To find the final volume after dilution, we can use the dilution equation:
\[ C_1 V_1 = C_2 V_2 \]
where:
- \( C_1 \) is the initial concentration,
- \( V_1 \) is the initial volume,
- \( C_2 \) is the final concentration,
- \( V_2 \) is the final volume.
From the problem statement:
- \( C_1 = 915.0 \) mM,
- \( V_1 = 86.3 \) mL,
- \( C_2 = 628.0 \) mM.
We need to find \( V_2 \).
Substituting the values into the equation:
\[ (915.0 , \text{mM}) \times (86.3 , \text{mL}) = (628.0 , \text{mM}) \times V_2 \]
Calculating the left side:
\[ 915.0 , \text{mM} \times 86.3 , \text{mL} = 78974.5 , \text{mM \cdot mL} \]
Now, substituting this value back into the equation:
\[ 78974.5 , \text{mM \cdot mL} = 628.0 , \text{mM} \times V_2 \]
Now, solve for \( V_2 \):
\[ V_2 = \frac{78974.5 , \text{mM \cdot mL}}{628.0 , \text{mM}} \approx 125.8 , \text{mL} \]
Now, convert \( V_2 \) from mL to liters:
\[ V_2 = \frac{125.8 , \text{mL}}{1000} = 0.1258 , \text{L} \]
Finally, we need to ensure our answer has the correct number of significant figures. Since our values have up to four significant figures, we will round \( 0.1258 \) L to three significant figures, which gives:
\[ \boxed{0.126} , \text{L} \]