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