Answer:
To find the concentration (molarity) of the final solution when a 12.5 mL portion of the stock solution is diluted to a total volume of 0.600 L, you can use the formula for dilution:
\[ M_1V_1 = M_2V_2 \]
Where:
- \( M_1 \) is the initial concentration (stock solution concentration).
- \( V_1 \) is the initial volume (12.5 mL, which should be converted to liters).
- \( M_2 \) is the final concentration (what we want to find).
- \( V_2 \) is the final volume (0.600 L).
First, convert the initial volume to liters:
\[ V_1 = 12.5 \, \text{mL} = 0.0125 \, \text{L} \]
Now, plug the values into the formula:
\[ M_1 \cdot 0.0125 \, \text{L} = M_2 \cdot 0.600 \, \text{L} \]
Rearrange the formula to solve for \( M_2 \):
\[ M_2 = \frac{M_1 \cdot 0.0125 \, \text{L}}{0.600 \, \text{L}} \]
Now, plug in the values:
\[ M_2 = \frac{M_1 \cdot 0.0125}{0.600} \]
To express the molarity to three significant figures, you'll need to know the initial concentration (M1) of the stock solution. Let's assume a hypothetical value of M1 = 0.150 M (but use the actual value from your experiment if available).
\[ M_2 = \frac{0.150 \cdot 0.0125}{0.600} \]
Now, calculate M2:
\[ M_2 = \frac{0.001875}{0.600} \]
\[ M_2 = 0.003125 \, \text{M} \]
Rounded to three significant figures, the concentration of the final solution is approximately 0.00312 M.