Question

Which of the following aqueous solutions contains the greatest number of ions?

a. 400. mL of 0.10 M NaCl;
b. 300. mL of 0.10 M CaCl2;
c. 200. mL of 0.10 M FeCl3

Answer 1

Answer : The correct option is, (b) 300. mL of 0.10 M CaCl₂

Explanation :

We have to calculate the number of moles of ions for the following aqueous solutions.

(a)  400. mL of 0.10 M NaCl

$NaCl\rightarrow Na^++Cl^-$

NaCl dissociates to give 2 moles of ions.

$\text{Moles of ion}=\text{Molarity of }NaCl\times \text{Volume of solution (in L)}$

Volume of solution = 400. mL = 0.400 L

$\text{Moles of ion}=0.10M\times 0.400L=0.04mol$

Total moles of ion = 0.04 mol × 2 = 0.08 mol

(b)  300. mL of 0.10 M CaCl₂

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

CaCl₂ dissociates to give 3 moles of ions.

$\text{Moles of ion}=\text{Molarity of }CaCl_2\times \text{Volume of solution (in L)}$

Volume of solution = 300. mL = 0.300 L

$\text{Moles of ion}=0.10M\times 0.300L=0.03mol$

Total moles of ion = 0.03 mol × 3 = 0.09 mol

(c)  200. mL of 0.10 M FeCl₃

$FeCl_3\rightarrow Fe^{3+}+3Cl^-$

FeCl₃ dissociates to give 4 moles of ions.

$\text{Moles of ion}=\text{Molarity of }FeCl_3\times \text{Volume of solution (in L)}$

Volume of solution = 200. mL = 0.200 L

$\text{Moles of ion}=0.10M\times 0.200L=0.02mol$

Total moles of ion = 0.02 mol × 4 = 0.08 mol

From this we conclude that, $CaCl_2$ aqueous solutions contains the greatest number of ions.

Hence, the correct option is, (b) 300. mL of 0.10 M CaCl₂

Answer 2

The aqueous solution that contains the greatest number of ions is b. 300. mL of 0.10 M CaCl₂.

a. The number of moles in 400. mL of 0.10 M NaCl is:

$0.400 L \times \frac{0.10mol}{L} = 0.040mol$

Each mole of NaCl contains 2 moles of ions (1 Na⁺ and 1 Cl⁻). The moles of ions in 0.040 moles of NaCl are:

$0.040molNaCl \times \frac{2molIons}{1molNaCl} = 0.080 mol Ions$

b. The number of moles in 300. mL of 0.10 M CaCl₂ is:

$0.300 L \times \frac{0.10mol}{L} = 0.030mol$

Each mole of CaCl₂ contains 3 moles of ions (1 Ca²⁺ and 2 Cl⁻). The moles of ions in 0.030 moles of CaCl₂ are:

$0.030molNaCl \times \frac{3molIons}{1molNaCl} = 0.090 mol Ions$

c. The number of moles in 200. mL of 0.10 M FeCl₃ is:

$0.200 L \times \frac{0.10mol}{L} = 0.020mol$

Each mole of FeCl₃ contains 4 moles of ions (1 Fe³⁺ and 3 Cl⁻). The moles of ions in 0.020 moles of FeCl₃ are:

$0.020molNaCl \times \frac{4molIons}{1molNaCl} = 0.080 mol Ions$

The aqueous solution that contains the greatest number of ions is b. 300. mL of 0.10 M CaCl₂.

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