Question

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 1

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