JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 24

Chemistry Objective Questions and Answers: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: The equivalent mass of sulphur in SCl2 is
(a) 8 g/mol
(b) 16 g/mol
(c) 64.8 g/mol
(d) 32 g/mol
Show Answer & Explanation

Answer: (b) 16 g/mol

Explanation:
1. Atomic mass of S = 32 and its valency in SCl2 is 2.
2. Equivalent mass = 32/2 = 16.

Teacher's Note:
1. Equivalent mass = atomic mass / valency.

Question 2: The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with ZnSO4.7H2O. The atomic mass of M is
(a) 40.3
(b) 36.3
(c) 24.3
(d) 11.3
Show Answer & Explanation

Answer: (c) 24.3

Explanation:
1. Isomorphous with ZnSO4.7H2O, so the formula is MSO4.7H2O with molar mass = M + 96 + 126 = M + 222.
2. M/(M + 222) = 0.0987, so M = 0.0987 × 222 / 0.9013.
3. M = 24.3 (magnesium).

Teacher's Note:
1. Isomorphous salts have the same formula type, which fixes the formula of the unknown salt.

Question 3: When 100 mL of 1 M NaOH solution and 10 mL of 10 N H2SO4 solution are mixed, the resulting solution will be
(a) Alkaline
(b) Acidic
(c) Strongly acidic
(d) Neutral
Show Answer & Explanation

Answer: (d) Neutral

Explanation:
1. 1 M NaOH = 1 N, so 100 mL has 100 milli-equivalents.
2. 10 mL of 10 N H2SO4 = 100 milli-equivalents.
3. Equal equivalents of acid and base, so the solution is neutral.

Teacher's Note:
1. Compare milli-equivalents (N × V in mL), not volumes.

Question 4: Boron has two stable isotopes, 10B (19%) and 11B (81%). The atomic mass that should appear for boron in the periodic table is
(a) 10.8
(b) 10.2
(c) 11.2
(d) 10.0
Show Answer & Explanation

Answer: (a) 10.8

Explanation:
1. Average atomic mass = (19 × 10 + 81 × 11)/100.
2. = (190 + 891)/100 = 10.81, i.e. about 10.8.

Teacher's Note:
1. The periodic-table atomic mass is the abundance-weighted average of the isotopic masses.

Question 5: What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed together?
(a) 0.1 M
(b) 0.2 M
(c) 0.05 M
(d) 0.25 M
Show Answer & Explanation

Answer: (c) 0.05 M

Explanation:
1. AgNO3 + NaCl → AgCl (precipitate) + NaNO3. Nitrate stays in solution.
2. Moles of NO3− stay the same but the total volume doubles.
3. Concentration = 0.1/2 = 0.05 M.

Teacher's Note:
1. Mixing equal volumes halves every concentration that is not removed by precipitation.

Question 6: The total number of atoms represented by the compound CuSO4.5H2O is
(a) 27
(b) 21
(c) 5
(d) 8
Show Answer & Explanation

Answer: (b) 21

Explanation:
1. CuSO4 has 1 + 1 + 4 = 6 atoms.
2. 5H2O has 5 × 3 = 15 atoms.
3. Total = 6 + 15 = 21 atoms.

Teacher's Note:
1. Count the water of crystallisation as well.

Question 7: 74.5 g of a metallic chloride contain 35.5 g of chlorine. The equivalent weight 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. Equivalent weight of the metal is the mass that combines with 35.5 g of chlorine.
3. So it is 39.0.

Teacher's Note:
1. Equivalent weight of a metal = mass of metal combining with 35.5 g of chlorine (or 8 g of oxygen, or 1 g of hydrogen).

Question 8: The number of atoms in 4.25 g of NH3 is approximately
(a) 1 × 1023
(b) 2 × 1023
(c) 4 × 1023
(d) 6 × 1023
Show Answer & Explanation

Answer: (d) 6 × 1023

Explanation:
1. Moles of NH3 = 4.25/17 = 0.25 mol.
2. Each molecule has 4 atoms, so moles of atoms = 0.25 × 4 = 1 mol.
3. Number of atoms = 6.02 × 1023.

Teacher's Note:
1. Moles × atomicity gives the moles of atoms.

Question 9: The equivalent weight of crystalline oxalic acid (H2C2O4.2H2O) is
(a) 30
(b) 63
(c) 53
(d) 45
Show Answer & Explanation

Answer: (b) 63

Explanation:
1. Molar mass of H2C2O4.2H2O = 90 + 36 = 126.
2. Oxalic acid is dibasic (n = 2).
3. Equivalent weight = 126/2 = 63.

Teacher's Note:
1. Include water of crystallisation in the molar mass.

Question 10: 1.25 g of a solid dibasic acid is completely neutralised by 25 mL of 0.25 M Ba(OH)2 solution. The molar mass of the acid is
(a) 100
(b) 150
(c) 120
(d) 200
Show Answer & Explanation

Answer: (d) 200

Explanation:
1. Moles of Ba(OH)2 = 0.25 × 25/1000 = 6.25 × 10−3 mol, giving 12.5 meq of base.
2. Dibasic acid: 1 mol reacts with 1 mol of Ba(OH)2, so moles of acid = 6.25 × 10−3.
3. Molar mass = 1.25/6.25 × 10−3 = 200 g/mol.

Teacher's Note:
1. Check basicity and acidity before converting to moles.

Question 11: The oxide of a metal contains 32% oxygen. Its equivalent weight is
(a) 34
(b) 32
(c) 17
(d) 8
Show Answer & Explanation

Answer: (c) 17

Explanation:
1. In 100 g of oxide: oxygen = 32 g, metal = 68 g.
2. Equivalent weight of metal = (68/32) × 8.
3. = 17.

Teacher's Note:
1. Equivalent weight = (mass of metal / mass of oxygen) × 8.

Question 12: The mass of a molecule of water is
(a) 3 × 10−26 kg
(b) 3 × 10−25 kg
(c) 1.5 × 10−26 kg
(d) 2.5 × 10−26 kg
Show Answer & Explanation

Answer: (a) 3 × 10−26 kg

Explanation:
1. Molar mass of water = 18 g = 0.018 kg.
2. Mass of one molecule = 0.018/6.022 × 1023 = 3 × 10−26 kg.

Teacher's Note:
1. Divide the molar mass by Avogadro's number to get the mass of a single particle.

Question 13: Assuming complete decomposition, the volume of CO2 released at STP on heating 9.85 g of BaCO3 (Ba = 137) is
(a) 0.84 L
(b) 2.24 L
(c) 4.06 L
(d) 1.12 L
Show Answer & Explanation

Answer: (d) 1.12 L

Explanation:
1. Molar mass of BaCO3 = 137 + 12 + 48 = 197 g/mol.
2. Moles = 9.85/197 = 0.05 mol.
3. BaCO3 → BaO + CO2 gives 0.05 mol CO2 = 0.05 × 22.4 = 1.12 L.

Teacher's Note:
1. One mole of any gas occupies 22.4 L at STP.

Question 14: If N_A is Avogadro's number, the number of valence electrons in 4.2 g of nitride ions (N3−) is
(a) 2.4 N_A
(b) 4.2 N_A
(c) 1.6 N_A
(d) 3.2 N_A
Show Answer & Explanation

Answer: (a) 2.4 N_A

Explanation:
1. Moles of N3− = 4.2/14 = 0.3 mol.
2. Each N3− has 8 valence electrons (it has a complete octet).
3. Total = 0.3 × 8 = 2.4 mol of electrons = 2.4 N_A.

Teacher's Note:
1. The nitride ion has gained three electrons to complete its octet.

Question 15: The specific heat of a metal is 0.16. Its approximate atomic weight would be
(a) 32
(b) 16
(c) 40
(d) 64
Show Answer & Explanation

Answer: (c) 40

Explanation:
1. By Dulong and Petit's law, atomic weight × specific heat ≈ 6.4.
2. Atomic weight = 6.4/0.16 = 40.

Teacher's Note:
1. Dulong–Petit gives only an approximate atomic weight.

Question 16: The weight of one molecule of the compound C60H122 is
(a) 1.4 × 10−21 g
(b) 1.09 × 10−21 g
(c) 5.025 × 1023 g
(d) 16.023 × 1023 g
Show Answer & Explanation

Answer: (a) 1.4 × 10−21 g

Explanation:
1. Molar mass = 60 × 12 + 122 × 1 = 842 g/mol.
2. Mass of one molecule = 842/6.022 × 1023 = 1.4 × 10−21 g.

Teacher's Note:
1. A mass of 1023 g is absurd for a single molecule; check the order of magnitude.

Question 17: The mass of oxygen required for the complete combustion of 2.8 kg of ethylene (C2H4) is
(a) 2.8 kg
(b) 6.4 kg
(c) 9.6 kg
(d) 96 kg
Show Answer & Explanation

Answer: (c) 9.6 kg

Explanation:
1. C2H4 + 3O2 → 2CO2 + 2H2O.
2. 28 kg of C2H4 needs 96 kg of O2, so 2.8 kg needs 9.6 kg.

Teacher's Note:
1. Use mole ratios from the balanced equation; kg and kmol scale together.

Question 18: Caffeine has a molecular weight of 194. If it contains 28.9% by mass of nitrogen, the number of nitrogen atoms in one molecule of caffeine is
(a) 4
(b) 6
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. Mass of nitrogen per mole = 0.289 × 194 = 56 g.
2. Number of N atoms = 56/14 = 4.

Teacher's Note:
1. Mass of an element per mole divided by its atomic mass gives atoms per molecule.

Question 19: A 400 mg iron capsule contains 100 mg of ferrous fumarate, (CHCOO)2Fe. The percentage of iron present in the capsule is approximately
(a) 33%
(b) 25%
(c) 14%
(d) 8%
Show Answer & Explanation

Answer: (d) 8%

Explanation:
1. Molar mass of (CHCOO)2Fe = 2 × 57 + 56 = 170 (C4H2O4Fe).
2. Fraction of Fe in the fumarate = 56/170 = 0.33, so Fe = 33 mg in 100 mg.
3. Percentage in the capsule = 33/400 × 100 ≈ 8%.

Teacher's Note:
1. Note that the percentage is asked for the whole capsule, not for the fumarate alone.

Question 20: The element whose atom has a mass of 10.86 × 10−26 kg is
(a) Boron
(b) Calcium
(c) Silver
(d) Zinc
Show Answer & Explanation

Answer: (d) Zinc

Explanation:
1. Molar mass = 10.86 × 10−26 kg × 6.022 × 1023 = 65.4 × 10−3 kg = 65.4 g/mol.
2. The element with atomic mass 65.4 is zinc.

Teacher's Note:
1. Multiply the atomic mass by Avogadro's number to get the molar mass.

Question 21: The number of gram atoms of oxygen present in 0.3 mol of (COOH)2.2H2O is
(a) 0.6
(b) 1.8
(c) 1.2
(d) 3.6
Show Answer & Explanation

Answer: (b) 1.8

Explanation:
1. One formula unit of (COOH)2.2H2O has 4 + 2 = 6 oxygen atoms.
2. Gram atoms of O = 0.3 × 6 = 1.8.

Teacher's Note:
1. A gram atom is one mole of atoms.

Question 22: A gaseous mixture contains CH4 and C2H6 in equimolecular proportion. The weight of 2.24 L of this mixture at NTP is
(a) 4.6 g
(b) 1.6 g
(c) 2.3 g
(d) 23 g
Show Answer & Explanation

Answer: (c) 2.3 g

Explanation:
1. 2.24 L at NTP = 0.1 mol.
2. Average molar mass = (16 + 30)/2 = 23 g/mol.
3. Mass = 0.1 × 23 = 2.3 g.

Teacher's Note:
1. With equal moles, the average molar mass is the simple mean.

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