Answer:
[tex]M_{acid}=26.804g/mol[/tex]
Explanation:
Hello,
In this case, by knowing that the used NaOH equals in moles the acid (monoprotic) as shown below, during the titration:
[tex]n_{acid}=n_{NaOH}[/tex]
By knowing the volume and the concentration of the NaOH, one obtains:
[tex]n_{acid}=0.05238L*0.396\frac{mol}{L}=0.0207mol[/tex]
Thus, the molar mass of the acid is computed based on the previously computed moles and the given mass as follows:
[tex]M_{acid}=\frac{m_{acid}}{n_{acid}}=\frac{0.556g}{0.0207mol}=26.804g/mol[/tex]
Best regards.