Question

What volume in milliliters of 0.250 M KI is needed to make 100.0 mL of a solution of 0.100 mol/L? When converting to liters, keep four decimal places.

Answer 1

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.