Apowder contains benzoic acid mixed with starch. a student finds that a 2.505 g sample of this powder requires 35.3 ml of a 0.107m sodium hydroxide solution for the acid to be neutralized. what is the percent by mass of benzoic acid in the powder?

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.

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c. is the answer

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