Respuesta :
Answer : The correct option is, (A) 300.0 mL of 0.10 M CaCl₂
Explanation :
(a) 300.0 mL of 0.10 M CaCl₂
The dissociation reaction will be:
[tex]CaCl_2(aq)\rightarrow Ca^{2+}(aq)+2Cl^-(aq)[/tex]
To calculate the number of moles of CaCl₂ as :
[tex]\text{Moles of }CaCl_2=\text{Concentration of }CaCl_2\times \text{Volume of solution in L}[/tex]
[tex]\text{Moles of }CaCl_2=0.10M\times 0.3L=0.03mol[/tex]
As, 1 mole of CaCl₂ contains ions = 3
So, 0.03 mole of CaCl₂ contains ions = 0.03 × 3 = 0.09 ions.
(b) 400.0 mL of 0.10 M NaCl
The dissociation reaction will be:
[tex]NaCl(aq)\rightarrow Na^{+}(aq)+Cl^-(aq)[/tex]
To calculate the number of moles of NaCl as :
[tex]\text{Moles of }NaCl=\text{Concentration of }NaCl\times \text{Volume of solution in L}[/tex]
[tex]\text{Moles of }NaCl=0.10M\times 0.4L=0.04mol[/tex]
As, 1 mole of NaCl contains ions = 2
So, 0.04 mole of NaCl contains ions = 0.04 × 2 = 0.08 ions.
(c) 200.0 mL of 0.10 M FeCl₃
The dissociation reaction will be:
[tex]FeCl_3(aq)\rightarrow Fe^{3+}(aq)+3Cl^-(aq)[/tex]
To calculate the number of moles of FeCl₃ as :
[tex]\text{Moles of }FeCl_3=\text{Concentration of }FeCl_3\times \text{Volume of solution in L}[/tex]
[tex]\text{Moles of }FeCl_3=0.10M\times 0.2L=0.02mol[/tex]
As, 1 mole of FeCl₃ contains ions = 4
So, 0.02 mole of FeCl₃ contains ions = 0.02 × 4 = 0.08 ions.
(d) 800.0 mL of 0.10 M sucrose
The dissociation of sucrose is not possible because it is a non-electrolyte.
To calculate the number of moles of sucrose as :
[tex]\text{Moles of }sucrose=\text{Concentration of }sucrose\times \text{Volume of solution in L}[/tex]
[tex]\text{Moles of }sucrose=0.10M\times 0.8L=0.08mol[/tex]
As, 1 mole of sucrose contains ions = 1
So, 0.08 mole of sucrose contains ions = 0.08 × 1 = 0.08 ions.
Hence, the solutions contains the greatest number of ions is, 300.0 mL of 0.10 M CaCl₂
