`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)`. The volume `(mL)` of stock solution required to prepare a `200 mL` solution of `0.4 M HCl` is :

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the weight of HCl in 100 g of the solution
Given that the solution is 29.2% (w/w) HCl, this means that in 100 g of the solution, there are 29.2 g of HCl.

### Step 2: Calculate the number of moles of HCl
To find the number of moles of HCl, we use the formula:
\[ \text{Number of moles} = \frac{\text{Weight (g)}}{\text{Molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of HCl} = \frac{29.2 \, \text{g}}{36.5 \, \text{g/mol}} \approx 0.800 \, \text{mol} \]

### Step 3: Calculate the volume of the solution in liters
To find the volume of the solution, we use the density of the solution:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the values:
\[ \text{Volume} = \frac{100 \, \text{g}}{1.25 \, \text{g/mL}} = 80 \, \text{mL} = 0.080 \, \text{L} \]

### Step 4: Calculate the molarity of the stock solution
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
Substituting the values:
\[ \text{Molarity of stock solution} = \frac{0.800 \, \text{mol}}{0.080 \, \text{L}} = 10 \, \text{M} \]

### Step 5: Use the dilution equation to find the volume of stock solution needed
Using the dilution equation \(M_1V_1 = M_2V_2\):
- \(M_1 = 10 \, \text{M}\) (molarity of stock solution)
- \(M_2 = 0.4 \, \text{M}\) (molarity of diluted solution)
- \(V_2 = 200 \, \text{mL}\) (volume of diluted solution)
Let \(V_1\) be the volume of the stock solution we need to find.

Substituting the values into the equation:
\[ 10 \, \text{M} \cdot V_1 = 0.4 \, \text{M} \cdot 200 \, \text{mL} \]
\[ 10 \, V_1 = 80 \]
\[ V_1 = \frac{80}{10} = 8 \, \text{mL} \]

### Final Answer
The volume of stock solution required to prepare a 200 mL solution of 0.4 M HCl is **8 mL**.

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