JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 41

Multiple Choice Questions (MCQs) for JEE Chemistry: Some Basic Concepts Of Chemistry

Access targeted multiple-choice questions for Some Basic Concepts Of Chemistry designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Some Basic Concepts Of Chemistry MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: 4 g of hydrogen are ignited with 4 g of oxygen. The weight of water formed is:
(a) 0.5 g
(b) 3.5 g
(c) 4.5 g
(d) 2.5 g
Show Answer & Explanation

Answer: (c) 4.5 g

Explanation:
1. Moles: H2 = 2, O2 = 0.125, so O2 is limiting.
2. Water = 2 × 0.125 = 0.25 mol = 4.5 g.
3. Answer: (c)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 2: 10 mL of a gaseous hydrocarbon on combustion gives 40 mL of CO2(g) and 50 mL of H2O (vapour). The hydrocarbon is:
(a) C4H5
(b) C8H10
(c) C4H8
(d) C4H10
Show Answer & Explanation

Answer: (d) C4H10

Explanation:
1. Per mL of hydrocarbon: 4 mL CO2 (4 C) and 5 mL H2O (10 H).
2. Formula C4H10.
3. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 3: 500 mL of a gaseous hydrocarbon when burnt in excess of O2 gave 2.5 L of CO2 and 3.0 L of water vapour under the same conditions. The molecular formula of the hydrocarbon is:
(a) C4H8
(b) C4H10
(c) C5H10
(d) C5H12
Show Answer & Explanation

Answer: (d) C5H12

Explanation:
1. Per volume of hydrocarbon: CO2 = 2.5/0.5 = 5, so 5 C; H2O = 3.0/0.5 = 6, so 12 H.
2. Formula C5H12.
3. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 4: The molecular weight of a tribasic acid is W. Its equivalent weight will be:
(a) W/2
(b) W/3
(c) W
(d) 3W
Show Answer & Explanation

Answer: (b) W/3

Explanation:
1. Equivalent weight = molar mass/basicity = W/3.
2. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 5: Which property of an element is NOT variable?
(a) Valency
(b) Atomic weight
(c) Equivalent weight
(d) None of these
Show Answer & Explanation

Answer: (b) Atomic weight

Explanation:
1. An element may show different valencies, and its equivalent weight changes with valency.
2. Its atomic weight is fixed.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 6: 0.45 g of an acid (molecular wt. = 90) was exactly neutralised by 20 mL of 0.5 N NaOH. The basicity of the acid is:
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Equivalents of NaOH = 0.020 × 0.5 = 0.01.
2. Moles of acid = 0.45/90 = 0.005.
3. Basicity = 0.01/0.005 = 2.
4. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 7: 2 g of a base whose equivalent weight is 40 reacts with 3 g of an acid. The equivalent weight of the acid is:
(a) 40
(b) 60
(c) 10
(d) 80
Show Answer & Explanation

Answer: (b) 60

Explanation:
1. Equivalents of base = 2/40 = 0.05.
2. Equivalents of acid are equal, so E = 3/0.05 = 60.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 8: 0.84 g of a metal carbonate reacts exactly with 40 mL of N/2 H2SO4. The equivalent weight of the metal carbonate is:
(a) 84
(b) 64
(c) 42
(d) 32
Show Answer & Explanation

Answer: (c) 42

Explanation:
1. Milliequivalents of acid = 40 × 0.5 = 20 meq = 0.02 eq.
2. Equivalent weight = 0.84/0.02 = 42.
3. Answer: (c)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 9: H3PO4 is a tribasic acid and one of its salts is NaH2PO4. What volume of 1 M NaOH solution should be added to 12 g of NaH2PO4 to convert it into Na3PO4? (at. wt. P = 31)
(a) 100 mL
(b) 200 mL
(c) 80 mL
(d) 300 mL
Show Answer & Explanation

Answer: (b) 200 mL

Explanation:
1. M(NaH2PO4) = 120, so 12 g = 0.1 mol.
2. NaH2PO4 + 2NaOH → Na3PO4 + 2H2O needs 0.2 mol NaOH = 200 mL of 1 M.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 10: When an element forms an oxide in which oxygen is 20% of the oxide by mass, the equivalent mass of the element will be:
(a) 32
(b) 40
(c) 60
(d) 128
Show Answer & Explanation

Answer: (a) 32

Explanation:
1. Metal 80 g combines with 20 g O.
2. Equivalent mass = 80/20 × 8 = 32.
3. Answer: (a)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 11: 2.8 g of iron displaces 3.2 g of copper from a solution of copper sulphate. If the equivalent mass of iron is 28, then the equivalent mass of copper will be:
(a) 16
(b) 32
(c) 48
(d) 64
Show Answer & Explanation

Answer: (b) 32

Explanation:
1. Equal equivalents are involved: 2.8/28 = 0.1 eq.
2. E(Cu) = 3.2/0.1 = 32.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 12: A metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete reduction 3.15 g of the oxide have yielded 1.05 g of the metal. We may conclude that:
(a) Atomic weight of the metal is 4
(b) Equivalent weight of the metal is 8
(c) Equivalent weight of the metal is 4
(d) Atomic weight of the metal is 8
Show Answer & Explanation

Answer: (c) Equivalent weight of the metal is 4

Explanation:
1. Oxygen = 3.15 − 1.05 = 2.10 g.
2. Equivalent weight of metal = 1.05/2.10 × 8 = 4.
3. Answer: (c)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 13: If 2.4 g of a metal displaces 1.12 L of hydrogen at normal temperature and pressure (STP), the equivalent weight of the metal would be:
(a) 12
(b) 24
(c) 1.2 × 11.2
(d) 1.2 ÷ 11.2
Show Answer & Explanation

Answer: (b) 24

Explanation:
1. 11.2 L of H2 at STP corresponds to 1 equivalent.
2. E = 2.4 × 11.2/1.12 = 24.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 14: 1.6 g of Ca and 2.60 g of Zn, when treated with an acid in excess separately, produced the same amount of hydrogen. If the equivalent weight of Zn is 32.6, the equivalent weight of Ca is:
(a) 10
(b) 20
(c) 40
(d) 5
Show Answer & Explanation

Answer: (b) 20

Explanation:
1. Equal hydrogen means equal equivalents: 1.6/E = 2.60/32.6.
2. E = 1.6 × 32.6/2.60 = 20.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 15: 74.5 g of a metallic chloride contains 35.5 g of chlorine. The equivalent mass of the metal is:
(a) 19.5
(b) 35.5
(c) 39.0
(d) 78.0
Show Answer & Explanation

Answer: (c) 39.0

Explanation:
1. Metal = 74.5 − 35.5 = 39 g combines with 35.5 g Cl (equivalent weight 35.5).
2. E(metal) = 39.
3. Answer: (c)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 16: The carbonate of a metal is isomorphous with MgCO3 and contains 6.091% of carbon. The atomic weight of the metal is nearly:
(a) 48
(b) 68.5
(c) 137
(d) 120
Show Answer & Explanation

Answer: (c) 137

Explanation:
1. Formula MCO3: molar mass = 12/0.06091 = 197.
2. Atomic weight of M = 197 − 60 = 137.
3. Answer: (c)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 17: The atomic weight of a metal (M) is 27 and its equivalent weight is 9. The formula of its chloride will be:
(a) MCl
(b) MCl2
(c) M3Cl
(d) MCl3
Show Answer & Explanation

Answer: (d) MCl3

Explanation:
1. Valency = atomic weight/equivalent weight = 27/9 = 3.
2. The chloride is MCl3.
3. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 18: The vapour density of a gas is 16. The ratio of specific heat at constant pressure to that at constant volume is 1.4. Its atomic weight is:
(a) 8
(b) 16
(c) 24
(d) 32
Show Answer & Explanation

Answer: (b) 16

Explanation:
1. Molar mass = 2 × 16 = 32.
2. γ = 1.4 means the gas is diatomic, so the atomic weight = 32/2 = 16.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 19: One litre of a certain gas weighs 1.16 g at STP. The gas may possibly be:
(a) C2H2
(b) CO
(c) O2
(d) NH3
Show Answer & Explanation

Answer: (a) C2H2

Explanation:
1. Molar mass = 1.16 × 22.4 = 26 g mol−1.
2. C2H2 has M = 26.
3. Answer: (a)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 20: The oxide of an element possesses the molecular formula M2O3. If the equivalent mass of the metal is 9, the molecular mass of the oxide will be:
(a) 27
(b) 75
(c) 102
(d) 18
Show Answer & Explanation

Answer: (c) 102

Explanation:
1. Valency = 3 (M2O3), so atomic mass of M = 9 × 3 = 27.
2. Molecular mass = 2 × 27 + 3 × 16 = 102.
3. Answer: (c)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

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