Question

16. Which of the following aqueous solutions contains the greatest number of 10ns.

a) 400.0 mL of 0.10 M NaCl
b) 300.0 mL of 0.10 M CaCl2
Y 200.0 mL of 0.10 M FeCl3
d) 200.0 mL of 0.10 M KB

Answer 1

`\boxed{\text{b) 300.0 mL of 0.10 mol/L CaCl}_(2)}`

Step-by-step explanation:

a) 400.0 mL of 0.10 M NaCl

(i) Moles of NaCl

`\text{ Moles of NaCl } = \text{0.4000 L NaCl} * \frac{\text{0.10 mol NaCl}}{\text{1 L NaCl}} = \text{0.040 mol NaCl}`

(ii) Moles of ions

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

We get 2 mol of ions from 1 mol of NaCl

`\text{Moles of ions } = \text{0.040 mol NaCl} * \frac{\text{2 mol ions}}{\text{1 mol NaCl}} = \text{0.080 mol ions}`

b) 300.0 mL of 0.10 M CaCl₂

(i) Moles of CaCl₂

`\text{ Moles of CaCl}_(2) =\text{0.3000 L CaCl}_(2) * \frac{\text{0.10 mol CaCl}_(2)}{\text{1 L CaCl}_(2)} = \text{0.030 mol CaCl}_(2)`

(ii) Moles of ions

CaCl₂(s) ⟶ Ca²⁺(aq) + 2Cl⁻(aq)

We get 3 mol of ions from 1 mol of CaCl₂

`\text{Moles of ions } = \text{0.030 mol CaCl}_(2) * \frac{\text{3 mol ions}}{\text{1 mol CaCl}_(2)} = \text{0.090 mol ions}`

c) 200.0 mL of 0.10 M FeCl₃

(i) Moles of FeCl₃

`\text{ Moles of FeCl}_(3) =\text{0.2000 L FeCl}_(3) * \frac{\text{0.10 mol FeCl}_(3)}{\text{1 L FeCl}_(3)} = \text{0.020 mol FeCl}_(3)`

(ii) Moles of ions

FeCl₂(s) ⟶Fe³⁺(aq) + 3Cl⁻(aq)

We get 4 mol of ions from 1 mol of FeCl₃

`\text{ Moles of FeCl}_(3) =\text{0.2000 L FeCl}_(3) * \frac{\text{0.10 mol FeCl}_(3)}{\text{1 L FeCl}_(3)} = \text{0.020 mol FeCl}_(3)`

d) 200.0 mL of 0.10 M KBr

(i) Moles of KBr

`\text{ Moles of KBr} = \text{0.2000 L KBr} * \frac{\text{0.10 mol KBr }}{\text{1 L KBr}} = \text{0.020 mol KBr}`

(ii) Moles of ions

KBr(s) ⟶ K⁺(aq) + Br⁻

We get 2 mol of ions from 1 mol of KBr

`\text{Moles of ions } = \text{0.020 mol KBr} * \frac{\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)}`