(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.

Molar mass of sodium hydroxide = 40
No.of moles = \(\frac { Mass }{ Molar\ mass } \)
No.of moles (n) = \(\frac { 120 }{ 40 } \)
= 3 moles

NO₃^-
x + 3(-2) = -1
x - 6 = -1
x = -1 + 6
x = +5
Oxidation number of N in NO₃^- is +5

Molar mass of NH₃ = (1 x 14 + 3 x 1) = 17g mol^-1
Mass of 0.2 mol of NH₃ = 0.2 mol x 17g mol^-1
= 3.4 g

Molecular mass of silver (Ag) = 107.87 u
Molar mass of Ag = 107.87 g mol^-1
\(\therefore\) Mass of 1 atom of Ag = \(\frac { Molar\ mass }{ Avogadro's\ number } \)
= \(\frac { 107.87g\ { mol }^{ -1 } }{ 6.023\times { 10 }^{ 23 }\ { mol }^{ -1 } } \)
= 17.91 x 10^-23g
Mass of 1 atom of Ag = 17.91 x 10^-23 g.

Molar mass = density x Molar volume
= 1.965 x 2.24 x 10^-2
= 4.4016 x 10^-2 kg/mol
= 4.4016 x 10^-2 x 10³ g/mol
= 44.016 g/mol

(i) The entire quantity of ammonia is consumed in the reaction. So ammonia is the Iimiting reagent. Some quantity of CO₂ remains unreacted, so CO₂ is the excess reagent.
(ii) Quantity of urea formed = number of moles of urea formed x molar mass of urea
= 19 moles x 60 g mol^-1
= 1140 g = 1.14 kg
Excess reagent leftover at the end of the reaction is carbon dioxide.
Amount of carbon dioxide leftover
= number of moles of CO₂ left over x molar mass of CO₂   
= 7 moles x 44 g mol^-1 
= 308 g

Step - 1 : To find atoms undergoing change in O.N.
K₂Cr₂O₇ + HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 2 : To find the total increase and decrease in O.N.
K₂Cr₂O7 + CrI₃  (decrease of 3 unit / atom = Total decrease = 6 units / 2 atom)
HI ⟶ I₂ (increase of 1 unit / atom = total increase 1 x 6 = 6)
Step - 3 : To balance the total increase and decrease in O.N, multiply HI by 6.,
K₂Cr₂O₇ + 6 HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 4 : To balance all atoms other than '0' and 'H'
K₂Cr₂O₇ + 6 HI ⟶ 2KI + 2Crl₃ + H₂O + 3I₂
This makes 14 iodine atoms on RHS. (These iodide ions do not undergo any change in O.N). Hence to balance the iodine atoms add 8HI to LHS.
i.e., K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + H₂O + 3I₂
These, the oxygen atoms are balanced by making 7H₂O as RHS.
K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + 7H₂O + 3I₂
The hydrogen atoms are balanced by themselves.
Hence the balanced equation is
K₂Cr₂O₇ + 14HI ⟶ 2KI + 2Crl₃ + 7H₂O + 3I₂

Half reactions are:
\(\overset { +7 }{ M } { nO }_{ 4 }^{ - }\longrightarrow { Mn }^{ 2+ }\)
and \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }\)
(1) \(\Rightarrow \) \({ MnO }_{ 4 }^{ - }+{ 8H }^{ - }+5e^{ - }\longrightarrow { Mn }^{ 2+ }+{ 4H }_{ 2 }O\)
(2) \(\Rightarrow \) \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+{ 2e }^{ - }\)

ii) 

iii) 

iv) 

(b) Both assertion and reason are correct but reason is not the correct explanation for assertion

Gaseous state of matter at very high temperature containing gaseous ions and free electron is referred to as the Plasma state. e.g. Lightning.

Molar mass of HNO₃ = 1 + 14 + 3 x 16 = 63
Basicity of HNO₃ = 1
Equivalent mass of HNO₃ = \(\frac { 63 }{ 1 } =63g\) eq^-1