SOLUTIONS

NCERT Expected Questions 2025 : Example- 1, 4, 5, 7, 9, 10,11 Intext – 3(b), 4, 6, 7, 9,10,11, 12

Exercise – 6, 8, 11, 13, 15, 16,17, 18, 19, 20, 21, 22, 32
BINARY SOLUTION = SOLUTE (2,B-less no. of moles) + SOLVENT(1, A)
THEORY(Prepare Short Notes of Concepts – Use Key Words)
1.Define: Solution, Molarity, Boiling Pt., Freezing Pt., Vapour pressure, Molality(independent of T), Mole
fraction, Ebullioscopic Constant, Cryoscopic Constant,Concentration of Solution.
Molarity : is the no of moles of solute dissolved in per litre of solution.

Molality : is the no of moles of solute dissolved in per Kg of solvent.

Ideal Solutions: are the solutions which obey Raoult’s Law over the entire range of concentration.

Osmosis: is the process of flow of solvent molecules from pure solvent to the solution through a semipermeable
membrane.(Write the applications)

Isotonic solutions:are the 2 different solutions having the same osmotic pressure at same Temperature.

Hypertonic & Hypotonic solutions:A solution which has more osmotic pressure than other solution is hypertonic &
similarly the solution having less osmotic pressure than the other solution is called hypotonic.

Azeotropes-Binary mixtures having the same composition in liquid & vapour phase & boil at a constant temperature.
Types-1.Min. boiling azeotrope = Substances which show a large +ve deviation from Raoults Law-
Eg.95%Ethanol(by Volume)
2.Max. boiling Azeotrope = Substances which show a large -ve deviation from Raoults Law-Eg.68% HNO3(By
Mass).

Solubility- of a substance is its maximum amount that can be dissolved in a specified amount of solvent.

Saturated solution-solution in which no more solute can be dissolved at the same T and P.

Raoult’s law For a solution of volatile liquids, the partial vapour pressure of each component in the solution is
directly proportional to its mole fraction.

Henry’s law(The partial pressure of the gas in vapour phase(p) is proportional to the mole fraction of the gas(x)
in the solution) & its applications(To increase solubility of CO2 in carbonated drinks, To avoid bends, Anoxia).

Colligative properties(C.P’s)- The properties of solutions which depend on the number of solute particles and are
independent of their chemical nature.

Semipermeable membrane(SPM) = is the membrane which allows the solvent molecules to pass through it but prevents
the passage of solute molecules through it.

Reverse Osmosis=is the process in which solvent molecules flows through SPM from the solution towards pure solvent
(in reverse dir. of osmosis) when pressure greater than osmotic pressure is applied on the solution side.

Osmotic Pressure=is the external pressure which must be applied on the solution in order to stop the flow of the solvent
molecules into the solution through a SPM.

2. Ideal & non-ideal solutions – (+ve & -ve deviation with examples & diagram)
IDEAL(obey Raoult’s Law) NON-IDEAL(+VE)(does NON-IDEAL(-VE)(does
not obey Raoult’s law) not obey Raoult’s Law)
Interactions: A-A, B-B=A-B A-A, B-B >A-B A-A ,B-B<A-B
△mixH. = 0 △mixH. > 0 △mixH. < 0
△mixV. = 0 △mixV. > 0 △mixV. < 0
n-Hexane + n-Heptane Ethanol + Acetone Chloroform + Acetone
 FORMULAS(Learn with meaning of each term and conversion factors)
1. Mass ( in g ) 1 cm3(cc) = 1 mL ; 1
No. of moles(mol)= ;( w/M)
(
Molar mass gmol−1 ) dm3 = 1 L

2. No.of moles of solute(mol) unit = mol L–1;

Molarity(M) = ; (n2/V)
Volume (L ) of solution Molarity Dilution
Equation: M1V1=M2V2
3. No.of moles of solute ( mol) unit = mol kg–1
Molality(m) = ; (n2/W1)
Weight of solvent ( kg)
4. No.of moles of component No unit
Mole fraction ( x ) of component = x1 + x2 + x3 + ….. + xn =
Total no. of moles of allcomponents
1
5. Mass of component ppm
Parts per million (ppm) =  106
Total mass of all components
6. Henry’s law → m = KH  P ; KH = Henry’s law constant 1 atm = 101325 Pa =
m = Mass of gas dissolved; P = P of the gas in equilibrium with the 760 torr = 760mm Hg =
solvent 1.01325 bar
P = KH  x (P = KH x S) [1 bar=0.987 atm]
P = Partial pressure of the gas above the solution.
X = mole fraction of the gas in the solution S = Solubility NOTE: 1 litre of water
contains 55.5 mol .
7.   P =Partial Pressure
Raoult’s law (for volatile)→ P i  Partial pressure   xi (Mole fraction) P0 = Pressure of pure
  component
PA = PA x A
o
PB = PB xB ; PTotal = PA + PB (Ideal solution)
o
X =Mole fraction
8. Dalton’s Law: Pi = xi PTotal
PTotal = P1 + P2 +….
9. . Raoult’s law for non-volatile solute (R.L.V.P) P0 = Vapour pressure of
(key words in numerical: non-volatile, solid solute) pure solvent;
P − Ps
o
n2  n  Ps = Vapour pressure of
= ix 2 = if  dilute = 2  solution
P o
n1 + n2  n1 
i=Van’t Hoff factor
10. (Elevation of boiling Point) Tb = Tb − Tb0 ; Tb = i x Kb  m unit of Kb = K kg mol–1
m = Molality(molKg-1),Tb = B.pt of solution, Kb= Ebullioscopic constant i=Van’t Hoff factor
or Molal elevation constant, Tb0 = B. pt of pure solvent
11. .(Depression of Freezing Point) Tf = Tfo − Tf ; Tf = i x Kf  m m= molality(molKg-1),
Tf =F.pt of solution,
R  M1  Tb0
2
R  M1  Tf0
2

Kf = Kb = M1=Molar mass of Tf0 =F.pt of pure solvent,

1000   fusionH 1000   vapourH
Kf = Cryoscopic constant
solvent, △H = Enthalpy Tf = Freezing pt. of pure solvent or Molal depression
Tb = Boiling pt. of pure solvent R = Gas constant constant,
i=Van’t Hoff factor unit of Kf = K kg mol–1
12.
(Osmotic Pressure) 
= i CRT =>  = 2
n (mol)
V (L )
 R  T (in Kelvin) (unit of depends on
unit of R)
Units of R: R = 0.0821 L atm K–1 mol–1 , 0.083 L bar mol-1K-1 , i=Van’t Hoff factor
8.314 x 103PaL K-1 mol-1
13. Calculation of i(Van’t Hoff Factor) i=1(For non-electrolyte)
(M.M)C ( C.P )o i>1 = Dissociation
i= = I < 1 = Assoiciation
(M.M)O ( C.P )C M.M = Molar mass
Normal M.M = Total number of moles after association / dissociation P0 = V.P. of pure
i= component
abnormal M.M Total number of moles before association / dissociation
O = observed
C 2
Ka = Ka= Dissociation/Association constant; C=Molarity/Molality Ps = V.P. of solution
1−  C = calculated
 = Degree of dissociation/association C.P = Colligative property