To determine the volume of 0.250 M KI needed to make 100.0 mL of a solution with a concentration of 0.100 M, we can use the dilution formula:
\[ C_1V_1 = C_2V_2 \]
where:
- \( C_1 \) is the concentration of the stock solution (0.250 M),
- \( V_1 \) is the volume of the stock solution we want to find,
- \( C_2 \) is the concentration of the diluted solution (0.100 M),
- \( V_2 \) is the volume of the diluted solution (100.0 mL or 0.100 L).
First, we need to convert the volumes to liters for consistency (keeping four decimal places):
\[ V_2 = 100.0 , \text{mL} = 0.1000 , \text{L} \]
Now we can rearrange the dilution equation to solve for \( V_1 \):
\[ V_1 = \frac{C_2 \times V_2}{C_1} \]
Substituting in the known values:
\[ V_1 = \frac{0.100 , \text{mol/L} \times 0.1000 , \text{L}}{0.250 , \text{mol/L}} \]
Calculating this:
\[ V_1 = \frac{0.0100 , \text{mol}}{0.250 , \text{mol/L}} = 0.0400 , \text{L} \]
Now converting liters back to milliliters (multiply by 1000):
\[ V_1 = 0.0400 , \text{L} \times 1000 = 40.0 , \text{mL} \]
Thus, the volume of 0.250 M KI needed to make 100.0 mL of 0.100 M KI is \( \boxed{40.0} \) mL.