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complete the tables by determining the moles and numbers of ions of eac…

Question

complete the tables by determining the moles and numbers of ions of each type present in the following aqueous solutions.

note: reference the fundamental constants table for additional information.

part 1 of 2

\\(718.\text{ ml}\\) of \\(0.618\text{ m}\\) aluminum chloride. be sure your answer has the correct number of significant figures.

part 2 of 2

\\(79.0\text{ ml}\\) of \\(1.35\text{ m}\\) magnesium chloride. be sure your answer has the correct number of significant figures.

Explanation:

Calculate moles of solute for Part 1

Using the Molarity and Molarity Calculations knowledge points

$$ \text{Moles of } \text{AlCl}_3 = 0.718\text{ L} \times 0.618\text{ M} = 0.443724\text{ mol} $$
$$ \text{Moles of } \text{Al}^{3+} = 0.443724\text{ mol} \approx 0.444\text{ mol} $$
$$ \text{Moles of } \text{Cl}^- = 3 \times 0.443724\text{ mol} = 1.331172\text{ mol} \approx 1.33\text{ mol} $$

Calculate number of ions for Part 1

Using the Molarity Calculations knowledge point

$$ \text{Ions of } \text{Al}^{3+} = 0.443724\text{ mol} \times 6.022 \times 10^{23}\text{ /mol} \approx 2.67 \times 10^{23} $$
$$ \text{Ions of } \text{Cl}^- = 1.331172\text{ mol} \times 6.022 \times 10^{23}\text{ /mol} \approx 8.02 \times 10^{23} $$

Calculate moles of solute for Part 2

Using the Molarity and Molarity Calculations knowledge points

$$ \text{Moles of } \text{MgCl}_2 = 0.0790\text{ L} \times 1.35\text{ M} = 0.10665\text{ mol} $$
$$ \text{Moles of } \text{Mg}^{2+} = 0.10665\text{ mol} \approx 0.107\text{ mol} $$
$$ \text{Moles of } \text{Cl}^- = 2 \times 0.10665\text{ mol} = 0.2133\text{ mol} \approx 0.213\text{ mol} $$

Calculate number of ions for Part 2

Using the Molarity Calculations knowledge point

$$ \text{Ions of } \text{Mg}^{2+} = 0.10665\text{ mol} \times 6.022 \times 10^{23}\text{ /mol} \approx 6.42 \times 10^{22} $$
$$ \text{Ions of } \text{Cl}^- = 0.2133\text{ mol} \times 6.022 \times 10^{23}\text{ /mol} \approx 1.28 \times 10^{23} $$

Answer:

Question 1

AluminumChloride
ions\(2.67 \times 10^{23}\)\(8.02 \times 10^{23}\)

Question 2

MagnesiumChloride
ions\(6.42 \times 10^{22}\)\(1.28 \times 10^{23}\)