Certainly! To find the concentration of the chemist's working solution, we will use the principle of dilution, which is summarized by the formula:
[tex]\[ C_1 V_1 = C_2 V_2 \][/tex]
Where:
- [tex]\( C_1 \)[/tex] is the initial concentration of the stock solution.
- [tex]\( V_1 \)[/tex] is the volume of the stock solution used.
- [tex]\( C_2 \)[/tex] is the concentration of the working (diluted) solution.
- [tex]\( V_2 \)[/tex] is the final volume of the working solution.
Given data:
- Initial concentration ([tex]\( C_1 \)[/tex]) = 0.246 M
- Volume of stock solution ([tex]\( V_1 \)[/tex]) = 50.0 mL
- Final volume of the working solution ([tex]\( V_2 \)[/tex]) = 980.0 mL
We need to find the final concentration ([tex]\( C_2 \)[/tex]) of the working solution.
The formula to calculate the final concentration ([tex]\( C_2 \)[/tex]) is rearranged as follows:
[tex]\[ C_2 = \frac{C_1 V_1}{V_2} \][/tex]
Substitute the given values into the equation:
[tex]\[ C_2 = \frac{0.246 \, \text{M} \times 50.0 \, \text{mL}}{980.0 \, \text{mL}} \][/tex]
After performing the calculation:
[tex]\[ C_2 = \frac{12.3 \, \text{mL} \cdot \text{M}}{980.0 \, \text{mL}} \][/tex]
[tex]\[ C_2 \approx 0.012551 \, \text{M} \][/tex]
Rounding this result to 3 significant digits, we get:
[tex]\[ C_2 \approx 0.013 \, \text{M} \][/tex]
Therefore, the concentration of the chemist's working solution is:
[tex]\[ \boxed{0.013} \, \text{M} \][/tex]
