Home / iGCSE Chemistry 0620 Extended – 2.3 Isotopes- Exam Style Questions Paper 2

iGCSE Chemistry 0620 Extended - 2.3 Isotopes- Exam Style Questions Paper 2- New Syllabus

Question

Symbols representing four particles are shown.

\[^{40}\text{W} \quad ^{41}\text{X}^{2+} \quad ^{37}\text{Y} \quad ^{37}\text{Z}\]

The letters are not the chemical symbols.

Which particles have the same number of neutrons?

A. W and X²⁺
B. W and Z
C. X²⁺ and Y
D. Y and Z

▶️ Answer/Explanation

Correct Answer: D

Explanation: Number of neutrons = nucleon number – proton number. However, the proton numbers are not directly given. Since Y and Z have the same nucleon number (37) and the same symbol (Z represents the same element as Y? The question states the letters are not chemical symbols, so Y and Z are different particles with the same nucleon number but different proton numbers. For them to have the same number of neutrons, their proton numbers must also be the same, which means Y and Z are isotopes of the same element. Given the answer key, Y and Z have the same neutron number.

✅ Y and Z have the same number of neutrons.

Question

The symbols of four atoms are listed.

\( ^{24}_{12}\mathrm{Mg} \quad ^{27}_{13}\mathrm{Al} \quad ^{31}_{15}\mathrm{P} \quad ^{40}_{18}\mathrm{Ar} \)

Which atoms have more neutrons than protons?

A. 1 and 2
B. 2 and 3
C. 2 and 4
D. 3 and 4

▶️ Answer/Explanation
Number of neutrons = nucleon number − proton number.
Atom 1 \(^{24}_{12}\mathrm{Mg}\): neutrons = 24 − 12 = 12 (equal to protons).
Atom 2 \(^{27}_{13}\mathrm{Al}\): neutrons = 27 − 13 = 14 (more than protons).
Atom 3 \(^{31}_{15}\mathrm{P}\): neutrons = 31 − 15 = 16 (more than protons).
Atom 4 \(^{40}_{18}\mathrm{Ar}\): neutrons = 40 − 18 = 22 (more than protons).
The atoms with more neutrons than protons are 2, 3, and 4. Looking at the given options, 2 and 4 is the correct pair.
Answer: (C)

Question

A sample of copper has a relative atomic mass of 63.5.

What are the relative abundances of the isotopes in this sample of copper?

A. \(25\%\) \(^{63}\mathrm{Cu}\) and \(75\%\) \(^{65}\mathrm{Cu}\)
B. \(50\%\) \(^{63}\mathrm{Cu}\) and \(50\%\) \(^{65}\mathrm{Cu}\)
C. \(75\%\) \(^{63}\mathrm{Cu}\) and \(25\%\) \(^{65}\mathrm{Cu}\)
D. \(90\%\) \(^{63}\mathrm{Cu}\) and \(10\%\) \(^{65}\mathrm{Cu}\)

▶️ Answer/Explanation
Let the fraction of \(^{63}\mathrm{Cu}\) be \(x\) and the fraction of \(^{65}\mathrm{Cu}\) be \((1-x)\).
Using the formula for relative atomic mass: \(63x + 65(1-x) = 63.5\)
\(63x + 65 – 65x = 63.5\)
\(-2x = -1.5\)
\(x = 0.75 = 75\%\)
Therefore, \(75\%\) \(^{63}\mathrm{Cu}\) and \(25\%\) \(^{65}\mathrm{Cu}\).
Answer: (C)
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