An aqueous solution consisting of 5 M `BaCl_(2)`, 58.5% w/v NaCl solution & 2M `Na_(2)` X has a density of 1.949 gm/ml. Mark the option(s) which represent correct molarity (M) of the specified ion.
[Assume 100% dissociation of each salt and molecular mass of X`^(2–)` is 96]

Options

A. `[Cl^(-)]=20M`
B. `[Na^(+)]=14M`
C. [Total anions] = 20.5 M
D. [Total cations]=15 M

Answer

To solve the problem, we need to calculate the molarity of the specified ions in the given solution. Let's break it down step by step.

### Step 1: Calculate the molarity of Ba²⁺ and Cl⁻ from BaCl₂
BaCl₂ dissociates into Ba²⁺ and 2Cl⁻ ions. Given that the concentration of BaCl₂ is 5 M:

- Molarity of Ba²⁺ = 5 M
- Molarity of Cl⁻ = 2 × 5 M = 10 M

### Step 2: Calculate the molarity of Na⁺ and Cl⁻ from NaCl
The NaCl solution is given as 58.5% w/v. This means there are 58.5 grams of NaCl in 100 mL of solution.

1. Calculate the molar mass of NaCl:
- Na: 23 g/mol
- Cl: 35.5 g/mol
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

2. Calculate the molarity of NaCl:
\[ \text{Molarity} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]
\[ \text{Molarity of NaCl} = \frac{58.5 \text{ g}}{58.5 \text{ g/mol} \times 0.1 \text{ L}} = 10 \text{ M} \]

3. Since NaCl dissociates into Na⁺ and Cl⁻:
- Molarity of Na⁺ = 10 M
- Molarity of Cl⁻ = 10 M

### Step 3: Calculate the molarity of Na⁺ and X²⁻ from Na₂X
Na₂X dissociates into 2Na⁺ and X²⁻. Given that the concentration of Na₂X is 2 M:

- Molarity of Na⁺ from Na₂X = 2 × 2 M = 4 M
- Molarity of X²⁻ = 2 M

### Step 4: Total molarity of each ion
Now, we can find the total molarity for each ion:

1. **Total Molarity of Na⁺:**
\[ \text{Total Na⁺} = \text{Na⁺ from BaCl₂} + \text{Na⁺ from NaCl} + \text{Na⁺ from Na₂X} = 0 + 10 + 4 = 14 \text{ M} \]

2. **Total Molarity of Cl⁻:**
\[ \text{Total Cl⁻} = \text{Cl⁻ from BaCl₂} + \text{Cl⁻ from NaCl} = 10 + 10 = 20 \text{ M} \]

3. **Total Molarity of X²⁻:**
\[ \text{Total X²⁻} = 2 \text{ M} \]

4. **Total Cations:**
\[ \text{Total Cations} = \text{Ba²⁺} + \text{Na⁺} = 5 + 14 = 19 \text{ M} \]

### Final Summary of Molarities
- Molarity of Ba²⁺ = 5 M
- Molarity of Na⁺ = 14 M
- Molarity of Cl⁻ = 20 M
- Molarity of X²⁻ = 2 M
- Total Cations = 19 M