Answer:
Procedure (2)
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
[tex]\dfrac{5}{4000} = \dfrac{1}{800}[/tex]
Procedure (2)
For the first dialysis, the factor is
[tex]\dfrac{5}{1000} = \dfrac{1}{200}[/tex]
After a second dialysis, the original concentration of NaCl will be reduced by a factor of
[tex]\dfrac{1}{200} \times \dfrac{1}{200} = \dfrac{1}{40000}[/tex]
Procedure (2) is more efficient by a factor of
[tex]\dfrac{40000}{800} = \mathbf{50}[/tex]