Mole Concept Formula Sheet (Class 11 –
JEE)
—
Avogadro’s Law and Avogadro’s Number
NA = 6.022 × 1023
N
n=
NA
Mole Definition (General Formulas)
n=
m
M
V
(at STP)
22.4 L
V
RT
n=
, Vm =
Vm
P
n=
Molecular Formulas
Molecular formula = n × Empirical formula
n=
Molar mass
Empirical formula mass
Stoichiometry
n=
m
,
M
n=
V
,
22.4
n=
N
NA
Concentration Terms
Molarity (M)
M=
moles of solute
Volume of solution (L)
M=
10 × mass of solute (g)
Msolute × Vsolution (mL)
Molality (m)
m=
moles of solute
Mass of solvent (kg)
1
Normality (N)
N=
gram equivalents of solute
Volume of solution (L)
N = M × nf
Formality (F)
F =
moles of solute (formula units)
Volume of solution (L)
Mole Fraction
χA =
nA
nA + nB
Mass Percent
%w/w =
Mass of solute
× 100
Mass of solution
Volume Percent
%v/v =
Volume of solute
× 100
Volume of solution
Parts Per Million (ppm)
ppm =
Mass of solute
× 106
Mass of solution
Density Relation
m=
1000M
1000d − M × M2
Gas Laws
P V = nRT
V = 22.4n (at STP)
Equivalent Concept
neq =
m
,
Eq. mass
Eq. mass =
M
,
nf
neq = n × nf
Dilution Law
M1 V1 = M2 V2 ,
N1 V1 = N2 V2
2
Key Relations Summary
n=
V
N
m
=
=
M
22.4
NA
P V = nRT
N = M × nf
neq = n × nf
nA
χA =
nA + nB
M1 V1 = M2 V2 , N1 V1 = N2 V2
3