Question

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

Answer 1

Answer:

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

Explanation:

a) 400.0 mL of 0.10 M NaCl

(i) Moles of NaCl

$\text{ Moles of NaCl } = \text{0.4000 L NaCl} \times \dfrac{\text{0.10 mol NaCl}}{\text{1 L NaCl}} = \text{0.040 mol NaCl}$

(ii) Moles of ions

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

We get 2 mol of ions from 1 mol of NaCl

$\text{Moles of ions } = \text{0.040 mol NaCl} \times \dfrac{\text{2 mol ions}}{\text{1 mol NaCl}} = \text{0.080 mol ions}$

b) 300.0 mL of 0.10 M CaCl₂

(i) Moles of CaCl₂

$\text{ Moles of CaCl}_{2} =\text{0.3000 L CaCl}_{2} \times \dfrac{\text{0.10 mol CaCl}_{2}}{\text{1 L CaCl}_{2}} = \text{0.030 mol CaCl}_{2}$

(ii) Moles of ions

CaCl₂(s) ⟶ Ca²⁺(aq) + 2Cl⁻(aq)

We get 3 mol of ions from 1 mol of CaCl₂

$\text{Moles of ions } = \text{0.030 mol CaCl}_{2} \times \dfrac{\text{3 mol ions}}{\text{1 mol CaCl}_{2}} = \text{0.090 mol ions}$

c) 200.0 mL of 0.10 M FeCl₃  

(i) Moles of FeCl₃

$\text{ Moles of FeCl}_{3} =\text{0.2000 L FeCl}_{3} \times \dfrac{\text{0.10 mol FeCl}_{3}}{\text{1 L FeCl}_{3}} = \text{0.020 mol FeCl}_{3}$

(ii) Moles of ions

FeCl₂(s) ⟶Fe³⁺(aq) + 3Cl⁻(aq)

We get 4 mol of ions from 1 mol of  FeCl₃

$\text{ Moles of FeCl}_{3} =\text{0.2000 L FeCl}_{3} \times \dfrac{\text{0.10 mol FeCl}_{3}}{\text{1 L FeCl}_{3}} = \text{0.020 mol FeCl}_{3}$

d) 200.0 mL of 0.10 M KBr

(i) Moles of KBr

$\text{ Moles of KBr} = \text{0.2000 L KBr} \times \dfrac{\text{0.10 mol KBr }}{\text{1 L KBr}} = \text{0.020 mol KBr}$

(ii) Moles of ions

KBr(s) ⟶ K⁺(aq) + Br⁻

We get 2 mol of ions from 1 mol of KBr

$\text{Moles of ions } = \text{0.020 mol KBr} \times \dfrac{\text{2 mol ions}}{\text{1 mol KBr}} = \text{0.040 mol ions}\\\\\text{We get the most ions ions from }\boxed{\textbf{300.0 mL of 0.10 mol/L CaCl}_{2}}$