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Reviewed CSCA Chemistry question · Easy

What is the total number of chloride ions in 0.150 mol of aluminum chloride, AlCl3? (Avogadro constant = 6.02 × 10^23 mol^-1)

  1. 2.71 × 10^23
  2. 9.03 × 10^22
  3. 1.81 × 10^23
  4. 5.42 × 10^23
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Correct answer

A. 2.71 × 10^23

Principle or equation

Each formula unit of AlCl3 contains 3 chloride ions. Number of ions = moles × number of ions per formula unit × Avogadro constant.

Why this answer is correct

Moles of Cl- = 0.150 mol × 3 = 0.450 mol. Number of Cl- ions = 0.450 × 6.02 × 10^23 = 2.709 × 10^23, approximately 2.71 × 10^23.

Example

For 0.100 mol of CaCl2, moles of Cl- = 0.200 mol, ions = 1.20 × 10^23.

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