Available are `1L` of `0.1M NaCl` and `2L` of `0.2M CaCl_(2)` solutions. Using only these two solutions what maximum volume of a solution can be prepared having `[Cl^(-)]=0.34M` exactly. Both electrolytes are strong

Options

A. `2.5L`
B. `2.4L`
C. `2.3L`
D. None of these

Explanation

To solve the problem of preparing a solution with a specific concentration of chloride ions ([Cl⁻] = 0.34 M) using the given NaCl and CaCl₂ solutions, we can follow these steps:

### Step 1: Identify the contributions of Cl⁻ from each solution
- **NaCl** dissociates into Na⁺ and Cl⁻, contributing 1 mole of Cl⁻ per mole of NaCl.
- **CaCl₂** dissociates into Ca²⁺ and 2 Cl⁻, contributing 2 moles of Cl⁻ per mole of CaCl₂.

### Step 2: Define variables for the volumes used
Let:
- \( X \) = volume (in liters) of 0.1 M NaCl solution used.
- \( Y \) = volume (in liters) of 0.2 M CaCl₂ solution used.

### Step 3: Calculate the moles of Cl⁻ from each solution
- Moles of Cl⁻ from NaCl = \( 0.1 \times X \) (since 0.1 M NaCl gives 0.1 moles of NaCl per liter).
- Moles of Cl⁻ from CaCl₂ = \( 0.2 \times 2 \times Y = 0.4Y \) (since 0.2 M CaCl₂ gives 0.4 moles of Cl⁻ per liter).

### Step 4: Set up the equation for total moles of Cl⁻
The total moles of Cl⁻ in the new solution will be:
\[ \text{Total moles of Cl⁻} = 0.1X + 0.4Y \]

### Step 5: Set up the equation for the concentration of Cl⁻
The total volume of the new solution is \( X + Y \). The concentration of Cl⁻ in the new solution is given as 0.34 M, so we can set up the equation:
\[ \frac{0.1X + 0.4Y}{X + Y} = 0.34 \]

### Step 6: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 0.1X + 0.4Y = 0.34(X + Y) \]

### Step 7: Expand and rearrange the equation
Expanding the right side:
\[ 0.1X + 0.4Y = 0.34X + 0.34Y \]
Rearranging gives:
\[ 0.1X - 0.34X + 0.4Y - 0.34Y = 0 \]
\[ -0.24X + 0.06Y = 0 \]

### Step 8: Solve for Y in terms of X
From the equation:
\[ 0.06Y = 0.24X \]
\[ Y = 4X \]

### Step 9: Apply constraints on X and Y
We know that:
- The maximum volume of NaCl solution available is 1 L, so \( X \leq 1 \).
- The maximum volume of CaCl₂ solution available is 2 L, so \( Y \leq 2 \).

Substituting \( Y = 4X \) into the constraint \( Y \leq 2 \):
\[ 4X \leq 2 \implies X \leq 0.5 \]

### Step 10: Calculate the maximum volume
Using \( X = 0.5 \) L:
\[ Y = 4(0.5) = 2 \text{ L} \]
Thus, the total volume \( V \) is:
\[ V = X + Y = 0.5 + 2 = 2.5 \text{ L} \]

### Final Answer
The maximum volume of the solution that can be prepared is **2.5 L**.

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