Element M has a relative atomic mass of 24.32. It exists naturally as three isotopes: ^{24… — JAMB UTME

Element $M$ has a relative atomic mass of $24.32$ . It exists naturally as three isotopes: $^{24}M$ with a relative abundance of $79\%$ , $^{25}M$ with a relative abundance of $10\%$ , and an isotope $^{x}M$ with a relative abundance of $11\%$ . What is the mass number ( $x$ ) of the third isotope?

Answer: 26

Answer

The mass number of the third isotope is 26.

The relative atomic mass is the weighted average of isotopic mass numbers:

24.32 = \frac{(24 \times 79) + (25 \times 10) + (x \times 11)}{100}

. Simplifying gives

2432 = 1896 + 250 + 11x

, which simplifies to

11x = 286

, yielding

x = 26

Step-by-Step Solution

Set up the relative atomic mass equation based on percentage abundances.

\text{RAM} = (24 \times 0.79) + (25 \times 0.10) + (x \times 0.11)

The relative atomic mass of an element is the weighted average of the mass numbers of its naturally occurring isotopes.

Calculate the mass contributions of the first two isotopes.

24 \times 0.79 = 18.96

and

25 \times 0.10 = 2.50

, giving a combined sum of

21.46

Evaluating the contribution of known isotopes isolates the variable term.

Subtract the combined contribution from the given relative atomic mass to find the contribution of the third isotope.

0.11x = 24.32 - 21.46 = 2.86

The remaining mass contribution must come entirely from the third isotope.

Divide by the fractional abundance of the third isotope to solve for

x

x = \frac{2.86}{0.11} = 26

Dividing the mass contribution by fractional abundance yields the integer mass number.

Key Concept

Relative Atomic Mass Calculation from Isotopic Abundances

Estimated Time:1m 15s

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