Answer : The volume of stock solution needed are, 12.5 mL
Explanation :
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume of copper (II) chloride.
[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume of stock solution of copper (II) chloride.
We are given:
[tex]M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL[/tex]
Hence, the volume of stock solution needed are, 12.5 mL