JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 06

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

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Multiple Choice Questions

Question 1: 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. In 100 g oxide: O = 20 g and element = 80 g.
2. E = (80/20) × 8 = 32.

Teacher's Note:
1. Use E = (mass of element ÷ mass of oxygen) × 8.
2. Percentages can be used as masses in 100 g.

Question 2: If 1.2 g of a metal displaces 1.12 L of hydrogen at NTP, the equivalent mass of the metal would be
(a) 1.2 × 11.2
(b) 12
(c) 24
(d) 1.2 + 11.2
Show Answer & Explanation

Answer: (b) 12

Explanation:
1. 1.12 L of H2 = 0.1 equivalent (11.2 L per equivalent).
2. E = 1.2/0.1 = 12.

Teacher's Note:
1. Equivalents of hydrogen = volume ÷ 11.2 L.
2. E = metal mass ÷ equivalents.

Question 3: 1 g of hydrogen is found to combine with 80 g of bromine. 1 g of calcium (valency = 2) combines with 4 g of bromine. The equivalent weight of calcium is
(a) 10
(b) 20
(c) 40
(d) 80
Show Answer & Explanation

Answer: (b) 20

Explanation:
1. Equivalent weight of Br = 80.
2. 1 g of Ca combines with 4 g Br, so ECa = (1/4) × 80 = 20.

Teacher's Note:
1. Combining masses are in the ratio of equivalent weights.
2. Calcium: 40/2 = 20, which matches.

Question 4: 2.8 g of iron displaces 3.2 g of copper from copper sulphate solution. If the equivalent mass of iron is 28, 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. ECu = 3.2/0.1 = 32.

Teacher's Note:
1. A metal displaces the other in equal equivalents.
2. Cu in Cu2+ has E = 63.5/2 = about 32.

Question 5: 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) The atomic weight of the metal is 4
(b) The equivalent weight of the metal is 8
(c) The equivalent weight of the metal is 4
(d) The atomic weight of the metal is 8
Show Answer & Explanation

Answer: (c) The equivalent weight of the metal is 4

Explanation:
1. Oxygen in the oxide = 3.15 − 1.05 = 2.10 g.
2. E = (1.05/2.10) × 8 = 4.

Teacher's Note:
1. Mass lost in reduction is the oxygen.
2. Equivalent weight does not tell us the atomic weight directly.

Question 6: If m1 g of a metal A displaces m2 g of another metal B from its salt solution and the equivalent weight of B is E1, then the equivalent weight of A can be expressed as
(a) \( \frac{m_1}{m_2} \times E_2 \)
(b) \( \frac{m_2}{m_1} \times E_2 \)
(c) \( \frac{m_1}{m_2} \times E_1 \)
(d) \( \frac{m_2}{m_1} \times E_1 \)
Show Answer & Explanation

Answer: (c) \( \frac{m_1}{m_2} \times E_1 \)

Explanation:
1. Equal equivalents: m1/E_A = m2/E1.
2. Therefore E_A = (m1/m2) × E1.

Teacher's Note:
1. Displacement takes place in equal equivalents.
2. Cross-multiply to isolate the unknown.

Question 7: 14 g of element X combines with 16 g of oxygen. On the basis of this information, which of the following is a correct statement?
(a) The element X could have an atomic weight of 7 and its oxide is XO
(b) The element X could have an atomic weight of 14 and its oxide is X2O
(c) The element X could have an atomic weight of 7 and its oxide is X2O
(d) The element X could have an atomic weight of 14 and its oxide is XO2
Show Answer & Explanation

Answer: (c) The element X could have an atomic weight of 7 and its oxide is X2O

Explanation:
1. Equivalent weight of X = (14/16) × 8 = 7.
2. Atomic weight = 7 × valency: for X2O the valency is 1, giving atomic weight 7.
3. The other options give wrong equivalent weights (3.5 or 14).

Teacher's Note:
1. Find the equivalent weight first and then test each option.
2. Valency is 1 in X2O and 2 in XO.

Question 8: 2.4 g of a metal displaces 1.12 L of hydrogen at normal temperature and pressure. 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. 1.12 L of H2 = 0.1 eq.
2. E = 2.4/0.1 = 24.

Teacher's Note:
1. Divide by 0.1 equivalents.
2. Compare with question 80, where the mass was 1.2 g.

Question 9: 45 g of an acid of molecular weight 90 is neutralised by 200 mL of 5 N caustic potash. The basicity of the acid is
(a) 1
(b) 2
(c) 3
(d) None of these
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Equivalents of KOH = 5 × 200/1000 = 1 eq.
2. Moles of acid = 45/90 = 0.5 mol.
3. Basicity = 1/0.5 = 2.

Teacher's Note:
1. Basicity is equivalents per mole of acid.
2. Caustic potash is KOH.

Question 10: The weights of two elements which combine with one another are in the ratio of their
(a) Atomic weights
(b) Molecular weights
(c) Equivalent weights
(d) None of these
Show Answer & Explanation

Answer: (c) Equivalent weights

Explanation:
1. By the law of reciprocal or equivalent proportions, elements combine in the ratio of their equivalent weights.

Teacher's Note:
1. Equivalent weights decide combining ratios.
2. Atomic weights decide only when valencies are equal.

Question 11: The oxide of a metal has 32% oxygen. Its equivalent weight would be
(a) 34
(b) 32
(c) 17
(d) 16
Show Answer & Explanation

Answer: (c) 17

Explanation:
1. In 100 g: metal = 68 g and O = 32 g.
2. E = (68/32) × 8 = 17.

Teacher's Note:
1. Use the 100 g basis.
2. E = (metal mass ÷ oxygen mass) × 8.

Question 12: 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, what is the equivalent weight of Ca?
(a) 10
(b) 20
(c) 40
(d) 5
Show Answer & Explanation

Answer: (b) 20

Explanation:
1. Same hydrogen means equal equivalents: 1.6/ECa = 2.60/32.6.
2. ECa = 1.6 × 32.6/2.60 = 20.06, about 20.

Teacher's Note:
1. Equal H2 liberated means equal equivalents.
2. Calcium: 40/2 = 20.

Question 13: 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. Mass of metal = 74.5 − 35.5 = 39.0 g.
2. E = (39.0/35.5) × 35.5 = 39.0.

Teacher's Note:
1. Equivalent weight of chlorine is 35.5.
2. This is potassium chloride.

Question 14: The equivalent weight of an element is 4. Its chloride has vapour density 59.25. Then the valency of the element is
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Molar mass of chloride = 2 × 59.25 = 118.5.
2. Let valency be v: atomic mass = 4v and mass of chloride = 4v + 35.5v = 118.5, so v = 3.

Teacher's Note:
1. Chloride mass = atomic mass + 35.5 per valency unit.
2. Solve for the valency.

Question 15: The vapour density of a metal chloride is 77. The equivalent weight of the metal is 3. Its atomic weight will be
(a) 3
(b) 6
(c) 9
(d) 12
Show Answer & Explanation

Answer: (d) 12

Explanation:
1. Molar mass of chloride = 154.
2. With valency v: 3v + 35.5v = 154, so v = 4.
3. Atomic weight = 3 × 4 = 12.

Teacher's Note:
1. Valency = molar mass of chloride ÷ (E + 35.5).
2. Atomic weight = E × valency.

Question 16: The specific heat of a solid element is 0.1 cal g-1 °C-1 and its equivalent weight is 31.8. Its exact atomic weight is
(a) 31.8
(b) 63.6
(c) 318
(d) 95.4
Show Answer & Explanation

Answer: (b) 63.6

Explanation:
1. Approximate atomic weight = 6.4/0.1 = 64 (Dulong and Petit).
2. Valency = 64/31.8 = about 2, so exact atomic weight = 2 × 31.8 = 63.6.

Teacher's Note:
1. Dulong–Petit gives an approximate atomic weight.
2. Round the valency to a whole number and multiply by E.

Question 17: The specific heat of an element is 0.214 cal g-1 °C-1. The approximate atomic weight is
(a) 0.6
(b) 12
(c) 30
(d) 65
Show Answer & Explanation

Answer: (c) 30

Explanation:
1. Atomic weight ≈ 6.4/0.214 = 29.9, about 30.

Teacher's Note:
1. Atomic weight × specific heat ≈ 6.4.
2. Use this law only for solid elements.

Question 18: A metal M forms a sulphate which is isomorphous with MgSO4·7H2O. If 0.6538 g of metal M displaced 2.16 g of silver from silver nitrate solution, then the atomic weight of M is
(a) 32.61
(b) 56.82
(c) 65.38
(d) 74.58
Show Answer & Explanation

Answer: (c) 65.38

Explanation:
1. E_M = (0.6538/2.16) × 108 = 32.69.
2. The sulphate is isomorphous with MgSO4·7H2O, so M is divalent: atomic weight = 2 × 32.69 = 65.38.

Teacher's Note:
1. Isomorphism fixes the valency (MSO4 type, so 2).
2. The metal is zinc.

Question 19: 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. The carbonate is MCO3 with molar mass = 12/0.06091 = 197.
2. Atomic weight of M = 197 − 60 = 137.

Teacher's Note:
1. Isomorphous with MgCO3 means formula MCO3.
2. The metal is barium.

Question 20: 71 g of chlorine combines with a metal giving 111 g of its chloride. The chloride is isomorphous with MgCl2·6H2O. The atomic mass of the metal is
(a) 20
(b) 30
(c) 40
(d) 69
Show Answer & Explanation

Answer: (c) 40

Explanation:
1. Mass of metal = 111 − 71 = 40 g combining with 71 g of Cl (2 equivalents).
2. The chloride is MCl2, so the atomic mass is 40.

Teacher's Note:
1. Isomorphism gives the formula MCl2.
2. 71 g of Cl is 2 mol of Cl atoms.

Question 21: 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 = 27/9 = 3, so the chloride is MCl3.

Teacher's Note:
1. Valency = atomic weight ÷ equivalent weight.
2. The metal is aluminium.

Question 22: The chloride of a metal contains 71% chlorine by weight and the vapour density of it is 50. The atomic weight of the metal will be
(a) 29
(b) 58
(c) 35.5
(d) 71
Show Answer & Explanation

Answer: (a) 29

Explanation:
1. Molar mass = 2 × 50 = 100 g/mol.
2. Chlorine in 100 g = 71 g = 2 atoms, so the chloride is MCl2.
3. Metal = 100 − 71 = 29.

Teacher's Note:
1. Percentage × molar mass gives the mass of each element in one mole.
2. 71 g of chlorine means two Cl atoms.

Question 23: The specific heat of a metal M is 0.25. Its equivalent weight is 12. What is its correct atomic weight?
(a) 25.6
(b) 36
(c) 24
(d) 12
Show Answer & Explanation

Answer: (c) 24

Explanation:
1. Approximate atomic weight = 6.4/0.25 = 25.6.
2. Valency = 25.6/12 = about 2, so atomic weight = 2 × 12 = 24.

Teacher's Note:
1. Use the approximate value only to find the valency.
2. The exact atomic weight is valency × equivalent weight.

Question 24: The density of air is 0.001293 g mL-1. Its vapour density is
(a) 143
(b) 14.3
(c) 1.43
(d) 0.143
Show Answer & Explanation

Answer: (b) 14.3

Explanation:
1. Molar mass of air = 0.001293 × 22400 = 28.96 g/mol.
2. Vapour density = 28.96/2 = 14.48, which is nearest to 14.3 among the options.

Teacher's Note:
1. V.D. = molar mass ÷ 2.
2. The options are rounded values.

Question 25: The relative density of a volatile substance with respect to CH4 is 4. Its molecular weight would be
(a) 8
(b) 32
(c) 64
(d) 128
Show Answer & Explanation

Answer: (c) 64

Explanation:
1. Relative density = molar mass of substance ÷ molar mass of CH4.
2. Molar mass = 4 × 16 = 64.

Teacher's Note:
1. Relative density compares molar masses.
2. CH4 has molar mass 16.

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

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