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Answer:
[tex]6.40\; \rm mL[/tex].
Explanation:
After the solution is prepared, the volumetric flask is expected to contain [tex]50.0\; \rm mL[/tex] of the [tex]0.0800\; \rm mol \cdot L^{-1}[/tex] solution
The concentrations of the solutions are given in the unit moles-per-liter. Hence, convert the unit of the volume of the volumetric flask to liters:
[tex]\begin{aligned} & 50.0\; \rm mL \\ =\; & 50.0\; {\rm mL} \times \frac{1\; \rm L}{10^{3}\; \rm mL} \\ =\; & 0.0500\; \rm L\end{aligned}[/tex].
Calculate the number of moles of the solute in the volumetric flask:
[tex]\begin{aligned} n &= c(\text{flask}) \cdot V(\text{flask}) \\ &= 0.0800\; \rm mol \cdot L^{-1} \times 0.0500\; \rm L \\ &= 4.00\times 10^{-3}\; \rm mol \end{aligned}[/tex].
All these [tex]4.00\times 10^{-3}\; \rm mol[/tex] of the solute would have come from the stock solution that was initially added to the flask.
The concentration of the stock solution is [tex]0.625\; \rm mol \cdot L^{-1}[/tex]. Calculate the volume of the stock solution that would contain [tex]4.00\times 10^{-3}\; \rm mol[/tex] of the solute:
[tex]\begin{aligned}V &= \frac{n}{c} \\ &= \frac{4.00\times 10^{-3}\; \rm mol}{0.625\; \rm mol \cdot L^{-1}} \\ &= 6.40 \times 10^{-3}\; \rm L\end{aligned}[/tex].
Convert the unit of the required stock solution to milliliters:
[tex]\displaystyle 6.40\times 10^{-3}\; \rm L \times \frac{10^{3}\; \rm mL}{1\; \rm L} = 6.40\; \rm mL[/tex].