Answer:
Moles of NaOH that reacted: [tex]n_{NaOH}=1.64*10^{-3}mol[/tex]
Explanation:
During the  titration, all moles of NaOH added in the solution, react with the acetylsalicylic acid neutralizing each other.
The titration required 15.62 ml (0.01562 L) of the 0.1052 M NaOH solution. The moles of NaOH in that volume are:
[tex]n_{NaOH}=0.01562 L * \frac{0.1052mol}{L}[/tex]
[tex]n_{NaOH}=1.64*10^{-3}mol[/tex]
