Answer:
[tex]50\; \rm mL[/tex] of the stock solution would be required.
Explanation:
Assume that a solution of volume [tex]V[/tex] contains a solute with a concentration of [tex]c[/tex]. The quantity [tex]n[/tex] of that solute in this solution would be:
[tex]n = c \cdot V[/tex].
For the solution that needs to be prepared, [tex]c = 0.20\; \rm M = 0.20\; \rm mol \cdot L^{-1}[/tex]. The volume of this solution is [tex]V = 250.0\; \rm mL[/tex]. Calculate the quantity of the solute (magnesium chloride) in the required solution:
[tex]\begin{aligned}n &= c \cdot V \\ &= 0.20\; \rm mol \cdot L^{-1} \times 250.0\; \rm mL \\ &= 0.20\; \rm mol \cdot L^{-1} \times 0.2500\; \rm L \\ &= 0.050\; \rm mol\end{aligned}[/tex].
Rearrange the equation [tex]n = c \cdot V[/tex] to find an expression of volume [tex]V[/tex], given the concentration [tex]c[/tex] and quantity [tex]n[/tex] of the solute:
[tex]\displaystyle V= \frac{n}{c}[/tex].
Concentration of the solute in the stock solution: [tex]c(\text{stock}) = 1.00\; \rm M = 1.00\; \rm mol \cdot L^{-1}[/tex].
Quantity of the solute required: [tex]n = 0.050\; \rm mol[/tex].
Calculate the volume of the stock solution that would contain the required [tex]n = 0.050\; \rm mol[/tex] of the magnesium chloride solute:
[tex]\begin{aligned}& V(\text{stock}) \\ &= \frac{n}{c(\text{stock})} \\ &= \frac{0.050\; \rm mol}{1.00\; \rm mol \cdot L^{-1}} \\ &= 0.050\; \rm L \\ &= 50\; \rm mL\end{aligned}[/tex].
