Home / iGCSE / Coordinated Sciences / C3.3 The mole and the Avogadro constant Paper 2

CIE iGCSE Co-Ordinated Science C3.3 The mole and the Avogadro constant Exam Style Questions Paper 2

CIE iGCSE Co-Ordinated Science Paper 2 - All Chapters

Question

Which quantities, when measured at room temperature and pressure, contain one mole of oxygen molecules?

  1. 16 g of oxygen molecules
  2. 12 dm³ of oxygen molecules
  3. the Avogadro constant number of oxygen molecules
  4. 24 dm³ of oxygen molecules

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

▶️ Answer/Explanation
One mole of oxygen molecules (O₂) has a mass of 32 g, not 16 g (16 g is half a mole). At r.t.p., one mole of any gas occupies 24 dm³. The Avogadro constant number of molecules is exactly one mole. Therefore, 3 and 4 are correct. Answer D.
✅ Answer: (D)

Question

Which statement about the Avogadro constant is correct?

A. It is the volume of 1 mole of a gas at room temperature and pressure.
B. It is the mass of 1 mole of a substance.
C. It is the number of particles in 1 mole of a substance.
D. It has a value of 24 dm³.

▶️ Answer/Explanation
The Avogadro constant is defined as the number of particles (atoms, molecules, ions, or electrons) present in one mole of any substance. Its value is \(6.02 \times 10^{23}\) per mole. The volume of 1 mole of a gas at room temperature and pressure is 24 dm³ (molar gas volume), which is different from the Avogadro constant. The mass of 1 mole of a substance is the molar mass.
✅ Answer: (C)

Question

When magnesium carbonate reacts with dilute hydrochloric acid, carbon dioxide gas is released.

The equation for this reaction is shown.

\(\text{MgCO}_3 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\)

Which volume of carbon dioxide, collected at room temperature and pressure, is released when 4.2 g of magnesium carbonate reacts with excess dilute hydrochloric acid?

A. 1.2 dm3
B. 2.4 dm3
C. 4.8 dm3
D. 12 dm3

▶️ Answer/Explanation
The molar mass of \(\text{MgCO}_3\) is \(24+12+48=84\text{ g/mol}\), so \(\frac{4.2}{84}=0.05\) mol reacts.
The equation shows a 1:1 mole ratio between \(\text{MgCO}_3\) and \(\text{CO}_2\), so 0.05 mol of \(\text{CO}_2\) is produced.
Using the molar gas volume at r.t.p. of \(24\text{ dm}^3/\text{mol}\): \(0.05 \times 24 = 1.2\text{ dm}^3\).
✅ Answer: (A)

Question

Which statements about the mole are correct?

  1. One mole of 12C contains twice as many atoms as one mole of 24Mg.
  2. One mole of 12C has a mass of 12 g.
  3. One mole of C contains Avogadro’s number of atoms.
  4. One mole of oxygen gas at room temperature and pressure occupies 32 dm3.

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

▶️ Answer/Explanation
Statement 2 is correct: one mole of any element has a mass in grams equal to its relative atomic mass, so one mole of \(^{12}\text{C}\) has a mass of 12 g.
Statement 3 is correct: one mole of any substance, including carbon, contains the Avogadro constant, \(6.02 \times 10^{23}\) particles.
Statement 1 is wrong (equal moles of any element contain the same number of atoms) and statement 4 is wrong (one mole of any gas at r.t.p. occupies 24 dm3, not 32 dm3).
✅ Answer: (C)

Question

\(1\,\text{g}\) of hydrogen contains \(6 \times 10^{23}\) atoms.

The relative atomic mass of helium is \(4\).

How many atoms does \(1\,\text{g}\) of helium contain?

A. \(1.5 \times 10^{23}\)
B. \(3 \times 10^{23}\)
C. \(6 \times 10^{23}\)
D. \(2.4 \times 10^{24}\)

▶️ Answer/Explanation
\(1\,\text{g}\) of hydrogen (\(A_r = 1\)) is exactly \(1\) mole, containing \(6 \times 10^{23}\) atoms.
\(1\,\text{g}\) of helium (\(A_r = 4\)) is only \(\frac{1}{4}\) of a mole.
So the number of atoms \(= 6 \times 10^{23} \div 4 = 1.5 \times 10^{23}\).
✅ Answer: (A)
More CIE iGCSE Co-Ordinated Science Paper 2 Questions..
Scroll to Top