You are given one litre of 0.183 M HCI and one litre of 0.381 M HCI. What is the maximum volume of 0.243 M HCI which you can make from these two solution? (Assume that no water is added).

Options

A. 1921 ml
B. 1435 ml
C. 928 ml
D. 2017 ml

Answer

To solve the problem of finding the maximum volume of 0.243 M HCl that can be prepared from 1 liter of 0.183 M HCl and 1 liter of 0.381 M HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let \( V_1 = 1 \) L (volume of 0.183 M HCl)
- Let \( V_2 = x \) L (volume of 0.381 M HCl to be added)
- The target molarity \( M_3 = 0.243 \) M

2. **Set Up the Equation:**
- The total moles of HCl in the final solution can be expressed as:
\[ M_3 \times (V_1 + V_2) = M_1 \times V_1 + M_2 \times V_2 \]
- Substituting the known values:
\[ 0.243 \times (1 + x) = 0.183 \times 1 + 0.381 \times x \]

3. **Expand and Rearrange the Equation:**
- Expanding the left side:
\[ 0.243 + 0.243x = 0.183 + 0.381x \]
- Rearranging gives:
\[ 0.243x - 0.381x = 0.183 - 0.243 \]
\[ -0.138x = -0.06 \]

4. **Solve for \( x \):**
- Dividing both sides by -0.138:
\[ x = \frac{0.06}{0.138} \approx 0.4347 \text{ L} \]

5. **Calculate the Total Volume:**
- The total volume of the final solution \( V_3 \) is:
\[ V_3 = V_1 + V_2 = 1 + x = 1 + 0.4347 \approx 1.4347 \text{ L} \]

6. **Convert to Milliliters:**
- To express the volume in milliliters:
\[ V_3 = 1.4347 \text{ L} \times 1000 \text{ mL/L} \approx 1434.7 \text{ mL} \]

7. **Final Answer:**
- The maximum volume of 0.243 M HCl that can be prepared is approximately **1435 mL**.