Answer:
1) 97.6%
2) No the purity is not acceptable because the standard is 99.5% purity.
3) Yes I will repeat the titration experiment to confirm my result.
Step-by-step explanation:
Equation of the reaction;
C7H6O2(aq) + NaOH(aq) ---------> C7H5ONa(aq) + H2O(aq)
From the information provided;
Number of moles of NaOH reacted = concentration × volume = 20.15/1000 × 0.500 = 0.01 moles
From the reaction equation;
1 mole of C7H6O2 reacts with 1 mole of NaOH
Hence 0.01 moles of C7H6O2 will react with 0.01 moles of NaOH
Mass of C7H6O2 reacted = number of moles of C7H6O2 × molar mass of C7H6O2
Molar mass of C7H6O2 = 122.12 g/mol
Mass of C7H6O2 reacted = 0.01 moles × 122.12 g/mol = 1.22 g
Percentage by mass of pure C7H6O2 in the impure sample = 1.22/1.250 × 100 = 97.6 %