A compound with molecular formula $C_4{H}_{10}{O}_3$   is converted by the action of acetyl chloride to a compound of molecular mass 190. The number of $-OH$  groups in the original compound $\left(C_4{H}_{10}{O}_3\right)$   is

In acetylation replacement of  H  of O-H   occurs by  $C{H}_{3}CO-group$

i.e., H atom of mass 1 amu is lost and an acetyl group of mass 43 amu is added. Thus,  there is a net gain of  $43-1=42$  amu for every acetyl group introduced. Mass of  difference of final product and original compound is  $190-106=84.$  Hence, number of  $-OH$  groups  $=\frac{84}{42}=2.$