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Unit 2 (Solutions)
A solution is a homogenous mixture of two or more chemically non-reacting substance whose
composition can be varies within certain limits.




In homogenous mixture all the parties are of molecular size up to 10-19 m diameter. Mixture
cannot be separated by any of the physical methods like filtration, setting or centrifuging.
Every solution is made up of solvent and one or more solutes.
A solution containing only solute dissolved in a solvent called binary solution.
Solution = solvent + solute.
Large Amount
Small Amount
Types of solution:
solute
Solvent
Solid
Liquid
Gas
Solid
Liquid
Gas
Solid
Solid
Solid
Solid
Liquid
Liquid
Liquid
Gas
Liquid
Gas
Gas
Gas
Example
Brace, German silver, bronze
Hydrated salt
Dissolve gas in mineral, H2 in P6.
Salt, glucose in water.
Methanol or ethanol in water
O2 in water.
Iodine vapours in air camphor in
N2 gas.
Humidity in air, chlorofoam
mixed with N2 gas
Air (O2 + N2 )
Molarity: Molarity of a Solution is define as no. of moles of he solute dissolved per litre (dm3) of
solution. It is denoted by M.
𝑀=
No. of germ equivqlent of solute
Volume of solution in litres
Molarity is an intensive poperty of solution. It is not depends on amount of solution.
Normality:Normality of solution is define as the number of gram equivalent of the solute dissolved per
litre (dm3) of given solution. It is denoted by ‘N’.
N=
No. of gram equivalent of solute
volume of solution in litres
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Mass fraction: It is define as the mass of the given component per unit mass solution> it XA and XB
denote the mass fraction of the two component A and B respectively. When wA g ofone component A is
mixed with wB g of second component B is a binary solution.
xA =
wA
wB
and xB =
wA + wB
wA + wB
𝑥𝐴 + 𝑥𝐵 = 1
Part per million (PPM):
for very dilute solution, when a very small quantity of a solute is present in a large quantity of a
solution, the concentration of the mass in the solute is expressed in term of PPM. It is define as the mass
of solute present in one milliom (1016) part of solution.
𝑃𝑃𝑀(𝐴) =
mass of A
× 1016
mass of solution
The pollution of atmosphere is also reported in PPM but it is expressed in term of volue rather than
mass.
Molality: Molality of a solution is defined as thee No. of moles the solute dissolved in 1000 grams (1kg)
of the solvent. It is denoted by’m’.
m=
No. of moles of solute
No. of kilograms of solvent
molarity =
solubility × 10
molecular mass of solute
Solution process involves three steps


Involves the separation of solvent molecules.
Entail the separation of solvent molecules. These steps require energy to break attractive
intermolecular force. Therefore they are endothermic.
 The solvent and solute molecules mix. This process may be exothermic or endothermic.
∆𝐻𝑠𝑜𝑙 = ∆𝐻1 + ∆𝐻2 + ∆𝐻3
 If solute – solvent attraction is stronger than solvent – solvent attraction and solute –
solute attraction then solution process will be favorable. It is exothermic(∆𝐻𝑠𝑜𝑙 < 0).
 If solute- solvent attraction is weaker than solvent – solvent attraction and solute –
solute interaction them he process is endothermic(∆𝐻𝑠𝑜𝑙 > 0 ).
1. Solution of gases in liquids (solubility of gases): The solubility of a gas in liquid is expressed in terms
of absorption coefficient. ‘It is defined as the column of gas in ml that can be dissolved by 1ml of a
liquid solvent temperature of the experiment at one atmospheric pressure’.
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If V is the volume of the gas dissolved, reduced to STP, V is the volume of solvent and P is the pressure
of gas in atmospheres, then absorption coefficient ∝ is given by
∝=
𝑣
𝑉𝑃
v = volume of gas dissolved in liquid
V= volume of solvent
The following factors affect the solubilities of gases in liquid:
1. Temperature: The solubility of most of gases in liquid decreases with increase of temperature as
dissolution is an exothermic process. When water is heated in a beaker, bubbles of air are
formed on the sides of the glass before water boils. As temperature raises the dissolved air
molecule begin to “boil out “.
2. Pressure: the effect of pressure on the solubility of gas in liquid is given by Henry’s law, which
states that “the solubility of a gas in a liquid is directly proportional to pressure of gas over the
solution definite temperature. Thus m is the mass of gas dissolved per unit volume of solvent
and P is the pressure of gas in equilibrium with the solution
𝑚∝𝑃
m = kP
Where k is proportionality constant
𝑚1
𝑚2
=
𝑃1
𝑃2
If solubility of the gas is known at one particular pressure, then it can be calculated at other
pressures using the above relation.
When mixture of two or more non reacting gases is brought in contact with a solvent each and
constituent own partial pressure.
Henry’s law can be applied to each individual gas independent of the presence of other gas.
“At the constant temperature the amount of a given gas that dissolves in a given type and volume of
liquid pressure of that gas in equilibrium with that liquid “
Or
“The mass of a gas which will dissolve into a solution id directly proportional to the partial pressure
of that gas above the solution.
𝑥 ∝𝑃
X= mole fraction of the gas in solution
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P= partial pressure of gas.
X = K’P
Or
P=
1
𝑥
𝐾1
𝐾ℎ 𝑥 { 𝐾𝐻 =
1
}
𝐾1
𝐾𝐻 is called Henry’s law constant.
different value of K H at definite temperature for a given solvent
Limitation of Henry’s law:
1. The pressure is not too high.
2. The temperature is not very low.
3. The gas does not chemically combine with solvent.
Note:
1. Gases are less soluble in aqueous solution of electrolytes-than in pure water. It is called salting
out effect.
2. Non-electrolytes such as sugar if present in solution also reduce the solubility of gases in water.
Application :
1. Producing carbonated beverages like soft drink, ber etc.
2. The solubility of gases increases with increasing pressure. The deep sea divers who breathe
Compressed air must be concerned about solubility of gases in their blood.
3. Generally the scuba divers us the tank filled with 11.7% He, 56.2% 𝑁 2 ,32.1% 𝑂2 .
4. At high altitudes, partial pressure of oxygen is low, therefore concentration of dissolved oxygen in
blood and tissue of people living there is also low. At low concentration of 𝑂2 in blood of the people
suffer from anoxia.
Q 1. If N2 gas is bubbled through water at 293 k, how many mill moles that N2 exerts a partial pressure
of 0.987 bars. Given that Henry’s law constant for N2 at 293 k is 76.48 bars.
Sol. According to Henry’s law
𝑃𝑁2 = 𝑘𝐻 × 𝑥𝑁2
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𝑥𝑁2 =
𝑃𝑁2
0.987
=
𝑏𝑎𝑟 = 1.29 × 10−5
𝑘4
76480
If N moles of N2 are present in 1 L of water -(55.5 moles)
𝑥𝑁2 =
𝑛
𝑛
(𝑎𝑠 𝑛 ≪ 55.5)
=
𝑛 + 55.5 55.5
𝑛
= 1.29 × 10−5
55.5
𝑛 = 1.29 × 10−5 × 55.5 𝑚𝑜𝑙
n= 0.716 milli moles
Q 2. At what partial pressure oxygen will have solubility of 0.05g L-1 in water at 293k? Henry’s constant
(𝑘𝐻 ) for o2 in water at 293k is 34.86k bar. Assume the density of the solution to be same as that of
solvent.
𝑛𝐻
Sol.
2𝑂
𝑛𝑂 2 =
𝑥𝑂 2 =
= 55.5 𝑚𝑜𝑙𝑒𝑠
0.05 𝑔
= 1.56 × 10−3 𝑚𝑜𝑙𝑒
32
𝑛𝑜 2
𝑛𝑜2 +𝑛𝐻2 𝑜
=
1.56 × 10−3
1.56 × 10−3 + 55.5
𝑥𝑂2 = 2.89 × 10−5
By Henry’s law,
𝑃𝑂2 = 𝑘𝐻 × 𝑥𝑂2
= (34.86 × 103 ) × (2.81 × 10−5 )
=0.98 bar
Vapour pressure of the solution and Raoult’s law.
When a small amount of a non-volatile solute is added to the liquid (solvent) to form a solution, it is
observed that vapours of the solvent above the solution are present as usual but their number is less I
comparison to when a pure solvent is taken under similar condition.
“Vapour presure of the solution is less than the vapour pressure of pure solvent.” This is explained on
the basis of surface area of liquid (solvent) from which evaporation occurs. In the case of a solution a
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part of the liquid surface is occupied by no-volatile solute particles. Therfore evaporation of the liquid
occurs from a lesser surface area as show.
Thus it is always true that whenever a non-volatile solute is added to a solvent the vapour pressure of
the resultant solution will be less thann the vapour pressure of solvent.
Raoult’s law: The french scientist francois maria raoult carried out a series of experiment to study
vapour pressure of binary solution. In 1886
1. Solutions conatining non volatile solutes: The vapour pressure of a solutions is due to the
solvent ony as solute being a non-volatile component has no contribution. Acccoording to
Raoult’s law: The vapour pressure of a solution conataining non – volatile solute is proportional
to mole fraction of the solvent.
PA ∝ xA or,
PA=kxa
Where k is proportinally constant
For pure liquid(solvent)
XA=1 then k become equal to vapour presuure of pure solvent which is equl to 𝑃𝐴 ° ,thus PA=𝑃𝐴 ° × 𝑋𝐴
Vapour pressure of solution = vapour pressure of pure solvent X mole fraction of solvent
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2. Raoult’s law for binary solutions of volatile liquid:
Let a solution consist of two miscible volatile liquid A and B
According to this law,” the partial pressure of any volatile constituent of a solution at a costant
temperature is equal to the mole fraction o f that constituent in the solution”. Let a mixture
(solution) be prepared by mixing nA moles of liquid A and nB moles of liquid B.
Let PA and PB be the partial pressure of two constituent A and B in solution and P°A and PB° are
vapour presures in pure state respectively.
Thus according to Raoult’s law,
PA=xAPA° and
PB=xBP°B
If P is the total pressure of system at the same temperature,then
P=PA+PB
P=xAP°A + xBP°B
Relation between Dalton’s law and Raoult law:
The compostion of the vapour is equilibrium with solution can be calculated by applying Dalton’s
law of partial pressure.
Let mole fraction of vapours A and B be yA and yB respectively, Let PA and PB be the partial pressures
of vapour A and B and total pressure is P.
PA = YAP…………………….1
PB = YBP……………………..2
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PA= xAP°A…………………………………3
PB = xBP°B……………………4
Eq 1. and Eq 3.
YAP = xAP°A ,or
YA =
𝑥𝐴 𝑃°𝐴 𝑃𝐴
=
𝑃
𝑃
Similarily eq 2. And eq 4.
YB =
𝑥𝐵 𝑃°𝐵
𝑃
=
𝑃𝐵
𝑃
In general for a solution containing a number of volatile components (liquid), for any component i
Pi = yi × PTOTAL
Characteristic of an Ideal Solution
Ideal solution obeyig Raoult’s law
i.
Volume change on mixing the two component should be zero, ∆Vimmix = 0.
𝑉𝑠𝑜𝑙𝑣𝑒𝑛𝑡 + 𝑉𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑉𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
ii.
Heat change on mixing to component to form a solution should be zero
∆𝐻𝑚𝑖𝑥 = 0
Neither heat is evolved nor absorbed when te solution is prepared.
iii.
There should be no chemica reaction between solvent and solute.
Solvent + Solvent  Soution.
H2O + NH3 NH4OH
H2O + CO2  H2CO3
Non ideal solution
H2O + CaO Ca(OH)2
iv.
Solute molecule should not disassociate in the ideal solution.
NaCl  Na+ + Cl2—
Aqueous
Non ideal.
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H2SO4  2H+ + SO4
Example of ideal Solution:
a)
b)
c)
d)
e)
f)
Benzene and toluene
Carbon tetrachloride and silicion tetrachloride
N-Hexane and n-Heptane
Ethylene dibromide and ethylene dichloride.
Chorobenzene and bromobenzene.
Ethyle chloride and ethyle bromide
P = PA + PB
{PA + PB = 1}
It is clear that the vapour pressure of solution of different composition in case of ideal solution lies b/w
vapour pressure of pure component (𝑃𝐴 ° & 𝑃𝐵 °)
P = PA+PB = 𝑃𝐴 °𝑥𝐴 + 𝑃𝐵 °𝑥𝐵
= 𝑃𝐴 °(1 − 𝑥𝐵 ) + 𝑃𝐵 °𝑥𝐵
= 𝑥𝐵 (𝑃𝐵 ° − 𝑃𝐴 °) + 𝑃𝐴 °
This equation is type y = mx + c in which m is the slope and c is the intercept of y axis.
Q.1. the vapour pressure of pure liquid A and pure liquid B at 20℃ are 22 and 75mm Hg respectively. A
solution is prepared by mixing equal moles of A and B. assuming the solution to be ideal, calculated the
vapour pressure of solution
Sol. Given 𝑃𝐴 ° = 22𝑚𝑚 𝐻𝑔
𝑃𝐵 ° = 75𝑚𝑚 𝐻𝑔
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Since equal moles of both liquids are mixed.
1
Hence, xA = xB = 2 (Given)
P= PA + PB
= 𝑃𝐴 °𝑥𝐴 + 𝑃𝐵 °𝑥𝐵
1
2
1
2
22 × + 75 × = 48.5 𝑚𝑚 𝐻𝑔.
Q.2. Methanol and ethanol from nearly ideal solution at 300K. a solution is made by mixing 32g
methanol and 23 g ethanol at 300 K. calculate the partial pressure of its constituent and the total
pressure of solution. Given at 300K, 𝑃°𝐶𝐻 𝑂𝐻 = 90𝑚𝑚 𝐻𝑔 , 𝑃°𝐶2 𝐻5 𝑂𝐻 = 51𝑚𝑚 𝐻𝑔
3
Sol. Let 𝐶𝐻3 𝑂𝐻 is liquid A and C2H5OH is liquid B.
𝑤
32
23
nA=𝑚𝐴 = 32 = 1
nB =46 = 0.5
𝐴
xA = 𝑛
𝑛𝐴
2
𝐴 +𝑛𝐵
2
2
= 1+0.5 = 3
1
xB =1-3 = 3m
P = PA +PB
= P°A + xA + P°B xB
2
1
= 90 × + 51 ×
3
3
P = 60+17 =77mm.Hg
𝑃𝐴 = 60mm , PB = 17mm
Collegative properties:
The properties depend on the ratio of solute to solvent in the solution and get numerically large or
smaller if the ratio changes.
Collegative properties depends on the number of particle found in the solute an on the solvent in a
solution. Collegative properties do not depend on the identity of solute or solvent.
Properties:
I.
II.
Lowering in vapour pressure.
Elevation in boiling point depression in freezing point
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III.
Osmotic pressure
Collegative properties or the properties of dilute solution that is why these are termed as colligative
properties of dilute solution.
Properties for the collegative properties:
I.
II.
III.
The solution should be very dilute.
The solute should be non-volatile.
The solute should not dissociate or associate in solution.
Relative lowering in vapour pressure:
When a non-volatile solute is added to a solvent, its vapour pressure is lowered; the vapour pressure of
solution is always less than vapour pressure of solvent at some temperature. It 𝑃0 is the vapour pressure
of solution at a given temperature.
𝑃0 − 𝑃 =Lowering in vapour pressure = ∆𝑃
And
𝑃0 −𝑃
𝑃0
=Relative lowering in vapour
Pressure =
∆𝑃
𝑃0
According to Raoult’s law:
𝑃 = 𝑃°𝑥𝐴
(𝑥𝐴 = mole of solvent)
∆𝑃 = 𝑃° − 𝑃
= 𝑃° − 𝑃°𝑥𝐴
= 𝑃°(1 − 𝑥𝐴 ) 𝑜𝑟
𝑥𝐵 =
𝑛𝐵
𝑛𝐴 + 𝑛𝐵
∆𝑃
𝑛𝐵
=
0
𝑃
𝑛𝐴 + 𝑛𝐵
P°
−1
P0 −P
P0 −P
P
∆𝑃
= (1 − 𝑥𝐴 ) = 𝑥𝐵 = mole fraction od solute
𝑃°
=
n
= nA
B
or
P0
nA + nB nA
=
=
+1
0
P −P
nB
nB
nA
nB
wB = weight of solution
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Or,
𝑃0 −𝑃
𝑃
=
𝑛𝐴
𝑛𝐵
P0 − P wB mA
=
×
P
mB wB
𝑛𝐵 =
𝑤𝐵
𝑚𝐵
𝑛𝐴 =
𝑤𝐴
𝑚𝐵
P0 = vapour pressure of solution.
This relationship is useful in the determination of molecular mass of the dissolved by measuring relative
lowering in vapour pressure.
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