Which of the following have equal mass of `Cl^(ɵ)` ions in `1.0 L` of each of the following solution?

Options

A. `5% NaCl` (density `= 1.07 g mL^(-1)`)
B. `5% KCl (d = 1.06 g mL^(-1))`
C. `58.5 g NaCl`
D. `55.5 g BaCl_(2)`

Explanation

To solve the problem of determining which solutions have equal masses of chloride ions in 1.0 L of each solution, we will analyze each option step by step.

### Step 1: Calculate the molarity of the 5% sodium chloride solution.

**Given:**
- Percentage by weight = 5%
- Density = 1.07 g/mL
- Molecular weight of NaCl = 58.5 g/mol

**Formula for molarity:**
\[ \text{Molarity} = \frac{\text{Percentage by weight} \times 10 \times \text{Density}}{\text{Molecular weight of solute}} \]

**Calculation:**
\[ \text{Molarity} = \frac{5 \times 10 \times 1.07}{58.5} = 0.95 \text{ M} \]

Since we have 1 L of solution, the moles of NaCl will be equal to the molarity:
\[ \text{Moles of NaCl} = 0.95 \text{ moles} \]
Since 1 mole of NaCl produces 1 mole of Cl⁻ ions:
\[ \text{Moles of Cl⁻} = 0.95 \text{ moles} \]

### Step 2: Calculate the molarity of the 5% potassium chloride solution.

**Given:**
- Percentage by weight = 5%
- Density = 1.06 g/mL
- Molecular weight of KCl = 74.7 g/mol

**Calculation:**
\[ \text{Molarity} = \frac{5 \times 10 \times 1.06}{74.7} = 0.77 \text{ M} \]
\[ \text{Moles of KCl} = 0.77 \text{ moles} \]
Since 1 mole of KCl produces 1 mole of Cl⁻ ions:
\[ \text{Moles of Cl⁻} = 0.77 \text{ moles} \]

### Step 3: Calculate the molarity of 58.5 g of sodium chloride.

**Given:**
- Given mass = 58.5 g
- Molecular weight of NaCl = 58.5 g/mol

**Calculation:**
\[ \text{Molarity} = \frac{58.5}{58.5} = 1 \text{ M} \]
\[ \text{Moles of NaCl} = 1 \text{ mole} \]
Since 1 mole of NaCl produces 1 mole of Cl⁻ ions:
\[ \text{Moles of Cl⁻} = 1 \text{ mole} \]

### Step 4: Calculate the molarity of 55.5 g of barium chloride.

**Given:**
- Given mass = 55.5 g
- Molecular weight of BaCl₂ = 137 + (2 × 35.5) = 111 g/mol

**Calculation:**
\[ \text{Molarity} = \frac{55.5}{111} = 0.5 \text{ M} \]
\[ \text{Moles of BaCl₂} = 0.5 \text{ moles} \]
Since 1 mole of BaCl₂ produces 2 moles of Cl⁻ ions:
\[ \text{Moles of Cl⁻} = 0.5 \times 2 = 1 \text{ mole} \]

### Summary of Moles of Cl⁻ Ions:
- Option A (5% NaCl): 0.95 moles of Cl⁻
- Option B (5% KCl): 0.77 moles of Cl⁻
- Option C (58.5 g NaCl): 1 mole of Cl⁻
- Option D (55.5 g BaCl₂): 1 mole of Cl⁻

### Conclusion:
The solutions that have equal masses of chloride ions are Option C (58.5 g NaCl) and Option D (55.5 g BaCl₂), as both contain 1 mole of Cl⁻ ions.

### Final Answer:
**Options C and D have equal mass of Cl⁻ ions.**

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