13.83 × 10⁻³ moles of chloride ions, Cl- ion present in 10.2 ml of 0.452 M AlCl₃ solution.
Volume of AlCl₃ solution = 10.2 ml or 0.0102 L
Molarity of AlCl₃ solution = 0.452 M
Number of moles of chloride ions (Cl⁻) = ?
Write the balanced ionic equation
AlCl₃ → Al³⁺ + 3Cl⁻
Calculate the number of moles of AlCl₃
number of moles = molarity / volume L
number of moles of AlCl₃ = 0.452 M × 0.0102 L
number of moles of AlCl₃ = 0.00461 mol
From the balanced ionic equation, it is clear that 1 mole of AlCl₃ will produce 3 moles of Cl⁻ ion
So multiply the number of moles of AlCl₃ by 3 to get the number of moles of Cl⁻
number of moles of Cl⁻ = 0.00461 × 3
number of moles of Cl⁻ = 0.01383 mol or 13.83 × 10⁻³ moles
You can also learn about ionic equations from the following question:
https://brainly.com/question/15467502
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