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
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
The volume of 0.0850 M NaOH is required to titrate 25.0 ml of 0.0720 M is 21.176 ml.
<h3>What are normality and molarity?</h3>Normality is defined as the number of equivalents per litre of the solution
It is given by
Normality = number of equivalents / 1 L of solution
Molarity is defined as the number of moles of solute present in one litre of solution
Molarity = moles of solute / 1 litre of solution
Relationship between molarity and normality
Normality = Molarity × Acidity or Basicity of a salt
Here, the Acidity Of NaOH is 1 and the basicity of HBr is also 1.
Thus, Normality = Molarity
We know, N₁V₁ = N₂V₂
We can also write it as M₁V₁ = M₂V₂
V1 =
Substitute M₁ = 0.0850 M, M₂ = 0.0720 M, V₂= 25 ml
V1 = 21.176 ml
Hence, The volume of 0. 0850 m NaOH are required to titrate 25. 0 ml of 0. 0720 m is 21.176 ml.
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