The problem asks for the volume of a concentrated HCl stock solution required to prepare a diluted HCl solution. We can solve this by first finding the molarity of the stock solution and then using the dilution equation, $M_1V_1 = M_2V_2$.

Step 1: Calculate the Molarity of the stock HCl solution ($M_1$)

The stock solution is 29.2% (w/w) HCl with a density of 1.25 g/mL.

1. Interpret the concentration: 29.2% (w/w) means that 100 g of the solution contains 29.2 g of HCl.

2. Calculate moles of HCl: The molar mass of HCl is given as 36.5 g/mol. $\text{Moles of HCl} = \frac{\text{Mass of HCl}}{\text{Molar mass of HCl}} = \frac{29.2 \text{ g}}{36.5 \text{ g/mol}} = 0.8 \text{ mol}$

3. Calculate the volume of the solution: We are considering 100 g of the solution. The density is 1.25 g/mL. $\text{Volume of solution} = \frac{\text{Mass of solution}}{\text{Density of solution}} = \frac{100 \text{ g}}{1.25 \text{ g/mL}} = 80 \text{ mL}$

4. Convert volume to Liters: $80 \text{ mL} = 80 \times 10^{-3} \text{ L} = 0.080 \text{ L}$

5. Calculate Molarity ($M_1$): Molarity is moles of solute per liter of solution. $M_1 = \frac{\text{Moles of HCl}}{\text{Volume of solution in L}} = \frac{0.8 \text{ mol}}{0.080 \text{ L}} = 10 \text{ M}$

Alternatively, we can use the direct formula: $M = \frac{\% (w/w) \times \text{density} \times 10}{\text{Molar Mass}} = \frac{29.2 \times 1.25 \times 10}{36.5} = \frac{365}{36.5} = 10 \text{ M}$ So, the molarity of the stock solution is 10 M.

Step 2: Use the dilution equation to find the required volume ($V_1$)

We need to prepare a 200 mL solution of 0.4 M HCl. The dilution equation is: $M_1V_1 = M_2V_2$ Where:

• $M_1$ = Molarity of stock solution = 10 M
• $V_1$ = Volume of stock solution required (in mL)
• $M_2$ = Molarity of final solution = 0.4 M
• $V_2$ = Volume of final solution = 200 mL

Substitute the known values into the equation: $(10 \text{ M}) \times V_1 = (0.4 \text{ M}) \times (200 \text{ mL})$

Now, solve for $V_1$: $V_1 = \frac{(0.4 \text{ M}) \times (200 \text{ mL})}{10 \text{ M}}$ $V_1 = \frac{80}{10} \text{ mL}$ $V_1 = 8 \text{ mL}$

Therefore, 8 mL of the stock solution is required to prepare 200 mL of 0.4 M HCl solution.