SOLUTION

KEY CONCEPTS
Solution is the homogeneous mixture of two or more substances in which the components are uniformly
distributed into each other. The substances which make the solution are called components. Most of the
solutions are binary i.e., consists of two components out of which one is solute and other is solvent.
Ternary solution consists of three components
Solubility - Maximum amount of solute that can be dissolved in 100g of solvent at particular temp. to
make saturated solution.
Solid solutions are of 2 types -
1. Substitutional solid solution e.g. Brass (Components have almost similar size)
2. Interstitial solid solution e.g. steel (smaller component occupies the interstitial voids)
Expression of concentration of solution
1. Mass percentage= amount of solute present in 100grm solution.
Mass Percentage = ( )
× 100
( )
Percentage by volume is expressed as = ( )
× 100( Liquid- liquid solution)

2. Mole fraction it is the ratio of no. of moles of one of the components to the total no. of moles of all
components. It is expressed as ‘x’. For two component system made of A and B ,
XA=nA/(nA+nB), XB=nB/(nA+nB) , Sum of all the components is 1 ; XA+XB =1
.
3. Molarity (M) = ( )
It decreases with increase in temperature as volume of solution increases with temperature.
.
4. Molality (m) = ( )
No effect of change of temperature on molality as it is mass to mass ratio.
6. Parts per million (ppm) concentration of very dilute solution is expressed in ppm.
Ppm = × 10
Vapour pressure – It is defined as the pressure exerted by the vapour of liquid over the liquid in
equilibrium with liquid at particular temperature .vapour pressure of liquid depends upon nature of
liquid and temperature.
Raoult’s Law –
1. For the solution containing non-volatile solute the vapour pressure of the solution is directly
proportional to the mole fraction of solvent at particular temperature
P A XA
PA = P0A.XA
2. For the solution consisting of two miscible and volatile liquids the partial vapour pressure of
each component is directly proportional to its own mole fraction in the solution at particular
temperature.
PA=P0A. XA, PB=P0B .XB
And total vapour pressure is equal to sum of partial pressure. P total = PA + PB
Ideal solution – The solution which obeys Raoult’s law under all conditions of temperature and
concentration and during the preparation of which there is no change in enthalpy and volume on mixing
the component.
Conditions –
PA = P0A XA, PB = P0B.XB
∆𝐻Mix = 0, ∆𝑉mix = 0
This is only possible if A-B interaction is same as A-A and B-B interaction nearly ideal solution are –
1. Benzene and Toluene
2. Chlorobenzene and Bromobenzene
Very dilute solutions exhibit ideal behavior to greater extent.

Non-ideal solution –
(a) PA ≠ P0A.XA (b) PB ≠P0B.XB
(b) ∆𝐻mix≠ 0 (d) ∆𝑉mix≠ 0
For non-ideal solution the A-B interaction is different from A-A and B-B interactions
i. For solution showing positive deviation
PA > P0A, PB> P0B. XB
∆𝐻Mix = positive, ∆𝑉mix=positive (A-B interaction is weaker than A-A and B-B )
E.g. alcohol and water
ii. For the solution showing negative deviation
PA< P0A.XA, PB<P0B.XB
∆𝐻Mix= negative, ∆𝑉mix = negative
A-B interaction is stronger than A-A and B-B interactions
E.g. Chloroform and acetone, HCl and water
What is Azeotrope? – The mixture of liquids at particular composition which has constant boiling point
which behaves like a pure liquid and cannot be separated by simple distillation. Azeotropes are of two
types:
(a) minimum boiling Azeotrope (mixture which shows +ve deviations ) ex. alcohol and water
(b) maximum boiling Azeotrope (which shows –ve deviations) ex. acetone and chloroform
Colligative Properties - All those properties which depend on the number of solute particles
irrespective of their nature relative to the total number of particles present in the solution. Such
properties are called colligative properties (colligative: from Latin: co means together, ligare
means to bind).
These are: (1) relative lowering of vapour pressure of the solvent (2) depression of freezing point of the
solvent (3) elevation of boiling point of the solvent and (4) osmotic pressure of the solution.

Relative lowering in vapour pressure:

(PoA – PA )/ PoA = 𝜒 ; where 𝜒 = mole fraction of solute

Determination of molar mass of solute

MB=( WA× MA× PoA)/WA×(PoA –PA)

Elevation in Boiling Point

∆TB = Kb. m
Where ∆T B = T’B- ToB
Kb = molal elevation constant
M = molality
MB=(Kb×1000×WB)/∆TB×WA
Depression in Freezing Point:
∆Tf= kf. m
Where ∆Tf = Tf– T’f ;m = molality
Kf = molal depression constant
unit = K.kgmol-1
Osmotic Pressure
When two solutions of different concentrations are separated by semipermeable membrane, there is
spontaneous flow of solvent from dilute solution to concentrated solution. This phenomenon is Osmosis.
The minimum external pressure applied on concentrated solution to stop the flow of solvent is known as
Osmotic pressure. It is denoted by π and is expressed as
 V  nRT
n
 RT
V
  CRT

n = No. of moles; v = volume of solution (L)

R = 0.0821 L atm K-1 mol-1; T = temperature in kelvin.
Isotonic solutions have same osmotic pressure.
Solutions having comparatively higher osmotic pressure are Hypertonic Solutions and those having
comparatively lower osmotic pressure are called hypotonic solutions.
Blood fluid possesses osmotic pressure equal to that of 0.91% solutions of sodium chloride.
Abnormal molecular mass: Molar mass of solute in solution either less or more than the normal molar
mass is called abnormal molar mass. Abnormality in the molar mass arises either due to association or
dissociation of the solute particles in the solution. Such abnormality in the calculation of molecular mass
is overcome by Van’t Hoff factor.
Van’t Hoff factor(i) = Observed Colligative property = Normal molecular mass
Normal Colligative properties Observed molecular mass

Degree of association() = i-1

1/n - 1

Degree of dissociation() = (i-1) / (n-1)

Important Questions

Q1- What do you mean by Henry’s Law? The Henry’s Law constant for oxygen dissolved in water is
4.34×104atm at 25o C. If the partial pressure of oxygen in air is 0.2 atm, under atmospheric pressure
conditions. Calculate the concentration in moles per Litre of dissolved oxygen in water in equilibrium with
water air at 25o C.
Ans: Partial pressure of the gas is directly proportional to its mole fraction in solution at particular
temperature.
PA α XA ; KH = Henry’s Law constant
PA = KH ×XA
KH = 4.34×104 atm
pO2 = 0.2 atm
Xo2= pO2 / KH =0.2 / 4.34×104= 4.6×10-6

If we assume 1L solution = 1L water

n water = 1000/18 = 55.5
XO2 = nO2 /(nO2+ n H2O ) ≅ nO2 /nH2O

nO2 = 4.6 X 10-6 X 55.5 = 2.55 X 10-4mol

M = 2.55 X 10-4 M
Q.2. What is Vant Hoff factor?
Ans. It is the ratio of normal molecular mass to observed molecular mass .It is denoted as ‘i’
i = Theoretical Molar mass / observed molar mass
= Observed no. of particles / Actual no. of particles
= Observed colligative property/Theoretical colligative property
Q.3. What is the Vant Hoff factor in K4[Fe(CN)6] and BaCl2 ?
Ans 5 and 3
Q.4. Why the molecular mass becomes abnormal?
Ans. Due to association or dissociation of solute in given solvent.
Q.5. Define molarity?
Q.6. How molarity is related with percentage and density of solution ?
Ans. M = P x Percentage of density x 10/molar mass
Q.7. What role does the molecular interaction play in the solution of alcohol and water?
Ans. A – A and B – B type interactions are more than A – B type interactions. Positive deviation from
ideal behaviour is observed.
Q.8. What is Vant Hoff factor, how is it related with
a. degree of dissociation b. degree of association
Ans.a. α=i – 1/n-1 b. α = i -1 / 1/n -1
Q.9. Why is NaCl used to clear snow from roads?
Ans. It lowers f.p of water
Q10. Why is the boiling point of solution higher than pure liquid?
Ans. Due to lowering in V.P.
Some special Questions
Q1. Out of 1M and 1m aqueous solution which is more concentrated?
Ans. 1M as density of water is 1gm/ml
Q2. Henry’s law constant for two gases are 21.5 and 49.5 atm, which gas is more soluble?
Ans. KH is inversely proportional to solubility. Hence gas with KH = 21.5 atm is more soluble.
Q.3. Define azeotrope. Give an example of maximum boiling azeotrope.
Q.4. Calculate the volume of 75% of H2SO4 by weight (d=1.8 gm/ml) required to prepare 1L of 0.2M
solution.
Hint: M1 = P x d x 10 /98
M1 V1 = M2V2
14.5ml
Self Assessment

1. We have many types of water purifiers. Zero-B is based on disinfecting properties of iodine. UV
purifier is based on killing of bacteria by UV light. There are R.O. Purifiers, which are being used in
common households.
(a) What is the full form of R.O. Purifier?
(b) What is the function of Porous membrane?
(c) Which method of purification is more economical for countries using sea water, flash distillation
or reverse osmosis for getting potable water. Why?
2. Water is used as coolant in vehicles. In cars, ethylene glycol is used as coolant. In cold countries,
water gets frozen so it cannot act as coolant. Ethylene glycol is added to water so that its freezing
point is lower and it does not freeze. At hill station snow fall takes place. Clearing the snow from
the road is essential for smooth running of traffic-
(a) What is the role of ethylene glycol when added to water?
(b) Why does sprinkling of salt help in clearing snow covered roads in hilly areas? Explain the
phenomenon involved in the process.
(c) How is life of people affected by snow fall?
 SHORT ANSWERS (2 MARKS)

Q.1. How many grams of KCl should be added to 1kg of water to lower its freezing point to -8.0 0C (kf =
1.86 K kg /mol)
Ans. Since KCl dissociate in water completely L=2
∆𝑇𝑓 = 𝑖 𝑘𝑓 × 𝑚 ; m =∆𝑇𝑓/𝑖𝑘𝑓
m= 8 / 2X1.86 = 2.15mol/kg.
Grams of KCl= 2.15 X 74.5 = 160.2 g.
Q.2. With the help of diagram: show the elevator in boiling point colligative properties?
Q.3. what do you mean by colligative properties, which colligative property is used to determine molar
mass of polymer and why?
Q.4. Define reverse osmosis, write its one use.
Ans. Desalination of water.
Q.5. Why does an azeotropic mixture distills without any change in composition.
Hint: It has same composition of components in liquid and vapour phase.
Q.6. Under what condition Vant Hoff’s factor is
a. equal to 1 b. less than 1 c. more than 1
Q.8. Why is it advised to add ethylene glycol to water in a car radiator in hill station?
Hint: Anti- freeze.
Q 9. (a). Define the following terms.
3. Mole fraction
4. Ideal solution
(b)15 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution
frrezez at -0.34 0c . What is the molar mass of material? Kf for water= 1.86 K Kg mol-1 .
Ans. 182.35 g/mol
Q 10.(a) Explain the following :
1. Henry’s law about dissolution of a gas in a liquid.
2. Boling point elevation constant for a solvent
(b) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in in 500 g of
water. The solution has a boiling point of 100.42 0C . What mass of glycerol was dissolved to make
this solution?
Kb for water = 0.512 K Kg mol-1
(hint: Tb = KB x WB x 1000
M B x WA
Ans. 37.73 g
Q 11. 2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point
equal to 1.62 K .Kf for benzene is 4.9 K Kg mol-1. What is the percentage association of acid if it
forms dimer in solution. Ans. 99.2%
Q12. Osmotic pressure of a 0.0103 molar solution of an electrolyte is found to be 0.70 atm at 27 0C.
Calculate Vant Hoff factor.( R=0.082 L atom mol-1 K-1) Ans. 2.76