Answer is: the percent by mass of benzoic acid in the powder is 18.41%.
1) Balanced chemical reaction:
C₆H₅COOH(aq) + NaOH(aq) → C₆H₅COONa(aq) + H₂O(l).
V(NaOH) = 35.3 mL ÷ 1000 mL/L.
V(NaOH) = 0.0353 L; volume of the sodium hydroxide.
c(NaOH) = 0.107 mol/L; molarity of the sodium hydroxide.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0.0353 L · 0.107 mol/L.
n(NaOH) = 0.00378 mol; amount of the sodium hydroxide.
2) From balanced chemical reaction: n(NaOH) : n(C₆H₅COOH).
n(NaOH) = n(C₆H₅COOH).
n(C₆H₅COOH) = 0.0038 mol; amount of the benzoic acid.
M(C₆H₅COOH) = 122.12 g/mol; molar mass of the benzoic acid.
m(C₆H₅COOH) = n(C₆H₅COOH) · M(C₆H₅COOH).
m(C₆H₅COOH) = 0.0038 mol · 122.12 g/mol.
m(C₆H₅COOH) = 0.461 g; mass of the benzoic acid.
m(powder) = 2.505 g; mass of a powder.
ω(C₆H₅COOH) = m(C₆H₅COOH) ÷ m(powder) · 100%.
ω(C₆H₅COOH) = 0.461 g ÷ 2.505 g · 100%.
ω(C₆H₅COOH) = 18.41%; mass percentage of the benzoic acid in a powder.