JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 05

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Chapter-wise Objective Questions: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: The number of litres of air required to burn 8 litres of C2H2 is approximately
(a) 40
(b) 60
(c) 80
(d) 100
Show Answer & Explanation

Answer: (d) 100

Explanation:
1. C2H2 + 5/2 O2 → 2CO2 + H2O, so 8 L of C2H2 needs 20 L of O2.
2. Air has about 20% O2, so air = 20 × 5 = 100 L.

Teacher's Note:
1. Air contains about one fifth oxygen by volume.
2. Multiply the O2 volume by 5.

Question 2: If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is (3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl)
(a) 0.7
(b) 0.5
(c) 0.3
(d) 0.1
Show Answer & Explanation

Answer: (d) 0.1

Explanation:
1. From BaCl2: 0.5/3 = 0.167 mol; from Na3PO4: 0.2/2 = 0.1 mol.
2. The limiting reagent is Na3PO4, so the maximum product is 0.1 mol.

Teacher's Note:
1. The smaller of the two ratios decides the product.
2. Divide moles by the coefficient before comparing.

Question 3: If 8 mL of uncombined O2 remain after exploding O2 with 4 mL of hydrogen, the number of mL of O2 originally present was
(a) 12
(b) 2
(c) 10
(d) 4
Show Answer & Explanation

Answer: (c) 10

Explanation:
1. 2H2 + O2 → 2H2O: 4 mL of H2 uses 2 mL of O2.
2. Original O2 = 8 + 2 = 10 mL.

Teacher's Note:
1. Add the amount consumed to the amount left over.
2. Gas volumes are in the ratio of the coefficients.

Question 4: 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; O2 is limiting.
2. Water formed = 2 × 0.125 = 0.25 mol = 4.5 g.

Teacher's Note:
1. Find the limiting reagent first.
2. Check: 0.5 g of H2 reacts with 4 g of O2 to form 4.5 g of water.

Question 5: For the reaction A + 2B → C, 5 mol of A and 8 mol of B will produce
(a) 5 mole of C
(b) 4 mole of C
(c) 8 mole of C
(d) 13 mole of C
Show Answer & Explanation

Answer: (b) 4 mole of C

Explanation:
1. A would give 5 mol of C, B would give 8/2 = 4 mol.
2. B is the limiting reagent, so C = 4 mol.

Teacher's Note:
1. Divide moles by the coefficient and take the smaller.
2. The limiting reagent decides the yield.

Question 6: If 1.6 g of SO2 and 1.5 × 1022 molecules of H2S are mixed and allowed to remain in contact in a closed vessel until the reaction 2H2S + SO2 → 3S + 2H2O proceeds to completion, which of the following statements is true?
(a) Only S and H2O remain in the reaction vessel.
(b) H2S will remain in excess.
(c) SO2 will remain in excess.
(d) None of these
Show Answer & Explanation

Answer: (c) SO2 will remain in excess.

Explanation:
1. Moles SO2 = 1.6/64 = 0.025; moles H2S = 1.5 × 1022/6.02 × 1023 = 0.0249.
2. H2S needs only half as many moles of SO2 (0.0125 mol), so H2S is limiting.
3. SO2 is left in excess.

Teacher's Note:
1. Compare the available moles with the coefficients 2 : 1.
2. The reagent that runs out first is limiting.

Question 7: 12 L of H2 and 11.2 L of Cl2 are mixed and exploded. The composition by volume of the mixture is
(a) 24 L of HCl(g)
(b) 0.8 L Cl2 and 20.8 L HCl(g)
(c) 0.8 L H2 and 22.4 L HCl(g)
(d) 22.4 L HCl(g)
Show Answer & Explanation

Answer: (c) 0.8 L H2 and 22.4 L HCl(g)

Explanation:
1. H2 + Cl2 → 2HCl: 11.2 L Cl2 reacts with 11.2 L H2 to give 22.4 L HCl.
2. H2 left = 12 − 11.2 = 0.8 L.

Teacher's Note:
1. Chlorine is the limiting gas.
2. Unused hydrogen stays in the mixture.

Question 8: 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. 10 mL gives 40 mL CO2, so 1 molecule has 4 C atoms.
2. 10 mL gives 50 mL H2O, so 1 molecule has 5 × 2 = 10 H atoms.
3. The hydrocarbon is C4H10.

Teacher's Note:
1. Volume ratios give atoms per molecule directly.
2. Each H2O contains two H atoms.

Question 9: 500 mL of a gaseous hydrocarbon when burnt in excess of O2 gives 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. CO2/hydrocarbon = 2500/500 = 5, so 5 C atoms.
2. H2O/hydrocarbon = 3000/500 = 6, so 12 H atoms.
3. The formula is C5H12.

Teacher's Note:
1. Divide product volumes by the hydrocarbon volume.
2. Double the water ratio to get hydrogen atoms.

Question 10: The molecular weight of a tribasic acid is W. Its equivalent weight will be
(a) \( \frac{W}{2} \)
(b) \( \frac{W}{3} \)
(c) W
(d) 3W
Show Answer & Explanation

Answer: (b) \( \frac{W}{3} \)

Explanation:
1. Equivalent weight of an acid = molar mass ÷ basicity.
2. For a tribasic acid, basicity = 3, so E = W/3.

Teacher's Note:
1. Basicity is the number of replaceable H+ ions.
2. Divide the molar mass by it.

Question 11: A, E, M and n are the atomic weight, equivalent weight, molecular weight and valency of an element. The correct relation is
(a) A = E × n
(b) \( A = \frac{M}{E} \)
(c) \( A = \frac{M}{n} \)
(d) M = A × n
Show Answer & Explanation

Answer: (a) A = E × n

Explanation:
1. Equivalent weight = atomic weight ÷ valency, so A = E × n.

Teacher's Note:
1. This is the basic definition of equivalent weight.
2. Valency links atomic weight and equivalent weight.

Question 12: Sulphur forms two chlorides S2Cl2 and SCl2. The equivalent mass of sulphur in SCl2 is 16. The equivalent weight of sulphur in S2Cl2 is
(a) 8
(b) 16
(c) 32
(d) 64
Show Answer & Explanation

Answer: (c) 32

Explanation:
1. In SCl2 sulphur has valency 2, and equivalent weight = 32/2 = 16.
2. In S2Cl2 each S has valency 1, so equivalent weight = 32/1 = 32.

Teacher's Note:
1. Equivalent weight changes with the valency shown in the compound.
2. Find the valency from the number of Cl atoms per S atom.

Question 13: If the equivalent weight of S in SO2 is 8, the equivalent weight of S in SO3 is
(a) \( \frac{8 \times 2}{3} \)
(b) \( \frac{8 \times 3}{2} \)
(c) 8 × 2 × 3
(d) \( \frac{2 \times 3}{8} \)
Show Answer & Explanation

Answer: (a) \( \frac{8 \times 2}{3} \)

Explanation:
1. In SO2, S has valency 4, so E = 32/4 = 8.
2. In SO3, S has valency 6, so E = 32/6 = 16/3 = (8 × 2)/3.

Teacher's Note:
1. Equivalent weight is inversely proportional to valency.
2. ESO3 = 8 × 4/6 = 8 × 2/3.

Question 14: 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. Valency and equivalent weight vary with the compound formed.
2. Atomic weight of an element is fixed.

Teacher's Note:
1. Atomic weight is a constant of the element.
2. Valency can change, for example Fe2+ and Fe3+.

Question 15: One g equivalent of a substance is present in
(a) 0.25 mol of O2
(b) 0.5 mol of O2
(c) 1.00 mol of O2
(d) 8.00 mol of O2
Show Answer & Explanation

Answer: (a) 0.25 mol of O2

Explanation:
1. Equivalent weight of oxygen = 8 g.
2. 1 g equivalent = 8 g of O = 0.25 mol of O2 (32 g).

Teacher's Note:
1. One g equivalent of O is 8 g.
2. 8 g is a quarter of the molar mass of O2.

Question 16: In a compound A_xB_y,
(a) Mole of A = mole of B = mole of A_xB_y
(b) eq. of A = eq. of B = eq. of A_xB_y
(c) yx mole of A = yx mole of B = (x + y) × mole of A_xB_y
(d) y × mole of A = y × mole of B
Show Answer & Explanation

Answer: (b) eq. of A = eq. of B = eq. of A_xB_y

Explanation:
1. In any compound, equivalents of the combining elements are equal and equal to the equivalents of the compound.
2. So eq. of A = eq. of B = eq. of A_xB_y.

Teacher's Note:
1. This is the law of equivalent proportions in another form.
2. Moles do not combine in equal numbers, but equivalents do.

Question 17: 0.45 g of an acid (molecular weight 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 = 0.5 × 20/1000 = 0.01 eq.
2. Moles of acid = 0.45/90 = 0.005 mol.
3. Basicity = 0.01/0.005 = 2.

Teacher's Note:
1. Basicity = equivalents per mole of acid.
2. Normality × volume in litres gives equivalents.

Question 18: 0.5 g of a base was completely neutralised by 100 mL of 0.2 N acid. The equivalent weight of the base is
(a) 50
(b) 100
(c) 25
(d) 125
Show Answer & Explanation

Answer: (c) 25

Explanation:
1. Equivalents of acid = 0.2 × 100/1000 = 0.02 eq.
2. Equivalent weight of base = 0.5/0.02 = 25.

Teacher's Note:
1. At neutralisation, equivalents of acid equal equivalents of base.
2. E = mass ÷ equivalents.

Question 19: 0.126 g of an acid requires 20 mL of 0.1 N NaOH for complete neutralisation. The equivalent weight of the acid is
(a) 45
(b) 53
(c) 40
(d) 63
Show Answer & Explanation

Answer: (d) 63

Explanation:
1. Equivalents of NaOH = 0.1 × 20/1000 = 0.002 eq.
2. E = 0.126/0.002 = 63.

Teacher's Note:
1. Use E = mass/equivalents.
2. 63 is the equivalent weight of oxalic acid dihydrate.

Question 20: 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 eq.
2. Equivalents of acid are also 0.05, so E = 3/0.05 = 60.

Teacher's Note:
1. Equal equivalents react.
2. Divide mass by equivalents to get E.

Question 21: The equivalent weight of a divalent metal is 24. The volume of hydrogen liberated at STP by 12 g of the same metal when added to excess of acid solution is
(a) 2.8 litres
(b) 5.6 litres
(c) 11.2 litres
(d) 22.4 litres
Show Answer & Explanation

Answer: (b) 5.6 litres

Explanation:
1. Equivalents of metal = 12/24 = 0.5 eq.
2. 1 equivalent of H2 occupies 11.2 L at STP, so volume = 0.5 × 11.2 = 5.6 L.

Teacher's Note:
1. Equivalent volume of hydrogen at STP is 11.2 L.
2. The valency is given only as extra information.

Question 22: 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. Equivalents of acid = 0.5 × 40/1000 = 0.02 eq.
2. E = 0.84/0.02 = 42.

Teacher's Note:
1. Normality N/2 means 0.5 N.
2. E = mass ÷ equivalents.

Question 23: 1.0 g of a metal combines with 8.89 g of bromine. The equivalent weight of the metal is nearly (at. wt. of Br = 80)
(a) 8
(b) 9
(c) 10
(d) 7
Show Answer & Explanation

Answer: (b) 9

Explanation:
1. Equivalent weight of Br = 80 (valency 1).
2. E of metal = (1/8.89) × 80 = 9.0.

Teacher's Note:
1. Metal and bromine combine in the ratio of their equivalent weights.
2. The result is close to 9.

Question 24: 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. of P = 31)
(a) 100 mL
(b) 200 mL
(c) 80 mL
(d) 300 mL
Show Answer & Explanation

Answer: (b) 200 mL

Explanation:
1. Molar mass of NaH2PO4 = 120, so 12 g = 0.1 mol.
2. Converting NaH2PO4 to Na3PO4 replaces 2 H, needing 2 mol NaOH per mole = 0.2 mol.
3. Volume of 1 M NaOH = 0.2 L = 200 mL.

Teacher's Note:
1. Count the hydrogens that must be replaced.
2. Volume = moles ÷ molarity.

Question 25: 0.84 g of a metal hydride contains 0.04 g of hydrogen. The equivalent weight of the metal is
(a) 80
(b) 40
(c) 20
(d) 60
Show Answer & Explanation

Answer: (c) 20

Explanation:
1. Mass of metal = 0.84 − 0.04 = 0.80 g.
2. E = (0.80/0.04) × 1 = 20.

Teacher's Note:
1. Equivalent weight of hydrogen is 1.
2. Metal : hydrogen mass ratio equals the ratio of their equivalent weights.

Question 26: A1 g of an element gives A2 g of its oxide. The equivalent mass of the element is
(a) \( \frac{A_2 - A_1}{A_1} \times 8 \)
(b) \( \frac{A_2 - A_1}{A_2} \times 8 \)
(c) \( \frac{A_1}{A_2 - A_1} \times 8 \)
(d) \( (A_2 - A_1) \times 8 \)
Show Answer & Explanation

Answer: (c) \( \frac{A_1}{A_2 - A_1} \times 8 \)

Explanation:
1. Mass of oxygen combined = A2 − A1.
2. Equivalent mass = (mass of element/mass of oxygen) × 8 = \( \frac{A_1}{A_2 - A_1} \times 8 \).

Teacher's Note:
1. Equivalent weight of oxygen is 8.
2. Element mass goes in the numerator.

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