Slide 1 : 6.023 × 1023
in terms of particle 22.4 lts
of Gas at s.t.p. 1 gm. atom 1 gm. Molecule
of substance 1 gm. formula
of mass MOLE Mole concept can be calculated as …. For all your chemistry problems call 09356060065 ….
Slide 2 : Like dozen means 12 things
IT may be mango , grapes, hydrogen, men…..
It means 12 number
irrespective of their nature About Mole …. For all your chemistry problems call 09356060065 ….
Slide 3 : Likewise mole is just a number
That is 6.023 × 1023
Avogadro”s number
irrespective of there nature About Mole …. For all your chemistry problems call 09356060065 ….
Slide 4 : Mass of one particle M amu
M = molecular mass of a molecule
or atomic mass of an atom
or formula wt of ionic compound
or ionic wt of an ion
amu = atomic mass unit
Mass of one mole M grams
i. e. 1 amu = 1/N0 grams About Molecular mass … For all your chemistry problems call 09356060065 ….
Slide 5 : n N 22.4lt n = no. of moles W = wt. of particle M = mol. wt. of particle
N = number of particles N. = avogadro,s number 6.023 x 1023 V = Volume in lt N. V W M D C B A A B C D method to calculate mole …. For all your chemistry problems call 09356060065 ….
Slide 6 : Solute
(B) Solvent
(A) Solution
(AB) + = Binary solution Like sugar + water = sweet solution
!!!!! Represantation of solution …. For all your chemistry problems call 09356060065 ….
Slide 7 : Mole fraction can be calculated as …. 4Mole fraction. The mole fraction of any component in the solution is equal to the number of moles of that component divided by the total number of moles of all the components. For a solution containing nb moles of the solute dissolved in na moles of the solvent, 4Mole fraction of solute in the solution (Xb) = x1 + x2 = 1 4Mole fraction of solvent in the solution (Xa) = For all your chemistry problems call 09356060065 ….
Slide 8 : Mass percent or weight percent (w/w%). 4Mass percent of a solute in a solution is
the mass of the solute in grams
present in 100 g of the solution. For all your chemistry problems call 09356060065 …. mass of the solute present
mass percent = --------------------------------- × 100
mass of the solution
Slide 9 : The strength of a solution is defined as the amount of the solute in grams present per litre of the solution (i.e. g/L or g L-1). For all your chemistry problems call 09356060065 …. Strength… mass of the solute present
Strength = -------------------------------------
Vab in ml
Slide 10 : nb Wb x 1000 normality
Molarity = ---------- x 1000 = ----------------- = -------------
Vab in ml Mb x Vab in ml n factor
(A/B/e/V) It is defined as the number of moles of the solute present per litre of the solution.
It is represented by the symbol, M For all your chemistry problems call 09356060065 …. Molarity
Slide 11 : It is defined as the number of moles of the solute present
per kg of the solvent.
It is represented by the symbol, m. Molality For all your chemistry problems call 09356060065 …. nb Wb x 1000
Molality = ----------- x 1000 = -----------------
Wa in gm Mb x Wa in gm
Slide 12 : It is defined as the number of gram equivelents of the solute
present per litre of the solution.
It is represented by the symbol, N. Normality. For all your chemistry problems call 09356060065 …. no. of gm equivelents of solute
Normality = ------------------------------------- x 1000
Vab in ml = Molarity x n factor (A/B/e/V)
Slide 13 : Wa in gm
= ----------------
equivelent wt no. of gm equivelents… For all your chemistry problems call 09356060065 ….
Slide 14 : Molecular wt in gm
= ------------------ ----
n factor (A/B/e/V) equivelent wt. For all your chemistry problems call 09356060065 ….
Slide 15 : 4 B, Basicity is the number of displaceable H+ ions from one
molecule of the acid 4 (e.g., IN H3 PO4 3 H+ lost i.e. Basicity 3.) 4 (e.g., IN H2SO4 2 H+ lost i.e. Basicity 2.) 4 (e.g., IN HCl 1 H+ lost i.e. Basicity 1.). In n-factor B can be calculated as …. For all your chemistry problems call 09356060065 ….
Slide 16 : 4 (e.g., IN Ca(OH)2 2 OH- lost i.e. Acidity 2.). 4 A, Acidity is the number of displacement OH– ions from one
molecule of 4 (e.g., IN NaOH 1 OH- lost i.e. Acidity 1.). In n-factor A can be calculated as …. For all your chemistry problems call 09356060065 ….
Slide 17 : In n-factor e and V can be calculated as …. 4 e, is the number of electron exchanged in a salt 4 V, is the charge on an ion whether it is +ve or -ve 4 (e.g., IN CaCl2 2 e- lost by Ca and 2 e- gained by 2 Cl i.e. no. of e- exchange are 2). 4 (e.g., IN Al3+ valency 3, for O2- valency 2). For all your chemistry problems call 09356060065 ….
Slide 18 : 4 B, Basicity is the number of displaceable H+ ions from one
molecule of the acid 4 (e.g., IN Ca(OH)2 2 OH- lost i.e. Acidity 2.). 4 (e.g., IN H3 PO4 3 H+ lost i.e. Basicity 3.) 4 (e.g., IN H2SO4 2 H+ lost i.e. Basicity 2.) 4 A, Acidity is the number of displacement OH– ions from one
molecule of 4 (e.g., IN HCl 1 H+ lost i.e. Basicity 1.). 4 (e.g., IN NaOH 1 OH- lost i.e. Acidity 1.). n-factor can be calculated as …. 4 e, is the number of electron exchanged in a salt 4 V, is the charge on an ion whether it is +ve or -ve 4 (e.g., IN CaCl2 2 e- lost by Ca and 2 e- gained by 2 Cl i.e. no. of e- exchange are 2). 4 (e.g., IN Al3+ valency 3, for O2- valency 2). For all your chemistry problems call 09356060065 ….
Slide 19 : 1 No. of gram equivelent of 1st For any reaction, compound, nutralisation, etc …. No. of gram equivelent of 2nd = 2 Wt of 1st / Eq wt of 1st Wt of 2nd / Eq wt of 2nd = = 3 4 Wt of 1st x n factor (A/B/e/V)
mol wt Wt of 2nd x n factor (A/B/e/V)
mol wt moles of 1st x n factor moles of 2nd x n factor = For all your chemistry problems call 09356060065 ….
Slide 20 : For any reaction, compound, nutralisation, etc …. 5 6 Normality of1st x volume in lt Normality of2nd x volume in lt = M x n factor x volume in lt M x n factor x volume in lt = 7 M1 x V1 (iff n factor is same) M2 x V2 (iff n factor is same) = For all your chemistry problems call 09356060065 ….
Slide 21 : For any reaction, compound, nutralisation, etc …. 10 8 If acid-acid or base-base mixing │N1V1 + N2V2 │ =N3(V1+V2) If acid-base mixing │N1V1- N2V2 │ =N3(V1+V2) For all your chemistry problems call 09356060065 ….
Slide 22 : 1 No. of gram equivelent of 1st For any reaction, compound, nutralisation, etc …. No. of gram equivelent of 2nd = 2 Wt of 1st / Eq wt of 1st Wt of 2nd / Eq wt of 2nd = = 3 5 6 4 Wt of 1st x n factor (A/B/e/V)
mol wt Normality of1st x volume in lt Wt of 2nd x n factor (A/B/e/V)
mol wt moles of 1st x n factor moles of 2nd x n factor = Normality of2nd x volume in lt = M x n factor x volume in lt M x n factor x volume in lt = 10 7 8 M1 x V1 (iff n factor is same) M2 x V2 (iff n factor is same) = If acid-acid or base-base mixing │N1V1 + N2V2 │ =N3(V1+V2) If acid-base mixing │N1V1- N2V2 │ =N3(V1+V2) For all your chemistry problems call 09356060065 ….
Slide 23 : Hydrogen Displacement Method Some other methods …. For all your chemistry problems call 09356060065 ….
Slide 24 : (b) Oxide Method Some other methods …. For all your chemistry problems call 09356060065 ….
Slide 25 : Chloride Method Some other methods …. For all your chemistry problems call 09356060065 ….
Slide 26 : Double Decomposition Method Some other methods …. For all your chemistry problems call 09356060065 ….
Slide 27 : Hydrogen
Displacement
Method
(b) Oxide Method
(c) Chloride Method
(d) Double
Decomposition
Method Some other methods …. For all your chemistry problems call 09356060065 ….