Class 12th SOLUTIONS chapter ALL formulas

Class 12th SOLUTIONS chapter ALL formulas Based on CBSE book.

The formula of Mass % of a component is-

Mass % of a component =

Mass of the component in the solutionTotal mass of the‘solution×100

The formula of Parts per million is -

Parts per million=

Number of parts of the componentTotal number of parts of all components of the solution×106

The formula of Volume percentage of a component -

Volume percentage of a component=

Volume of the componentTotal volume of solution×100

The formula of Mole fraction of a component -

Mole fraction of a component =

Number of moles of the componentTotal number of moles of all the components

The formula of mole fraction of A -

xA=nAnA+nB

The formula of a solution containing i number of components-

xi=nin1+n2+...+ni=ni∑ni

The formula of sum of all the mole fractions-

x1 + x2 + .................. + xi = 1

The formula of Molarity-

Molarity=moles of soluteVolume of solution in liter

The formula of Molality -

Molality(m)=Moles of soluteMass of solvent in kg

A state of dynamic equilibrium is reached when-

Solute + Solvent ⇋ Solution

Partial vapour pressure of each component of the solution is directly proportional to its mole fraction

p1∝x1

The formula of Partial vapour pressure of component 1-

p1=p01x1

The formula of Partial vapour pressure of component 2-

p2=p02x2

The formula of The total pressures of the components of the solution-

ptotal=p1+p2

ptotal=x1p01+x2p02

=(1–x2)p01+x2p02

=p01+(p02–p01)x2

Dalton’s law of The formula of partial pressures-

p1=y1ptotal

p2=y2ptotal

In general

pi=yiptotal

The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero-

∆mixH=0, ∆mixV=0

The formulas of Relative Lowering of Vapour Pressure

∆p1=p01–p1=p01−p01x1

=p01(1–x1)

∆p1=x2p01 ...(1−x1=x2)

∆p1p01=p01−p1p01=x2

p01−p1p01=n2n1+n2(since x2=n2n1+n2)

p01−p1p01=n2n1

p01−p1p01=w2×M1w1×M2

The formulas of Elevation of Boiling Point -

∆Tb=Tb−T0b

∆Tb∝m or ∆Tb=Kbm

m=w2/M2w1/1000=1000×w2M2×w1

Tb=Kb×1000×w2M2×w1

M2=Kb×1000×w2Tb×w1

The formulas of Depression of Freezing Point

∆Tf=Tf−T0f

m=w2/M2w1/1000=1000×w2M2×w1

Tf=Kf×1000×w2M2×w1

M2=Kf×1000×w2Tf×w1

Kb=R×M1×T2b1000×∆vapH

Kf=R×M1×T2f1000×∆fusH

The formula of osmotic pressure-

Π = (n2 /V) R T

∏V=w2 R TM2 or M2=w2 R T∏V

The Formulas of the van’t Hoff factor-

i=Normal molar massAbnormal molar mass

i=Observed colligative propertyCalculated colligative property

i=Total number of moles of particles after association/dissociationNumber of moles of particles before association/dissociation

The Formula of Relative lowering of vapour pressure of solvent-

p01−p1p01=i.n2n1

The Formula of Elevation of Boiling point-

∆Tb=iKbm

The Formula of Depression of Freezing point-

∆Tf=iKfm

The Formula of Osmotic pressure of solution-

Π=in2RTV

Keypoint:-

Kн = The Henry’s law constant

n_A =Moles of A

x1 = Mole fraction

p01= The vapour pressure of pure component 1.

∆p1 = The vapour pressure of solvent

Ptotal = Total pressure

p1 = the vapour pressure of the solvent,

∆mixH = absorbed or evolved when the components are mixed.

∆mixV = the volume of solution

w1 and w2 = The masses

M1 and M2 = The molar masses

Tb = The boiling point of solution

T0b = The boiling point of pure solvent

∆Tb = The elevation of boiling point.

Kb = The Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).

Tf = The freezing point of solution

T0f = The freezing point of pure solvent

∆Tf = The depression in freezing point.

Kf = The Freezing Point Depression Constant or Molal Depression Constant or Cryoscopic Constant.

Π = The osmotic pressure

R = Gas constant

T = Temperature

V = Volume of a solution in litres.

i = The van’t Hoff factor.