Chemistry NSMATHCHEM Solutions

9. Effect of temperature on solubility of gas
According to Charles’ law, volume of a given mass of a gas directly varies with temperature. Hence
with rise in temperature, solvent in solution can’t accommodate gaseous solute, as a result gas
bubbles out. Thus solubility of gas in liquid decreases with increase in temperature.
Effect of pressure on the solubility of gases
Henry’s law
The solubility of a gas in a liquid at constant temperature is proportional to the
pressure of the gas above the solution.
Explanation
If S – Solubility of gas (mol/dm3).
P – Pressure of the gas in atmosphere,
According to Henry’s law.
S ∝ P
∴ S = K.P where ‘K’ is Henry’s constant
∴ S = K Where P = 1 atm,
Unit of K: mol dm-3 atm-1
Definition of K
It is the solubility of gas (mol/dm3), At 1 atm. pressure at reference temperature.
Physical Chemistry NSMATHCHEM Solutions

10. Effect of addition of soluble salt on solubility of gas
The solubility of dissolved gas is reduced by addition of soluble salt to the solution of gas. For example, addition of
table salt to carbonated soft drink decreases the solubility of CO2, thus it escape into atmosphere with
effervescence.
In Short
i) Solubility of solid solutes increases with temperature (except Na2SO4)
ii) Change of pressure has no effect on solubility of solids in liquids.
iii) Increase of external pressure increases the solubility of gas.
iv) Solubility of gas in liquid decreases with increase in temperature.
v) Solubility of dissolved gas is suppressed when a soluble salt is added to the solution of gas.
Vapour pressure of Solutions of Liquid in Liquuid:
The pressure exerted by the vapours in equilibrium with liquid at a given temperature is called vapour
pressure of liquid.
Vapour pressure of liquid depends upon following factors
Nature of liquid:
Volatile liquids having more vapour pressure than non-volatile liquid.
eg. dimethyl ether has more vapour pressure than ethyl alcohol.
Temperature:
Vapour pressure of solvent is directly proportional to the temperature. At constant temperature vapour
pressure remains constant.
At the boiling point of solvent its vapour pressure is equal to the atmospheric pressure.
Purity of liquid:
Pure liquid always has more vapour pressure than its solution. Physical Chemistry NSMATHCHEM Solutions

11. Raoult’s law (for volatile liquids)
In a solution of volatile liquids, the partial vapour pressure of each component in
the solution is directly proportional to its mole fraction at constant temperature.
Consider a solution containing two volatile components A and B with mole fractions x1 and x2
respectively.
Let P1
0
and P2
0
be the vapour pressure of the pure components A and B respectively. According to
Raoult’s law, the partial pressures P1 and P2 of the two components in the given solution are
given by.
P1 x1 and P2 x2
P1 = P1
0
x1 and P2 = P2
0
x2 ------ (1)
And total vapour pressure, PT of solutions of two volatile components is the sum of partial vapour
pressure of the two components.
Hence, PT = P1 + P2
= P1
0
x1 + P2
0
x2 --- from (1)
For binary solutions,
x1 + x2 =1 OR x2 = 1 – x1
∴ PT = P1
0
x1 + P2
0
(1-x1)
= P1
0
x1 + P2
0
- P2
0
x1
= P2
0
+ (P1
0
- P2
0
) x1
The solution which obeys Raoult’s law over the entire range of concentration is called an ideal
solution.
If a solution does not obey Raoult’s law, the solution is non-ideal. Physical Chemistry NSMATHCHEM Solutions

12. Composition of vapour in equilibrium with the
solution can determined by Dalton’s law.
Let,
y1 = mole fraction of A in vapour
y2 = mole fraction of B in vapour
P1 = partial pressure of A in vapour
P2 = partial pressure of B in vapour
P = total vapour pressure
∴ According to Daltons law,
P1 = y1P
P2 = y2P
Therefore we have to determine the values of y1 & y2
Composition of Vapour phase
Physical Chemistry NSMATHCHEM Solutions

13. Vapour pressure
Mole fractionX1 = 1
X2 = 0
X1 = 0
X2 = 1
PT/S = P1 + P2
P2 =𝑃2
0
X2
𝑃2
0
P1 =𝑃1
0
X1
𝑃1
0
Ideal deviation/ Ideal solution
1. Obey Raults law at every range of
concentration.
2. ∆Hmix = 0, Neither heat evolved nor
absorbed.
3. ∆𝑉 𝑚𝑖𝑥 = 0, Sum of volumes of component
is equal to total volume of solution.
4. PT/S = P1 + P2 = 𝑃1
0
X1 + 𝑃2
0
X2 (i.e. P1 =𝑃1
0
X1
; P2 =𝑃2
0
X2)
5. A-A, A-B, B-B interactions should be
same (i.e. A & B are identical in shape,
size and character)
6. Escaping tendency of A & B should be
same in pure liquids & in the solution.
7. Ex.
i. Dilute solutions
ii. Benzene + Toluene
iii. n-hexane + n-heptane
iv. Chlorobenzene + Bromobenzene
v. Ethyl bromide + Ethyl chloride
vi. N-butyl chloride + n-butyl bromide
Physical Chemistry NSMATHCHEM Solutions

14. X1 = 1
X2 = 0
X1 = 0
X2 = 1
PT/S = P1 + P2
P2 =𝑃2
0
X2
𝑃2
0
P1 =𝑃1
0
X1
𝑃1
0
Non-Ideal negative deviation
1. Do not obey Raults law.
2. ∆Hmix < 0, Exothermic dissolution; heat
is evolved.
3. ∆𝑉 𝑚𝑖𝑥 < 0,Volume is decreased after
dissolution.
4. P1 + P2 <𝑃1
0
X1 + 𝑃2
0
X2
P1 < 𝑃1
0
X1 ; P2 < 𝑃2
0
X2
5. A-B attractive force should be greater
than A-A & B-B interactions. (i.e. A & B
are different shape, size and character)
6. Escaping tendency of A & B is lowered
showing lower vapour pressure than
expected.
7. Ex.
i. acetone + aniline
ii. water + chloroform
iii. water + HCl
iv. Acetic acid + Pyridine2
v. chloroform + benzene
vi. chloroform + diethyl ether
Physical Chemistry NSMATHCHEM Solutions

15. X1 = 1
X2 = 0
X1 = 0
X2 = 1
PT/S = P1 + P2
P2 =𝑃2
0
X2
𝑃2
0
P1 =𝑃1
0
X1
𝑃1
0
Non-Ideal positive deviation
1. Do not obey Raults law.
2. ∆Hmix > 0, Endothermic dissolution; heat
is absorbed.
3. ∆𝑉 𝑚𝑖𝑥 > 0,Volume is increased after
dissolution.
4. P1 + P2 > 𝑃1
0
X1 + 𝑃2
0
X2
P1 > 𝑃1
0
X1 ; P2 >𝑃2
0
X2
5. A-B attractive force should be weaker
than A-A & B-B interactions. (i.e. A & B
are different shape, size and character)
6. A & B escape easily showing higher
vapour pressure than the expected.
7. Ex.
i. acetone + ethanol
ii. water + methanol
iii. water + ethanol
iv. acetone + CS2
v. acetone + benzene
vi. cyclohexane + ethanol
Physical Chemistry NSMATHCHEM Solutions

16. Colligative Properties Of Nonelectrolyte Solutions
The properties of dilute solutions (0.2M or less) which depends on the number of solute
particles and not on their nature are called colligative properties of the solution.
Important Conditions For Colligative Properties
a) The solution should be very dilute (0.2M or less).
b) The solute should be non – volatile whereas solvent is volatile.
c) The solute does not associated or dissociated in the solution.
As colligative properties are depend on number of solute particles, then observed value of colligative
properties are.
a) Normal value: When solute molecules neither associated nor dissociated.
b) Higher than normal value : When solute molecules undergoes dissociation in solution.
c) Lower than normal value : When solute molecules undergoes association in solution.
Eg. 2C6H5COOH ⇌ (C6H5COOH)2
Benzoic acid dimer
The Four Important Colligative Properties Are
1) Vapour pressure lowering (P1
0
-P)
2) Boiling point elevation (∆Tb)
3) Freezing point depression (∆Tf)
4) Osmotic pressure (𝜋)
Physical Chemistry NSMATHCHEM Solutions

17. Vapour Pressure Lowering (𝐏𝟏
𝟎
-P)
Vapour pressure: The pressure exerted by the vapours in equilibrium with liquid at a given
temperature is called vapour pressure of liquid.
Vapour pressure Lowering (𝐏𝟏
𝟎
-P)
The difference between the vapour pressure of pure solvent and that of its solution is called lowering of
vapour pressure.
If P1
0
= Vapour pressure of pure solvent
P = Vapour pressure of solution.
∴ Vapour pressure lowering is
ΔP = P1
0
-P
Raoult’s Law For A Solution Of Non-volatile Solute
Consider a solution of two components A(Volatile)
and B(non-volatile Solute)
x1 = Mole fraction of A; x2 = Mole fraction of B
P1
0
= V.P. Of pure component A;
P2
0
= V.P. of pure component B
= 0.
We know,
𝑃 𝑇/𝑆 = P2
0
+ (P1
0
- P2
0
) x1
𝑃 𝑇/𝑆 = 0 + (P1
0
- 0) x1 (∵ P2
0
= 0)
∴ 𝑃 𝑇/𝑆 = P1
0
x1 = 𝑃1
This is Raoult’s law
For binary solution
x1 + x2 =1
x1 = 1 - x2
𝑃𝑆 = P1
0
𝑥1
= P1
0
(1 - x2)
= P1
0
- P1
0
x2
P1
0
x2 = P1
0
−𝑃𝑆
P1
0
x2 = ∆P
∆P = P1
0
x2
Vapour pressure lowering depends on mole
fraction of solute (i.e. number of solute).
Therfore it is colligative property.
Physical Chemistry NSMATHCHEM Solutions

18. Relative (Fractional) lowering of vapour pressure
∆𝑷
𝐏 𝟏
𝟎
The ratio of the lowering of vapour pressure to the vapour pressure of pure solvent.
Therefore relative lowering of vapour pressure =
∆𝑃
P1
0 =
P1
0−𝑃
P1
0
∆P
P1
0 =
P1
0−P
P1
0 =
P1
0 x2
P1
0 = x2 …… (1)
Relative lowering of vapour pressure is also depends on mole fraction of solute (i.e. number of solute).
Therfore it is colligative property.
Molar mass of solute and relative lowering of vapour pressure
Let W2 g of solute of molar mass M2 be dissolved in W1 g of solvent of molar mass M1. Hence number
of moles of solvent, n1 and number of moles of solute n2, in solution are given as,
n1 =
W1
M1
and n2 =
W2
M2
…….(2)
The mole fraction of solute, x2 is given by,
X2 =
n2
n1+n2
=
W2/M2
W1/M1 + W2/M2
…….(3)
from equation (1),(2) and (3)
∆𝑃
𝑝1
0 =
𝑝1
0−𝑝
𝑝1
0 = x2 =
W2/M2
W1/M1 + W2/M2
For dilute solution n1 >> n2. Hence n2 may be neglected in comparison with n1 in equation (2) and
equation (3) takes the form.
∆𝑃
𝑝1
0 =
𝑛2
𝑛1
=
𝑊2/𝑀2
𝑊1/𝑀1
=
𝑊2 𝑀1
𝑊1 𝑀2
Physical Chemistry NSMATHCHEM Solutions

19. ΔP = P1
0
-P
∆P = P1
0
x2
∆𝑃
𝑝1
0 =
𝑊2 𝑀1
𝑊1 𝑀2
∆𝑃
𝑝1
0 = x2
=
W2/M2
W1/M1 + W2/M2
Formula
ΔP = V.P. Lowering
P1
0
= V.P. Of pure component(Solvent)
P = V.P. of Solution
X2 = Mole fraction of solute
W2 = Mass of solute; M2 = Molar Mass of solute
W1 = Mass of solvent; M1 = Molar Mass of solvent
∆𝑃
𝑝1
0 = Relative lowering of V.P.
𝑃
𝑝1
0 = Relative V.P.

20. Physical Chemistry NSMATHCHEM Solutions
In an experiment, 18.04 gm of mannitol
were dissolved in 100gm of water. The
v.p. of water was lowered by 0.309mm Hg
from 17.535 mm Hg. Calculate molar
mass of of mannitol.
The v.p. of 2.1% solution of non-volatile
solute in water at 373k is 755mm Hg.
Calculate the molar mass of solute.

21. Physical Chemistry NSMATHCHEM Solutions
Calculate the mass of solute (molar mass
40) which is dissolved in 114 X 10-3 kg of
octane to reduce its v.p. to 80%