An impure sample of benzoic acid (C 6H 5COOH, 122.12 g/mol) is titrated with 0.2099 M NaOH. A 1.021-g sample requires 35.73 mL of titrant to reach the endpoint. What is the percent by mass of benzoic acid in the sample? C6H5COOH(aq) + NaOH(aq) → NaC6H5COO(aq) + H2O(l)

The percent by mass of benzoic acid = 89.7 %

Explanation:

Step 1: Data given

Molarity of NaOH = 0.2099 M

Mass of the sample = 1.021 grams

Volume of NaOH = 35.73 mL = 0.03573 L

Step 2: The balanced equation

C6H5COOH(aq) + NaOH(aq) → NaC6H5COO(aq) + H2O(l)

Step 3: Calculate moles NaOH

Moles NaOH = molarity NaOH * volume NaOH

Moles NaOH = 0.2099 * 0.03573 L

Moles NaOH = 0.00750 moles

Step 3: Calculate moles benzoic acid

From your balanced equation - benzoic acid reacts in 1:1 molar ratio with NaOH  

The sample must have contained 0.00750 moles benzoic acid  

Mass benzoic acid in 0.00750 moles  = 0.00750 * 122.12 = 0.9159 g benzoic acid  

Mass % acid in sample =  0.9159/1.021 * 100 = 89.7% purity.

The percent by mass of benzoic acid = 89.7 %

i think   the answer is c hope i'm right

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