29.2% (w/w) HCl stock solution has a density of 1.25 g `mL^(-1)` . The molecular weight of HCl is 36.5 g `"mol"^(-1)`.
Find the Volume (V)(mL) of stock solution required to prepare a 500 mL solution of 0.4 M HCl.
Report your answer as `V//5`

Explanation

To solve the problem, we need to find the volume of a 29.2% (w/w) HCl stock solution required to prepare a 500 mL solution of 0.4 M HCl. We will follow these steps:

### Step 1: Calculate the Molarity of the Stock Solution
We know that the molarity (M) can be calculated using the formula:
\[ M = \frac{(\text{W/W percentage}) \times (\text{Density}) \times 10}{\text{Molar Mass}} \]
Where:
- W/W percentage = 29.2%
- Density = 1.25 g/mL
- Molar Mass of HCl = 36.5 g/mol

Substituting the values:
\[ M = \frac{(29.2) \times (1.25) \times 10}{36.5} \]

Calculating the numerator:
\[ 29.2 \times 1.25 = 36.5 \]
Then:
\[ M = \frac{36.5 \times 10}{36.5} = 10 \text{ M} \]

### Step 2: Use the Dilution Formula to Find the Volume of Stock Solution
We will use the dilution equation:
\[ M_1V_1 = M_2V_2 \]
Where:
- \(M_1\) = Molarity of the stock solution = 10 M
- \(M_2\) = Molarity of the diluted solution = 0.4 M
- \(V_2\) = Volume of the diluted solution = 500 mL
- \(V_1\) = Volume of the stock solution (what we need to find)

Rearranging the formula to solve for \(V_1\):
\[ V_1 = \frac{M_2 \times V_2}{M_1} \]

Substituting the values:
\[ V_1 = \frac{0.4 \times 500}{10} \]

Calculating:
\[ V_1 = \frac{200}{10} = 20 \text{ mL} \]

### Step 3: Report the Answer as \(V/5\)
We need to report the answer as \(V/5\):
\[ \frac{20}{5} = 4 \]

### Final Answer
The volume of stock solution required to prepare a 500 mL solution of 0.4 M HCl is \(4\).

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