Atomic & Molecular Masses — JEE Main Chemistry MCQs with Solutions
Free JEE Main Chemistry Atomic & Molecular Masses MCQs with step-by-step solutions (7 questions). Part of Some Basic Concepts of Chemistry. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Atomic & Molecular Masses · medium · theory
For sodium chloride the value 58.5 u is called its formula mass rather than its molecular mass because:
A. the atomic masses of sodium and chlorine are not known accurately
B. in the solid state NaCl exists as a three-dimensional array of Na⁺ and Cl⁻ ions, not as discrete NaCl molecules ✓ Correct
C. sodium chloride is held together by covalent bonds between Na and Cl atoms
D. sodium chloride is a heterogeneous mixture of sodium and chlorine
Solution: In solid NaCl one Na⁺ ion is surrounded by six Cl⁻ ions and vice versa, forming a giant ionic lattice with no separate NaCl molecule. Because there is no discrete molecule, the mass of the simplest formula unit (23 + 35.5 = 58.5 u) is called the formula mass, not the molecular mass.
Q2 — Atomic & Molecular Masses · medium · numerical
Boron occurs as two isotopes, ¹⁰B (abundance 20%) and ¹¹B (abundance 80%). The average atomic mass of boron is:
A. 10.8 u ✓ Correct
B. 11.0 u
C. 10.2 u
D. 10.5 u
Solution: Average atomic mass = (0.20 × 10) + (0.80 × 11) = 2.0 + 8.8 = 10.8 u. Taking a plain average of 10 and 11 gives 10.5 u; swapping the two abundances by mistake, (0.80 × 10) + (0.20 × 11) = 8.0 + 2.2 = 10.2 u.
Q3 — Atomic & Molecular Masses · hard · numerical
An element has two isotopes of atomic masses 63 u and 65 u, and its average atomic mass is 63.5 u. The percentage abundance of the lighter (63 u) isotope is:
A. 25%
B. 50%
C. 75% ✓ Correct
D. 65%
Solution: Let the fraction of the 63 u isotope be x. Then 63x + 65(1 − x) = 63.5 ⇒ 65 − 2x = 63.5 ⇒ 2x = 1.5 ⇒ x = 0.75, i.e. 75% (the element is copper, Cu = 63.5 u). The value 25% is the abundance of the heavier isotope, and 50% would give a plain average of 64 u, not 63.5 u.
Q4 — Atomic & Molecular Masses · hard · numerical
The molecular mass of ammonium sulphate, (NH₄)₂SO₄, using N = 14, H = 1, S = 32 and O = 16 (all in u), is:
A. 96 u
B. 124 u
C. 132 u ✓ Correct
D. 114 u
Solution: The subscript 2 outside the bracket doubles both N and H: (NH₄)₂SO₄ has 2 N, 8 H, 1 S and 4 O. Mass = 2(14) + 8(1) + 32 + 4(16) = 28 + 8 + 32 + 64 = 132 u. Using only one NH₄ group (14 + 4 + 32 + 64) gives 114 u; forgetting the 8 hydrogens (28 + 32 + 64) gives 124 u; counting only the sulphate part (32 + 64) gives 96 u.
Q5 — Atomic & Molecular Masses · medium · numerical
The molecular mass of a metal carbonate MCO₃ is found to be 84 u (C = 12 u, O = 16 u). The atomic mass of the metal M is:
A. 56 u
B. 44 u
C. 24 u ✓ Correct
D. 40 u
Solution: MCO₃ mass = M + 12 + (3 × 16) = M + 60. So M = 84 − 60 = 24 u (the metal is magnesium). A common slip is to treat the carbonate group as CO₂ (44 u), giving M = 84 − 44 = 40 u; using only one oxygen (M + 12 + 16 = 84) gives M = 56 u.
Q6 — Atomic & Molecular Masses · hard · numerical
Neon consists of three isotopes: ²⁰Ne (90.9%), ²¹Ne (0.3%) and ²²Ne (8.8%), with isotopic masses 20, 21 and 22 u respectively. The average atomic mass of neon is closest to:
A. 21.00 u
B. 20.00 u
C. 20.18 u ✓ Correct
D. 20.90 u
Solution: Average atomic mass = (0.909 × 20) + (0.003 × 21) + (0.088 × 22) = 18.18 + 0.063 + 1.936 = 20.18 u. A plain average of 20, 21 and 22 would give 21.00 u, and simply taking the most abundant isotope would give 20.00 u — both ignore the weighting by abundance.
Q7 — Atomic & Molecular Masses · hard · numerical
A binary oxide has the formula X₂O₃ and a molecular mass of 102 u (O = 16 u). The atomic mass of element X is:
A. 54 u
B. 51 u
C. 27 u ✓ Correct
D. 43 u
Solution: X₂O₃ mass = 2X + (3 × 16) = 2X + 48 = 102 ⇒ 2X = 54 ⇒ X = 27 u (the oxide is Al₂O₃). Forgetting to divide by 2 gives 54 u; using only one oxygen (2X + 16 = 102) gives 43 u; ignoring the oxygen entirely (2X = 102) gives 51 u.