You have a 5.0 M copper sulfate stock solution and a 2 M NaCl stock solution. You wish to prepare a solution with a final concentration of 0.25 M copper sulfate and 0.25 M NaCl containing 330 mL of water as a solvent. How many milliliters of the NaCl stock solution would the solution contain?
Answer
The volume of Final Solution = 330 mL =V3
C1 = Concentration of copper Sulphate =5.0 M
C2= Concentration of NaCl initially =2.0 M
C3= Concentration of NaCl in final solution= 0.25 M
use the formula
C2V2 =C3V3
where V2 is the volume of 2.0 M NaCl required
2* V2 = 0.25* 330
V2= 41.25 mL
hence 41.25 mL 0f 2.0 M NaCl would be required
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