Answer
$94.5$ g/mol
Work Step by Step
A monoprotic acid suggests that it only releases 1 proton (1 $H+$) in its dissociation process in water. Without knowing its identity, we can write the acid's chemical formula as HPr.
When let react with NaOH, the titration process can be written as:
$NaOH + HPr -> NaPr + H_{2}O$
As we can see, because NaOH also only releases 1 hydroxide ion (1 $OH-$), the react ratio is $NaOH:HPr=1:1$
$n_{NaOH}=C*V=0.0500M*0.02210l=1.11*10^{-3}(mol)=n_{HPr}$
$M_{HPr}=m_{HPr}/n_{HPr}=0.1044g/1.11*10^{-3}mol=94.5(g/mol)$
