2 M of 100 mL `Na_2 SO_4` is mixed with 3 M of 100 mL NaCl solution and 1 M of 200 mL `CaCl_2` solution . Then the ratio of the concentration of cation and anion is

Options

A. `1:1`
B. 2:1
C. 2:3
D. `1:2`

Answer

To solve the problem, we need to calculate the concentration of cations and anions from the given solutions and then find the ratio of these concentrations.

### Step-by-Step Solution:

1. **Identify the Components:**
- Sodium Sulfate (`Na2SO4`): Dissociates into 2 Na⁺ and SO₄²⁻.
- Sodium Chloride (`NaCl`): Dissociates into Na⁺ and Cl⁻.
- Calcium Chloride (`CaCl2`): Dissociates into Ca²⁺ and 2 Cl⁻.

2. **Calculate Moles of Each Component:**
- **For `Na2SO4`:**
- Concentration = 2 M, Volume = 100 mL = 0.1 L
- Moles of `Na2SO4` = 2 M × 0.1 L = 0.2 moles
- Cations: 2 moles of Na⁺ for each mole of `Na2SO4` = 2 × 0.2 = 0.4 moles of Na⁺
- Anions: 1 mole of SO₄²⁻ for each mole of `Na2SO4` = 1 × 0.2 = 0.2 moles of SO₄²⁻

- **For `NaCl`:**
- Concentration = 3 M, Volume = 100 mL = 0.1 L
- Moles of `NaCl` = 3 M × 0.1 L = 0.3 moles
- Cations: 1 mole of Na⁺ for each mole of `NaCl` = 1 × 0.3 = 0.3 moles of Na⁺
- Anions: 1 mole of Cl⁻ for each mole of `NaCl` = 1 × 0.3 = 0.3 moles of Cl⁻

- **For `CaCl2`:**
- Concentration = 1 M, Volume = 200 mL = 0.2 L
- Moles of `CaCl2` = 1 M × 0.2 L = 0.2 moles
- Cations: 1 mole of Ca²⁺ for each mole of `CaCl2` = 1 × 0.2 = 0.2 moles of Ca²⁺
- Anions: 2 moles of Cl⁻ for each mole of `CaCl2` = 2 × 0.2 = 0.4 moles of Cl⁻

3. **Total Moles of Cations:**
- Total Na⁺ = 0.4 (from `Na2SO4`) + 0.3 (from `NaCl`) = 0.7 moles
- Total Ca²⁺ = 0.2 (from `CaCl2`)
- Total Cations = 0.7 + 0.2 = 0.9 moles

4. **Total Moles of Anions:**
- Total SO₄²⁻ = 0.2 (from `Na2SO4`)
- Total Cl⁻ = 0.3 (from `NaCl`) + 0.4 (from `CaCl2`) = 0.7 moles
- Total Anions = 0.2 + 0.7 = 0.9 moles

5. **Calculate Concentrations:**
- Total Volume of the mixture = 100 mL + 100 mL + 200 mL = 400 mL = 0.4 L
- Concentration of Cations = Total moles of Cations / Total Volume = 0.9 moles / 0.4 L = 2.25 M
- Concentration of Anions = Total moles of Anions / Total Volume = 0.9 moles / 0.4 L = 2.25 M

6. **Ratio of Concentration of Cations to Anions:**
- Ratio = Concentration of Cations : Concentration of Anions = 2.25 M : 2.25 M = 1 : 1

### Final Answer:
The ratio of the concentration of cations to anions is **1:1**.