JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 26

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

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Multiple Choice Questions

Question 1: In the chemical scale, the relative mass of the isotopic mixture of oxygen atoms (\( \ce{^16O} \), \( \ce{^17O} \), \( \ce{^18O} \)) is assumed to be equal to
(a) 16.002
(b) 16.00
(c) 17.00
(d) 11.00
Show Answer & Explanation

Answer: (b) 16.00

Explanation:
1. On the old chemical scale the average atomic mass of natural oxygen was fixed at exactly 16.00.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 2: For preparing a 0.1 N solution of a compound from its impure sample of known percentage purity, the weight of the substance required will be
(a) more than the theoretical weight
(b) less than the theoretical weight
(c) the same as the theoretical weight
(d) none of these
Show Answer & Explanation

Answer: (a) more than the theoretical weight

Explanation:
1. Only the pure fraction counts, so more sample than the theoretical weight is required to supply the same amount of pure compound.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 3: 1 mol of \( \ce{CH4} \) contains
(a) \( 6.02\times10^{23 \)} atoms of H
(b) 4 g atom of hydrogen
(c) \( 1.81\times10^{23 \)} molecules of \( \ce{CH4} \)
(d) 3.0 g of carbon
Show Answer & Explanation

Answer: (b) 4 g atom of hydrogen

Explanation:
1. 1 mol of \( \ce{CH4} \) has 4 mol of H atoms, i.e. 4 g atom of hydrogen.
2. It contains 12 g of carbon and \( 6.02\times10^{23 \)} molecules.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 4: When potassium permanganate is titrated against ferrous ammonium sulphate in acidic medium, the equivalent weight of potassium permanganate is
(a) molecular weight/10
(b) molecular weight/5
(c) molecular weight/2
(d) molecular weight
Show Answer & Explanation

Answer: (b) molecular weight/5

Explanation:
1. \( \ce{MnO4-} \) goes to \( \ce{Mn^2+} \), a gain of 5 electrons.
2. Equivalent weight = M/5.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 5: One litre of a gas at STP weighs 1.16 g. It can possibly be
(a) \( \ce{C2H2} \)
(b) CO
(c) \( \ce{O2} \)
(d) \( \ce{CH4} \)
Show Answer & Explanation

Answer: (a) \( \ce{C2H2} \)

Explanation:
1. Molar mass = 1.16 x 22.4 = 26 g/mol.
2. This matches \( \ce{C2H2} \) (26 g/mol).

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 6: The equivalent weight of an element is 4. Its chloride has a vapour density of 59.25. 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 x 59.25 = 118.5.
2. Atomic mass = 4n and chloride mass = 4n + 35.5n = 118.5, so n = 3.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 7: 1.24 g of phosphorus is present in 2.2 g of a phosphorus sulphide. The compound is
(a) \( \ce{P4S3} \)
(b) \( \ce{P2S2} \)
(c) \( \ce{PS2} \)
(d) \( \ce{P2S4} \)
Show Answer & Explanation

Answer: (a) \( \ce{P4S3} \)

Explanation:
1. Sulphur = 2.2 - 1.24 = 0.96 g.
2. Moles: P = 1.24/31 = 0.04, S = 0.96/32 = 0.03.
3. Ratio P:S = 4:3, so \( \ce{P4S3} \).

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 8: The atomic weights of two elements A and B are 40 and 80 respectively. If x g of A contains y atoms, the number of atoms in 2x g of B is
(a) y/2
(b) y/4
(c) y
(d) 2y
Show Answer & Explanation

Answer: (c) y

Explanation:
1. x g of A = x/40 mol. 2x g of B = 2x/80 = x/40 mol.
2. Equal moles contain equal numbers of atoms, so y.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 9: The weight of \( 1\times10^{22 \)} molecules of \( \ce{CuSO4.5H2O} \) is
(a) 41.59 g
(b) 415.9 g
(c) 4.159 g
(d) none of these
Show Answer & Explanation

Answer: (c) 4.159 g

Explanation:
1. Moles = \( 10^{22}/6.02\times10^{23 \)} = 0.0166.
2. Mass = 0.0166 x 249.5 = 4.14 g, closest to 4.159 g.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 10: Arrange the following in increasing order of mass (N = 14, O = 16, Cu = 63): I. 1 molecule of oxygen, II. 1 atom of nitrogen, III. \( 1\times10^{-10 \)} g molecular weight of oxygen, IV. \( 1\times10^{-10 \)} g atomic weight of copper
(a) II < I < III < IV
(b) IV < III < II < I
(c) II < III < I < IV
(d) III < IV < I < II
Show Answer & Explanation

Answer: (a) II < I < III < IV

Explanation:
1. I = 32 amu = \( 5.3\times10^{-23 \)} g; II = 14 amu = \( 2.3\times10^{-23 \)} g.
2. III = \( 10^{-10}\times32=3.2\times10^{-9 \)} g; IV = \( 10^{-10}\times63=6.3\times10^{-9 \)} g.
3. Order: II < I < III < IV.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 11: 1.520 g of the hydroxide of a metal on ignition gave 0.995 g of oxide. The equivalent weight of the metal is
(a) 1.520
(b) 0.995
(c) 19.00
(d) 9.00
Show Answer & Explanation

Answer: (d) 9.00

Explanation:
1. Hydroxide and oxide: (E + 17)/(E + 8) = 1.520/0.995 = 1.528.
2. E + 17 = 1.528E + 12.2 gives E = 9.0.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 12: How much coulomb charge is on 1 g ion of \( \ce{N^3-} \)?
(a) \( 5.2\times10^{6 \)} C
(b) \( 2.894\times10^{5 \)} C
(c) \( 6.6\times10^{6 \)} C
(d) \( 8.2\times10^{6 \)} C
Show Answer & Explanation

Answer: (b) \( 2.894\times10^{5 \)} C

Explanation:
1. 1 g ion = 1 mole of ions, each carrying 3 electronic charges.
2. Charge = 3 x 96500 = \( 2.895\times10^{5 \)} C.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 13: What should be the equivalent weight of phosphorous acid (\( \ce{H3PO3} \)) if P = 31, O = 16, H = 1?
(a) 82
(b) 41
(c) 20.5
(d) none of these
Show Answer & Explanation

Answer: (b) 41

Explanation:
1. \( \ce{H3PO3} \) has molar mass 82 and is dibasic (only two replaceable H).
2. Equivalent weight = 82/2 = 41.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 14: The number of molecules at NTP in 1 mL of an ideal gas will be
(a) \( 6\times10^{23 \)}
(b) \( 2.69\times10^{19 \)}
(c) \( 2.69\times10^{23 \)}
(d) none of these
Show Answer & Explanation

Answer: (b) \( 2.69\times10^{19 \)}

Explanation:
1. 1 mL = 1/22400 mol.
2. Molecules = \( 6.02\times10^{23}/22400=2.69\times10^{19 \)}.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 15: What volume of \( \ce{NH3} \) gas at STP would be needed to prepare 100 mL of 2.5 molal ammonium hydroxide solution (taking the solution as essentially 2.5 M)?
(a) 0.056 litres
(b) 0.56 litres
(c) 5.6 litres
(d) 11.2 litres
Show Answer & Explanation

Answer: (c) 5.6 litres

Explanation:
1. Moles needed = 0.1 L x 2.5 = 0.25 mol.
2. Volume = 0.25 x 22.4 = 5.6 L.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 16: If the density of water is 1 g \( cm^{-3 \)}, the volume occupied by one molecule of water is approximately
(a) 18 \( cm^3 \)
(b) 22400 \( cm^3 \)
(c) \( 6.02\times10^{-23 \)} \( cm^3 \)
(d) \( 3.0\times10^{-23 \)} \( cm^3 \)
Show Answer & Explanation

Answer: (d) \( 3.0\times10^{-23 \)} \( cm^3 \)

Explanation:
1. 18 g of water has volume 18 \( cm^3 \) and contains \( 6.02\times10^{23 \)} molecules.
2. One molecule: 18/\( 6.02\times10^{23 \)} = \( 3.0\times10^{-23 \)} \( cm^3 \).

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 17: One gram of hydrogen is found to combine with 80 g of bromine. One gram 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 is the mass combining with 80 g of bromine (1 equivalent).
2. Calcium: 1 g per 4 g of bromine, so 80 g of bromine needs 20 g of calcium.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 18: 100 mL of \( \ce{PH3} \) on decomposition produced phosphorus and hydrogen. The change in volume is
(a) 50 mL increase
(b) 500 mL decrease
(c) 900 mL decrease
(d) nil
Show Answer & Explanation

Answer: (a) 50 mL increase

Explanation:
1. \( \ce{4PH3 -> P4 + 6H2} \) (phosphorus is solid).
2. 400 mL of gas gives 600 mL of hydrogen, so 100 mL gives 150 mL: increase of 50 mL.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 19: 12 g of Mg (atomic mass 24) on reacting completely with acid gives hydrogen gas, the volume of which at STP would be
(a) 22.4 L
(b) 11.2 L
(c) 44.8 L
(d) 6.1 L
Show Answer & Explanation

Answer: (b) 11.2 L

Explanation:
1. 12 g Mg = 0.5 mol, giving 0.5 mol of \( \ce{H2} \).
2. Volume = 0.5 x 22.4 = 11.2 L.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 20: Which of the following has the least mass?
(a) 2 g atom of nitrogen
(b) \( 3\times10^{23 \)} atoms of C
(c) 1 mole of S
(d) 7.0 g of Ag
Show Answer & Explanation

Answer: (b) \( 3\times10^{23 \)} atoms of C

Explanation:
1. 2 g atom N = 28 g; \( 3\times10^{23 \)} atoms of C = 0.5 mol = 6 g; 1 mol S = 32 g; Ag = 7.0 g.
2. The least is 6 g.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 21: The volume of a gas at STP is \( 1.12\times10^{-7 \)} cc. The number of molecules in it is
(a) \( 3.01\times10^{20 \)}
(b) \( 3.01\times10^{12 \)}
(c) \( 3.01\times10^{23 \)}
(d) \( 3.01\times10^{24 \)}
Show Answer & Explanation

Answer: (b) \( 3.01\times10^{12 \)}

Explanation:
1. Moles = \( 1.12\times10^{-7}/22400=5\times10^{-12 \)}.
2. Molecules = \( 5\times10^{-12}\times6.02\times10^{23}=3.01\times10^{12 \)}.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

Question 22: The number of moles of oxygen in 1 L of air containing 21% oxygen by volume, in standard conditions, is
(a) 0.186 mol
(b) 0.21 mol
(c) 2.10 mol
(d) 0.0093 mol
Show Answer & Explanation

Answer: (d) 0.0093 mol

Explanation:
1. Oxygen volume = 0.21 L.
2. Moles = 0.21/22.4 = 0.0094 mol.

Teacher's Note:
1. Write the relevant definition or formula first, then substitute the data carefully.
2. Check units and the order of magnitude before choosing the option.

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Some Basic Concepts Of Chemistry Objective Questions & Solutions for JEE Chemistry

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