Answer:
Procedure (2)
Step-by-step explanation:
Assume the dialyses come to equilibrium in the allotted times.
Procedure (1)
If you are dialyzing 5 mL of sample against 4 L of water, the concentration of NaCl will be decreased by a factor of

Procedure (2)
For the first dialysis, the factor is

After a second dialysis, the original concentration of NaCl will be reduced by a factor of

Procedure (2) is more efficient by a factor of
