To balance O  and H atoms, H₂O and H⁺ are added.
MnO₄^- + 5Fe²⁺ + 8H⁺ \(\rightarrow\) 5Fe³⁺ + Mn²⁺ + 4H₂O

Volume of drop of water = 0.05 ml
Mass of a drop of water = Volume x Density
= 0.05 ml x 1.0 g / ml.
= 0.05 g
Molar mass of water (H₂O) = 18 g
18 g of water = 1 mole
0.05 g of water = \(\frac { 1 }{ 18 } \times 0.05\)
= 0.0028 mol.
No. of molecules present in one mole of water = 6.023 x 10²³
No. of molecules present in 0.0028 mole of water = \(\frac { 6.023\times { 10 }^{ 23 }\times 0.0028 }{ 1 } \)
= 1.68 x 10²¹ water molecules.

1 mole of MnO₂ = 55 + 32 = 87 g.
87 g of MnO₂ reacts with 4 moles of HCI. i.e. = 4 x 36.5 = 146 g of HCI.
\(\therefore\)  5 g of MnO₂ wilI react with \(\frac { 146 }{ 87 } \)  x 5.0 = 8.40 g.

(i) 1 mole of C₂H₆ contains 2 moles of Carbon atoms.
\(\therefore\) 3 moles of C₂H₆ wilI have 6 moles of Carbon atoms.
(ii) 1 mole of C₂H₆ contains 6 moles of Hydrogen atoms.
\(\therefore\) 3 moles of C₂H₆ wilI have 18 moles of Hydrogen atoms.
(iii) 1 mole of C₂H₆ contains 6.023 x 10²³number of molecules.
\(\therefore\) 3 moles of C₂H₆ will contain 3 x 6.023 x 10²³molecules.

\(\therefore\) The empirical formula is Fe₂O₃.

S + HNO₃  H₂SO₄ + NO₂ + H₂O

2. S + 6HNO₃ \(\rightarrow\)  H₂SO₄ + NO₂ + H₂O
3. Balance the equation (except O and H).
S + 6HNO₃ \(\rightarrow\) H₂SO₄ + 6NO₂ + H₂O
4. Balance O atoms by adding 2H₂O.
S + 6HNO₃ \(\rightarrow\) H₂SO₄ + 6NO₂ + 2H₂O

These are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.
Example: Zn releases electrons to Cu and Cu releases electrons to Silver and so on.
Zn(s) + Cu²⁺ \(\rightarrow\) Zn²⁺_(aq) + Cu_(s) (Here Zn - oxidised; Cu²⁺ - reduced)
Cu_(S) + 2Ag⁺ \(\rightarrow\) Cu²⁺_(aq)+ 2Ag_(s) (Here Cu - oxidised; Ag⁺ - reduced)

The reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions
Example: P₄ + 3NaOH + 3H₂O \(\rightarrow\) PH₃ + 3NaH₂PO₂
2HCHO + H₂O \(\rightarrow\) CH₃OH + HCOOH

The reactions in which one ion or atom in a compound is replaced (or substituted) by an ion or atom of the other element are called displacement reactions.
AB  + C \(\rightarrow\) AC + B
Example: Metal displacement
CuSO₄ + Zn \(\rightarrow\) ZnSO₄ + Cu
Example: Non-metal displacement
2KBr +C1₂ \(\rightarrow\)  2KCI + Br₂ 

Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction
A B \(\rightarrow\) A + B
Example: 2KClO₃ \(\rightarrow\) 2KCl + O₂
PCl₅ \(\rightarrow\) PCl₃ +Cl₂

When two or more substances combine to form a single substance, the reactions are called combination reactions
A + B \(\rightarrow\) C
Example: 2MG + O₂ \(\rightarrow\) 2MgO

Molecular mass and empirical formula are used to deduce molecular formula of the compound.
Steps to calculate molecular formula:
1. Empirical formula is found out from the percentage composition of elements
2. Empirical formula mass can be found from the empirical formula
3. Molecular mass is found out from the given data
4. Molecular formula = (Empirical formula)_n
5. where, n = \(\frac { molecular\ mass }{ empirical\ formula\ mass } \)

\(\therefore\) The empirical formula is N₂O₅.

In acid medium,
\(\overset { \overset { COOH }{ \underset { COOH }{ | } } }{ \overset { Oxalic\ acid }{ \left( 90\ g \right) } } +\underset { (16g) }{ \left[ O \right] } \longrightarrow { 2CO }_{ 2 }+{ H }_{ 2 }O\)
16 g of oxygen is used for oxidation of 90 g of oxalic acid.
\(\therefore\)  8 g of oxygen will oxidize = \(\frac { 90 }{ 16 } \times 8\) = 45 g eq^-1.
Equivalent mass of Anhydrous oxalic acid 45 g eq^-1.
Equivalent mass of \(\overset { COOH }{ \underset { COOH }{ | } } .{ 2H }_{ 2 }O=\frac { Molar\ mass }{ basicity } \)
= \(\frac { 126 }{ 2 } \)
= 63 g eq^-1.

Mass of metal = 0.6
Mass of metal oxide = 1 g
\(\therefore\)  Mass of oxygen = .1 - 0.6 = 0.4 g
0.4 g of oxygen combines with 0.6 g of metal.
\(\therefore\) 8 g of oxygen will combine with \(\frac { 0.6 }{ 0.4 } \times 8\)
Equivalent mass of the metal = 12 g eq^-1

Molar mass = \(\frac { weight\ of\ the\ substance\ \times \ Molar\ volume }{ Volume\ of\ the\ substance\ at\ STP } \)
Molar volume at STP = 2.24 x 10^-2 m³
= 22.4 L (or) 22400 cc
Molar mass of the gas at STP = \(\frac { 23.2\times 22.4 }{ 20 } \)
= 25.984 g

Molar mass of HCl = 36.5 g
The molar volume of 36.5 g (1 mole) of He 1 = 2.24 x 10^-2m³.
\(\therefore\) The volume of 146 g (4 moles) of HCI  = \(\frac { 2.24\times { 10 }^{ -2 } }{ 36.5 } \times 146\)
= 8.96 x 10^-2m³
No. of molecules in 146 g of HCI = 4 N
=  4 x Avogadro Number
=  4 x 6.023 x 10²³
=  24.092 x 10²³
=  2.4092 x 10²⁴ molecules. 

The atomic mass of Gold = 197 g mol^-1.
197 g of gold contains 6.023 x 10²³atoms of gold.
\(\therefore\) 1000 g of gold will contain = \(\frac { 1000\times 6.023\times { 10 }^{ 23 } }{ 197 } \)
= 3.055 x 10²⁴ atoms of Gold.

1 amu = 1.66056 x 10^-24
(a) The mass of He hum atom in amu  = \(\frac { 6.641\times { 10 }^{ -24 } }{ 1.66056\times { 10 }^{ -24 } } \) = 3.9992 amu
(b) The mass of Silver atom in amu = \(\frac { 1.790\times { 10 }^{ -22 } }{ 1.66056\times { 10 }^{ -24 } } \) = 107.79 amu

Step 1:
 
Step 2 : Mg + 2HNO₃ \(\rightarrow\) Mg(NO₃)₂ + NO₂ + H₂O
Step 3 . To balance the number of oxygen atoms and hydrogen atoms 2HNO₃ is multiplied by 2.
Mg + 4HNO₃ \(\rightarrow\) Mg(NO₃)₂ + 2NO₂ + H₂O
Step 4. To balance the number of hydrogen atoms, the H₂O molecule is multiplied by 2.
Mg + 4HNO₃ \(\rightarrow\) Mg(NO₃)₂ + 2NO₂ + 2H₂O

0.869 g of CuSO₄.aH₂O gave a residue of 0.556 g of Anhydrous CuSO₄.
\(\therefore\) Weight of a H₂O molecule = 0.869 - 0.556
= 0.313 g
Molecular weight of H₂O   =  (1 x 2) + 16 = 2 + 16 = 18
No. of moles of water = \(\frac { mass }{ Molecular\ mass } \)
CuSO₄.5H₂O - Molecular mass = 63.5 + 32 + 64 + 90.
= 249.5 g
249.5 g of CuSO₄.5H₂O on heating gives 159.5 g of CuSO₄.
0.869 g of CuSO₄.aH₂O on heating gives
 = \(\frac { 159.5 }{ 249.5 } \times 0.869\)
= 0.556 g of anhydrous CuSO₄.
\(\therefore\) a = 5

Hydrated ferrous sulphate = FeSO₄.7H₂O
Ferrous sulphate - Reducing agent
Ferrous sulphate reacts with an oxidising agent in acid medium according to the equation.
2FeSO₄ + H2SO₄ + [O] \(\rightarrow\) Fe₂(SO₄)₃ + H₂O
2 x 152g                  16g
16 parts by mass of oxygen oxidised 304 g of FeSO₄.
8 parts by mass of oxygen will oxidise \(\frac { 304 }{ 16 } \times 8\) Parts by mass of FeSO₄.
= 152
Equivalent mass of Ferrous sulphate (Anhydrous) = 152
Equivalent mass of crystalline Ferrous sulphate FeSO₄ 7H₂O
= 152 + 126 = 278

Anhydrous copper sulphate = CuSO₄
Molecular mass of CuSO4 = 63.5 + 32 + (16 x 4)
= 63.5 + 32 + 64
= 159.5 g
159.5 g of CuSO₄ contains 63.5 g of copper.
\(\therefore\) 100 g of CuSO₄ contains \(\frac { 63.5 }{ 159.5 } \times 100\) = 0.39811 x 100
= 39.81 g of Copper.

Equivalent mass of a salt:
It is defined as the number of parts by mass of the salt that is produced by the neutralization of one equivalent of an acid by a base. Therefore the equivalent mass of the salt is equal to its molar mass.

Differences among three physical states of matter (solid, liquid and gas) are as follows.

Mass of impure sample of sodium nitrate (NaNO₃) = 95 x 1000g
Mass of pure sample of sodium nitrate (NaNO₃) = 95 x 1000 x \(\frac { 70 }{ 100 } \)
= 66500 g
1 mole of pure NaNO₃ = 85 g
85 g NaNO₃ = 523 g of Na
66,500 of NaNO₃ = \(\frac { 23 }{ 85 } \) x 66500 g of Na
= 17994.1 gram
= 17.99 kg

Hydrated sodium carbonate has the formula Na₂CO₃.10H₂O
Molecular mass of Na₂CO₃.10H₂O = { 2 x atomic mass of Na + 1 x atomic mass of C + 3 x atomic mass of oxygen + 10 x molar mass of H₂O
= 2 x 23 + 12 + 3 x 16 + 10 x 18
= 46 + 12 + 48 + 180
= 286 u
2 NaOH + H₂CO₃ ⟶ Na₂CO₃ + 2H₂O
Equivalent mass of NaOH = Molar mass of NaOH
2 equivale it of NaOH = 1 equivalent of Na₂CO₃
1 equivalent of NaOH = 1/2 equivalent mass of Na₂CO₃
Equivalent mass = \(\frac { Molar\quad mass }{ 2 } \)
=\(\frac { 286 }{ 2 } \) = 143 g eq^-1

The balanced equation for the reaction is,
N₂ + 3H₂ \(\rightarrow\) 2NH₃
According to the equation,
2 x 22400 cm³ of NH₃ is produced by 1 mol of N₂ = 22400 cm³
100 cm³ of NH₃ will be produced by = \({22400\over2\times22400}\times100=50cm^3\)
Similarly, 2 x 22400 cm³ of NH₃ are produced by = 3 x 22400 cm³ of H₂
100cm³ of NH₃ Will be produced by =\({3\times22400\over2\times22400}\times100=150cm^3\)

The balanced chemical equation for the reaction is
Mg + 2HCI \(\rightarrow\) MgCl₂ + H2
Atomic mass of Mg = 24g mol^-1
According to the equation,
1 mol of Mg (24 g) will liberate 1 mol of H₂ = 22.4L at STP
24g of Mg = 22.4 L of H₂ at STP
\(0.12g \ of \ Mg={22.4\over24}\times0.12g \ of H_2 \ at STP\)
= 0.112 L or
= 112 cm³ at STP

Molar mass of A g₂CO₃ = \(\{2\times atomic\ mass\ of \ Ag+atomic\ mass\ of\ C +3\times atomic\ mass \ of\ 'O'\)
= 2 x 108 + 12 + 3 x 16
= 276 u
Molar mass of AgO = atomic mass of Ag+ atomic mass of '0'
= 108 + 16 = 124g mol^-1
When silver carbonate is heated, carbon dioxide leaves as gas and silver oxide remains as a residue.
According to the equation,
One mol of Ag₂C0₃leaves = 2 mol of AgO as residue
276 g of Ag₂0 = 124g of AgO
\(11.04 \ of \ Ag_2O={124\over276}\times 11.04g \ of \ AgO\)
= 4.96g

No. of mol = \(\frac { Weight\ of\ the\ substance\ in\ gram }{ Gram\ molecular\ mass } \)
Step - 1 : Find the gram molecular weight of the atom.
Mass of 6.3023 x 10²³ atoms = 1 mole
= 1 gram molecular weight
Mass of one atom of the element = 6.626 x 10^-23 g
Mass of 6.023 x 10²³ atoms = 6.626 x 10^-23 x 6.023 x 10²³
= 40.1
Therefore, the gram molecular weight of the element is 40.1g
Step -2 : To find the number of mol in 0.320 kg
Number of mol = \(\frac { 0.320\times 1000g }{ 40.1g } \)
= 7.98 ≃ 8 mol.

The empirical formula is C₃H₄0₃
Empirical formula mass = 3 x 12 + 4 x 1 + 3 x 16 = 88
Molar mass = 176g mol^-1
n = \(\frac { molar\quad mass }{ Empirical\quad formula } =\frac { 176 }{ 88 } =2\)
Molecular formula = (Empirical formula)_n
= (C₃H₄O₃)₂
= C₆H₈O₆

The empirical formula is CH
The empirical formula mass = 12 + 1 = 13
To find the molecular mass:
10 L of the gas at STP weigh = 11.6g
22.4L of the gas at STP will weigh = \(\frac { 11.6 }{ 10 } \times 22.4\)
\(=25.98=26\)
\(n=\frac { Molar\ name }{ Empirical\ formula\ mass } \)
= \(\frac { 26 }{ 13 } =2\)
Hence, the molecular formula = (Empirical formula)
(CH)₂ = C₂H₂
The molecular formula of the compound is C₂H₂

The empirical Formula is Na₂Co₃

1 mol of CH₄ = 22400 cm³ at NTP
0.5 mol of CH4 = 22400 x 0.5
= 11200 cm³ at NTP

Atomic mass of sodium = 23 a.m.u
1 mol of Na = 23g = 6.023 x 10²³ atoms
46 g = \({6.023\times 10^{23}\over 23}\times 46 \ atoms\)
= 12.046 x 10²³ atoms

Molecular formula of acetic acid is CH₃ COOH.
Molecular mass of acetic acid = 60
1 mol of acetic acid = 60g of acetic acid
= 6.023 x 10²³molecules
1000 g of acetic acid = \({6.023\times10^{23}\over 60}\times 1000\)
=1.0 x10²⁵ molecules.

Molecular formula of sulphur dioxide = S0₂
Molecular mass of sulphur dioxide = 64
1 mole of S0₂ = 64 g of S0₂ = 6.023 x 10²³molecules
\(100g of SO_2={6.023\times 10^{23}\over64}\times 100 \)
= 9.41 x 10²³ molecules.

Molecular mass of H₂0 = 18
gram molecular mass of H₂0 = 18g
1 mol of H₂0 = 18g of H₂0 = 6.023 x 10²³molecules of water
1.8g of H₂0 = \({6.023\times10^{23}\over18}\times1.8 \)
\(=6.023\times10^{22} molecules\)

Atomic mass of Hg = 200.59
(i) Gram atomic mass of Hg = 200.59 g
1 mol of Hg = 200g = 6.023 x 10²³atoms
\(10g={6.023\times 10^{23}\over200}\times 10\)
= 0.301 x 10²³
= 3.01 x 10²² atoms

(i) Mole is the SI unit to represent a specific amount of a substance.
(ii) One mole is the amount of substance of a system which contains as many elementary particles as there are in 12g of carbon -12-isotope.
(iii) Mole represents the mass in gram of 6.022 x 10²³entities (atom or molecule / ion).
(iv) 1 mole of any substance contains 6.023 x 10²³entities.

C₆H₁₂O₆ (glucose)
\(Molecular \ mass \ of \ C_6H_{12}O_6=\{{6\times atomic \ mass \ of \ C +12 \times \ atomic \ mass \ of \ H+6\times \ atomic \ mass \ of \ O}\)
= 6 \(\times\)12 + 12\(\times\)1 + 6\(\times\)16
= 72 + 12 + 96 = 180u

H₂C₂O₄.2H₂O(crystalline oxalic acid)
\(Molecular \ mass \ of \ H_2C_2O_4.2H_2O= \{{2\times atomic \ mass \ of \ H + 2 \times \ atomic \ mass \ of \ C+4\times \ atomic \ mass \ of \ o+2\times moecular \ mass \ of H_2O}\)
= 2 x 1 + 2 x 12 + 4 x 16 + 36
= 2 + 24 + 64 + 36 = 126 u. 

\(Molecular \ mass \ of \ NaOH=\{{1\times atomic \ mass \ of \ Na + 1 \times \ atomic \ mass \ of \ O+1\times \ atomic \ mass \ of \ H}\)
= 1 \(\times\) 23 + 1\(\times\)16 + 1\(\times\)1
= 23 + 16 + 1 = 40 u

\(Molecular \ mass \ of \ KMnO_4=\{{1\times atomic \ mass \ of \ K + 1 \times \ atomic \ mass \ of \ Mn +4 \times atomic \ mass \ of \ o}\)
atomic mass of K = 39.098 = 39
atomic mass of Mn = 54.938 = 55
atomic mass of 'O' = 15.999 = 16
\(\therefore\) Molecular mass of KMnO₄ = 39 + 55 + 4 x 16
=·158 u

\(Average \ atomic \ mass \ of \ Argon={36\times \% abundance of \ Ar^{36}+38\times \% abundance of Ar^{38} +40 \times \% abundance of Ar^{40}\over 100}\)
given % abundance of Ar³⁶ = 0.337
Ar³⁸ = 0.063
Af⁴⁰ = 99.6
Average atomic mass of Argon = \(36\times 0.337 +38\times 0.063+40\times 99.6\over 100\)
=\(12.132+2.394+3.984\over100\)
=\(3998.5\over100\) = 39.98 a.m.u

\(Average \ atomic \ mass \ of \ carbon ={12\times \%abundance \ of \ C^{12}+13 \times \% \ abundance \ of \ C^{13}+14\% abundance of c^{14} \over 100} \)
given % abundances Of C¹² = 98.892
C¹³ = 1.108
C^I4 = 2 x 10^-10
Since 2 x 10^-10is so small compound to 100, it can be neglected .
\(Average \ atomic \ mass \ of \ carbon ={12\times 98.892 +13 \times 1.108 \over 100} \)
= 12.01 a.m.u

(i) Oxygen: (average mass of oxygen atom = 2.656 x 1O^-23g
1 a.m.u = 1.6605 x 1O^-27kgor 1.6605 x 1O^-24gram)
(ii) Mass of an oxygen atom = 2.656 x 1O^-23g
The mass of oxygen atom In a.m.u = \({Average \ mass \ of \ oxygen \ atom \ in \ gram\over1.6605\times10^{-24}}\)
\(={2.656\times10^{-23}g\over1.6605 \times 10^{-24}g}\)
= 1.599 x10
= 15.99 a.m.u

The quantity \(1\over12\) mass of an atom of carbon -12 is known as the atomic mass unit. (a.m.u). The
actual mass of one atom ofcarbon-12 is 1.9914 x 10^-23g or 1.9914 x 10^-26kg.
Thus,
\(\boxed{1a.m.u={1.9924\times10^{-23}\over12}=1.66\times10^{-27}kg}\)
\(\therefore\)Atomic mass 0f an element =\({Mass \ of \ one \ atom \ of \ the \ element \over 1 a.m.u}\)

(i) Monoatomic molecules: Copper (Cu) and Gold (Au)
(ii) Diatomic molecule: Hydrogen (H₂) and Oxygen (O₂)
(iii) Polyatomic molecule: Phosphorous (P₄) and Sulphur (S₈)

(i) An element consists of only one type of atom ..An atom is the smallest electrically neutral - particle, being made up of fundamental particles, namely electrons, protons and neutrons.
(ii) Compounds are made up of molecules which contain two or more atoms of different elements. Properties of compounds are different.from those of their constituent elements. The constituents of a compound is present in a fixed ratio by weight.

Equalise the increase/decrease in Oxidation number by multiplying with suitable numbers.
5Sb³⁺+ 2Mn\({ O }_{ 4 }^{ - }\)\(\rightarrow \) Sb⁵⁺+ Mn²⁺
Balance all other atoms except O and H
5Sb³⁺+ 2Mn\({ O }_{ 4 }^{ - }\)\(\rightarrow \) 5Sb⁵⁺+ 2Mn²⁺
Balance Oxygen atom by adding H₂O on the side falling short of oxygen.
5Sb³⁺+ 2Mn\({ O }_{ 4 }^{ - }\)\(\rightarrow \) 5Sb⁵⁺+ 2Mn²⁺ +8H₂O
Balance hydrogen atom by adding H+ on the side falling short of hydrogen atoms.
5Sb³⁺+ 2Mn\({ O }_{ 4 }^{ - }\)\(\rightarrow \) 5Sb⁵⁺+ 2Mn²⁺ +8H₂O + 16H⁺

Molar mass of ethane, C₂H₆ = (2 x 12) + (6 x 1) = 30 g mol^-1
n = mass / molar mass = 9g / 30 g mol^-1 = 0.3 mole

Equate the total no. of electrons in the reactant side by cross multiplying.
\(\Rightarrow 3 \mathrm{As}_{2} \mathrm{~S}_{3}+28 \mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO}\)
Based on reactant side, balance the products
\(\Rightarrow 3 \mathrm{As}_{2} \mathrm{~S}_{3}+28 \mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow 6 \mathrm{H}_{3} \mathrm{AsO}_{4}+9 \mathrm{HSO}_{4}+28 \mathrm{NO}\)
Product side : 36 hydrogen atoms & gg Orygen atoms
Reactant side : 28 hydrogen atoms & 74 Orygen atoms
Difference is 8 hydrogen atoms & 4 oxygen atoms
∴ Add 4 H₂O molecule on the reactant side.
Balanced equation is,
\(3 \mathrm{As}_{2} \mathrm{~S}_{3}+28 \mathrm{HNO}_{3}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow 6 \cdot \mathrm{H}_{3} \mathrm{AsO}_{4}+9 \mathrm{H}_{2} \mathrm{SO}_{4}+28 \mathrm{NO}\)