2024 H2 Chem P1 Q7 — Atomic structure Solution

The answer is A (58.8).
\(A_r = \frac{(69.3 \times 58) + (26.7 \times 60) + (1.20 \times 61) + (3.80 \times 62) + (1.00 \times 64)}{69.3 + 26.7 + 1.20 + 3.80 + 1.00}\)
\(= \frac{4019.4 + 1602.0 + 73.2 + 235.6 + 64.0}{102.0} = \frac{5994.2}{102.0} = 58.8\)
The relative abundances add up to 102.0, not 100, so the total must be divided by the sum of the abundances, not by 100.
• B is wrong: 59.8 is 1.0 too high. Nickel-58 makes up 69.3 of the 102.0 total abundance, so the weighted mean must lie close to 58; the correct weighting gives 58.8.
• C is wrong: 59.9 comes from dividing 5994.2 by 100, wrongly treating the abundances as percentages.
• D is wrong: 61.0 is the simple average of the five isotope masses, \(\frac{58 + 60 + 61 + 62 + 64}{5}\), which ignores the abundances.