To neutralize a solution of NaOH and use the least volume of acid, you should select

I. 1.0M HCl
II. 1.0 M HC(2)H(3)O(2)
III. 1.0 M H(2)SO(4)

(A) I only
(B) II only
(C) III only
(D) I and II use equal and lesser volumes.
(E) I, II, and III use the same volumes of acids

(C) III only

because M=mol/L, and mol=grams/molar mass (dimensional analysis).

Assuming the same grams for each solution, a higher molar mass would yield a smaller number of moles when divided in the mol=grams/molar mass equation.

This same number of moles will be divided by the same number of liters (in the M=mol/L equation). The number of liters is therefore also a smaller value in order to match the number of moles to eventually get a molarity of 1.0

To determine which acid would require the least volume to neutralize a solution of NaOH, we need to consider the balanced chemical equation for the reaction between NaOH and each acid.

I. 1.0M HCl (Hydrochloric acid)
The balanced equation for the reaction between NaOH and HCl is:
NaOH + HCl -> NaCl + H2O

II. 1.0M HC2H3O2 (Acetic acid)
The balanced equation for the reaction between NaOH and HC2H3O2 is:
NaOH + HC2H3O2 -> NaC2H3O2 + H2O

III. 1.0M H2SO4 (Sulfuric acid)
The balanced equation for the reaction between NaOH and H2SO4 is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the balanced equations, we can see that for each mole of NaOH, we have a 1:1 stoichiometric ratio with HCl, a 1:1 stoichiometric ratio with HC2H3O2, and a 2:1 stoichiometric ratio with H2SO4.

Since the goal is to use the least volume of acid, we need to choose the acid that has the highest stoichiometric ratio with NaOH. In this case, it is H2SO4 with a 2:1 ratio.

Therefore, the correct answer would be:
(C) III only