Answer:
[tex]\boxed{\text{b) 300.0 mL of 0.10 mol/L CaCl}_{2}}[/tex]
Explanation:
a) 400.0 mL of 0.10 M NaCl
(i) Moles of NaCl
[tex]\text{ Moles of NaCl } = \text{0.4000 L NaCl} \times \dfrac{\text{0.10 mol NaCl}}{\text{1 L NaCl}} = \text{0.040 mol NaCl}[/tex]
(ii) Moles of ions
NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)
We get 2 mol of ions from 1 mol of NaCl
[tex]\text{Moles of ions } = \text{0.040 mol NaCl} \times \dfrac{\text{2 mol ions}}{\text{1 mol NaCl}} = \text{0.080 mol ions}[/tex]
b) 300.0 mL of 0.10 M CaCl₂
(i) Moles of CaCl₂
[tex]\text{ Moles of CaCl}_{2} =\text{0.3000 L CaCl}_{2} \times \dfrac{\text{0.10 mol CaCl}_{2}}{\text{1 L CaCl}_{2}} = \text{0.030 mol CaCl}_{2}[/tex]
(ii) Moles of ions
CaCl₂(s) ⟶ Ca²⁺(aq) + 2Cl⁻(aq)
We get 3 mol of ions from 1 mol of CaCl₂
[tex]\text{Moles of ions } = \text{0.030 mol CaCl}_{2} \times \dfrac{\text{3 mol ions}}{\text{1 mol CaCl}_{2}} = \text{0.090 mol ions}[/tex]
c) 200.0 mL of 0.10 M FeCl₃
(i) Moles of FeCl₃
[tex]\text{ Moles of FeCl}_{3} =\text{0.2000 L FeCl}_{3} \times \dfrac{\text{0.10 mol FeCl}_{3}}{\text{1 L FeCl}_{3}} = \text{0.020 mol FeCl}_{3}[/tex]
(ii) Moles of ions
FeCl₂(s) ⟶Fe³⁺(aq) + 3Cl⁻(aq)
We get 4 mol of ions from 1 mol of FeCl₃
[tex]\text{ Moles of FeCl}_{3} =\text{0.2000 L FeCl}_{3} \times \dfrac{\text{0.10 mol FeCl}_{3}}{\text{1 L FeCl}_{3}} = \text{0.020 mol FeCl}_{3}[/tex]
d) 200.0 mL of 0.10 M KBr
(i) Moles of KBr
[tex]\text{ Moles of KBr} = \text{0.2000 L KBr} \times \dfrac{\text{0.10 mol KBr }}{\text{1 L KBr}} = \text{0.020 mol KBr}[/tex]
(ii) Moles of ions
KBr(s) ⟶ K⁺(aq) + Br⁻
We get 2 mol of ions from 1 mol of KBr
[tex]\text{Moles of ions } = \text{0.020 mol KBr} \times \dfrac{\text{2 mol ions}}{\text{1 mol KBr}} = \text{0.040 mol ions}\\\\\text{We get the most ions ions from }\boxed{\textbf{300.0 mL of 0.10 mol/L CaCl}_{2}}[/tex]
