Answer
$M(HA)=229\frac{g}{mol}$
Work Step by Step
Since the acid is monoprotic (we can denote it by $HA$), the amount of acid required for the reaction equals the amount of sodium hydroxide:
$n(HA)=n(NaOH)=c(NaOH)\times V(NaOH) = 0.295\frac{mol}{dm^3}\times 0.02782dm^3=0.0082069mol$
Now we can obtain the molecular weight of the unknown acid:
$M(HA)=\frac{m(HA)}{n(HA)}=\frac{1.88g}{0.0082069mol}\approx 229\frac{g}{mol}$
