Question #1ae6a

The reason for this is that aqueous $K O H$ being alkaline in
nature gives $O {H}^{-}$
ions. These $O {H}^{-}$ ions then act as strong
molecule. Thus the leaving group leaves and $O {H}^{-}$ substitute the halogen atom of alkyl halide molecule, forming alcohol as a product. Consequently, aqueous $K O H$ gives substitution reactions.
On the other hand, alcoholic $K O H$(mostly having ethanol) produces ${C}_{2} {H}_{5} {O}^{-}$ ions which are stronger base than $O {H}^{-}$ ions. So these abstract the $\beta$- hydrogen of alkyl halide molecule, forming an alkene. That's the reason behind the statement; Alcoholic $K O H$ give elimination reactions.