JEE Advanced (Single Correct MCQs): Some Basic Concepts of Chemistry - JEE MCQ

4 Al + 3O₂ —→ 2 Al₂O₃
At. wt. of Al = 27 Thus 4 × 27 g of Al reacts with oxygen = 3 × 32 g
∴ 27 g of Al reacts with oxygen =

  = 24 g

No. of nitrogen atoms =

No. of oxygen atoms =

∴ Formula of compound is N₂O₅.

(a) 18 g of H₂O = 6.02 × 10²³ molecules of H₂O
∴ 36 g of H₂O = 2 × 6.02 × 10²³ molecules of H₂O  
= 12.04 × 10²³ molecules of H₂O
(b) 28 g of CO = 6.02 × 10²³ molecules of CO
(c) 46 g of C₂H₅OH = 6.02 × 10²³ molecules of C₂H₅OH
(d) 108 g of N₂O₅ = 6.02 × 10²³ molecules of N₂O₅
∴ 54 g of N₂O₅ = × 6.02 × 10²³ molecules of N₂O₅  
= 3.01 × 10²³ molecules of N₂O₅
∴ 36 g of water has highest number of molecules.

No. of e^– in C = 6 and in O = 8
∴ Total no. of e^– in CO₂ = 6 + 8 × 2 = 22

Let mass of oxygen = 1g,
Then mass of nitrogen = 4g Mol. wt. of N₂ = 28g, Mol. wt. of O₂ = 32g
28 g of N₂ has = 6.02 × 10²³ molecules of nitrogen
4 g of N₂ has =  molecules of nitrogen

molecules of nitrogen

 32 g of O₂ has = 6.02 × 10²³ molecules of oxygen

∴ 1g of O₂ has =  moleculesof oxygen

Thus, ratio of molecules of oxygen : nitrogen

 = 7 : 32

NOTE : Ag₂O is thermally unstable and decompose on heating liberating oxygen]
Mol. wt. of Ag₂CO₃ = 10⁸ × 2 + 12 + 16 × 3 = 276 g
∴ 276 g of Ag₂CO₃ on heating gives residue      = 2 × 10⁸ = 216 g of Ag
∴ 2.76 g of Ag₂CO₃ on heating gives =    = 2.16g of Ag

The change involved is 

i.e. it involves only one electron

Eq.wt =  = M [∵ Mol. wt. = M]

TIPS/Formulae : (i) Write balanced chemical equation for chemical change.
(ii) Find limiting reagent.
(iii) Amount of product formed will be determined by amount of limiting reagent.
The balanced equation is :

Limiting reagent is Na₃PO₄ (0.2 mol), BaCl₂ is in excess.
From the above equation : 2.0 moles of Na₃PO₄ yields Ba₃(PO₄)₂ = 1 mole
∴ 0.2 moles of Na₃PO₄ will yield  Ba₃(PO₄)₂ =  = 0.1 mol.

TIPS/Formulae : (i) Find oxidation state of N in N₂H₄.
(ii) Find change in oxidation number with the help of number of electrons given out during formation of compound Y.
N₂H₄ → Y + 10 e^–, Calculation of O.S. of N in
N₂H₄ : 2x + 4 = 0  ⇒  x = –2 The two nitrogen atoms will balance the charge of 10 e.
Hence oxidation state of N will increase by +5, i.e. from –2 to +3.

NOTE : The sum of oxidation states of all atoms in compound is zero.
Calculation of O.S. of  C in CH₂O. x + 2 + (–2) = 0  ⇒ x = 0

TIPS/Formulae : Molality =

A molal solution is one which contains one mole of solute per 1000 g of solvent.

For equivalent weight of MnSO₄ to be half of its molecular weight, change in oxidation state must be equal to 2. It is possible only when oxidation state of Mn in product is + 4. Since oxidation state of Mn in MnSO₄ is + 2. So, MnO₂ is correct answer.
In MnO₂, O.S. of Mn = +4
∴ Change in O.S. of Mn = +4 – (+2) = +2

TIPS/Formulae : (i) Volume of substance changes with temperature and mass is not effected by change in temperature.
(ii) Find expression which does not have volume term in it.
(a) Molarity – Moles of solute/volume of solution in L.
(b) Normality – gm equivalents of solute/volume of solution in L.
(c) Formality – gm formula wt./volume of solution in L.
(d) Molality – Moles of solute/mass of solvent in kg
∵ Molality does not involve volume term.
∴ It is independent of temperature.

2 + 2 (2 + x – 4) = 0
[∵ Ba(H₂PO₂)₂ is neutral molecule] or 2x – 2 = 0 ⇒ x = + 1

TIPS/Formulae : Balance the reaction by ion electron method.
Oxidation reaction : C₂ O ₄^-2 → 2CO₂ + 2e^–] × 5
Reduction reaction : -
MnO^-₄ + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O] × 2
Net reaction : 2 MnO ^-₄ + 16H⁺ + 5 C₂O₄^2- → 2Mn²⁺ + 10CO₂ + 8H₂O

TIPS/Formulae :
(i) H₃PO₃ is dibasic acid as it contains two –OH groups.

(ii) Normality = Molarity × basicity of acid.
(iii) Basicity of H₃PO₃ = 2
∴   Normality = 0.3 × 2 = 0.6

TIPS/Formulae : (i) Oxidation state of element in its free state is zero.
(ii) Sum of oxidation states of all atoms in compound is zero.
O.N. of S in  S₈ = 0;   O.N. of S in  S₂F₂ = + 1;
O.N. of S in  H₂S = –2;

TIPS/Formulae : (i) In an ion sum of oxidation states of all atoms is equal to charge on ion and in a compound sum of oxidation states of all atoms is always zero.
Oxidation state of Mn in MnO₄^– = + 7
Oxidation state of Cr in Cr(CN)₆^3– = + 3
Oxidation state of Ni in NiF₆^2– = + 4
Oxidation state of Cr in CrO₂Cl₂ = + 6

TIPS/Formulae : (i) In a disproportionation reaction same elemen t undergoes oxidation as well as reduction during the reaction.
(ii) In decomposition reaction a molecule breaks down to more than one atoms or molecules

It is disproportionation reaction because Cl is both oxidised (+ 1 to + 5) and reduced ( + 1 to – 1) during reaction.

TIPS/Formulae : Equivalents of H₂C₂O₄.2H₂O = Equivalents of NaOH (At equivalence point)
Strength of H₂C₂O₄ . 2H₂O (in g/L) =   = 25.2 g/L

Normality of H₂C₂O₄ . 2H₂O = 

Using normality equation : N₁V₁     =   N₂V₂
(H₂C₂O₄.2H₂O)     (NaOH)

0.4 × 10       =  0.1 × V₂ or V₂ = = 40 ml.

TIPS/Formulae : (i) Find change in oxidation number of Cr atom.
(ii) Eq. wt. =

In iodometry, K₂Cr₂O₇ liberates I₂ from iodides (NaI or KI).
Thus it is titrated with Na₂S₂O₃ solution.
2Na₂S₂O³⁺ I₂ → 2NaI + Na₂S₄O₆ O.N. of Cr changes from + 6 (in K₂Cr₂O₇) to +3. i.e. +3 change for each Cr atom
Cr₂ O7^-- + 14H⁺ + 6e^- → 2Cr³⁺+ 7H₂O
 Thus, one mole of K₂Cr₂O₇ accepts 6 mole of electrons.
∴   Equivalent weight =

TIPS/Formulae : (i) Mass of one electron = 9.108 × 10^–31 kg
(ii) 1 mole of electron = 6.023 × 10²³ electrons
Weight of 1 mole of electron = Mass of one electron × Avogadro Number = 9.108 × 10^–31  × 6.023 × 10²³ kg
∴ No. of moles of electrons in 1 kg

TIPS/Formulae : Atomic weight in gms = 6.023 × 10²³ atoms = 1 Mole atoms (i) Number of atoms in 24 g of C

=   × 6.023 × 10²³ = 2 × 6.023 × 10²³ atom

= 2 mole atoms

(ii) Number of atoms in 56 g of Fe

= × 6.023 × 10²³ = 6.023 × 10²³ atom   = 1 mole atoms

(iii) Number of atoms in 27 g of Al

=  × 6.023 × 10²³ = 6.023 × 10²³ atom   = 1 mole atoms

(iv) Number of atoms in 108 g of Ag

= × 6.023 × 10²³ = 6.023 × 10²³ atom   = 1 mole atoms

∴ 24 g of C has maximum number of atoms.

TIPS/Formulae : Write the reaction for chemical change during reaction and equate moles of products formed.
[Co(NH₃)₅SO₄] Br has ionisable Br^– ions & [Co(NH₃)₅ Br] SO₄ has ionisable SO₄^{– –} ion.

Given mixture X = 0.02 mol of [Co(NH₃)₅SO₄] Br and 0.02 mol of [Co(NH₃)₅Br] SO₄ Volume = 2 L
∴ Mixture X has 0.02 mol. of [Co(NH₃)₅SO₄] Br and 0.02 mol of [Co(NH₃)₅Br] SO₄ in 2 L of solution
∴ Conc. of [Co(NH₃)₅SO₄] Br and [Co(NH₃)₅Br]SO₄ = 0.01 mol/L for each of them.
(i) 1 L mixture of X + excess AgNO₃ → Y

∴ No. of moles of Y = 0.01

(ii) Also 1 L mixture of X + excess BaCl₂ → Z

∴ moles of Z = 0.01.

TIPS/Formulae : The highest O.S. of an element is equal to the number of its valence electrons  
(i) [Fe(CN)₆]^3–, O.N. of Fe = + 3,
[Co(CN)₆]^3–, O.N. of Co = + 3
(ii) CrO₂Cl₂, O.N. of Cr = +6, (Highest O.S. of Cr)
[MnO₄]^– O.N of Mn = + 7 (Highest O.S. of Mn)
(iii) TiO₃, O.N. of Ti = + 6, MnO₂ O.N. of Mn = + 4
(iv) [Co(CN)₆]^3–, O.N. of Co = + 3, MnO₃, O.N. of Mn = + 6