iGCSE Chemistry 0620 Extended - 3.3 The mole and the Avogadro constant- Exam Style Questions Paper 2- New Syllabus
▶️ Answer/Explanation
\(M_r\) of \(\mathrm{CaCO_3} = 40 + 12 + (3 \times 16) = 100\).
Moles of \(\mathrm{CaCO_3} = \frac{10.0}{100} = 0.10\,\text{mol}\).
From the equation, 1 mole of \(\mathrm{CaCO_3}\) produces 1 mole of \(\mathrm{CO_2}\) molecules.
Number of \(\mathrm{CO_2}\) molecules = \(0.10 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}\).
✅ Answer: (C)
Moles of \(\mathrm{CaCO_3} = \frac{10.0}{100} = 0.10\,\text{mol}\).
From the equation, 1 mole of \(\mathrm{CaCO_3}\) produces 1 mole of \(\mathrm{CO_2}\) molecules.
Number of \(\mathrm{CO_2}\) molecules = \(0.10 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}\).
✅ Answer: (C)
▶️ Answer/Explanation
Molar mass: \( \mathrm{C_2H_5OH} = 46 \), \( \mathrm{CO_2} = 44 \), \( \mathrm{H_2O} = 18 \).
Option D: 184g ethanol = \(184 / 46 = 4\) moles. From equation, 1 mole ethanol \(\rightarrow\) 3 moles water. 4 moles ethanol \(\rightarrow 12 \times 18 = 216\) g water.
✅ Answer: (D)
Option D: 184g ethanol = \(184 / 46 = 4\) moles. From equation, 1 mole ethanol \(\rightarrow\) 3 moles water. 4 moles ethanol \(\rightarrow 12 \times 18 = 216\) g water.
✅ Answer: (D)
▶️ Answer/Explanation
Correct Answer: C
Explanation: The Avogadro constant (\(N_A = 6.02 \times 10^{23}\)) defines the number of particles (atoms, molecules, or ions) in one mole of any substance. Argon is a monatomic gas, so 1.00 mol of argon contains \(6.02 \times 10^{23}\) atoms.
✅ \(6.02 \times 10^{23}\) atoms.