Answer
KOH(aq)
Work Step by Step
Usually when a pure solid dissolves in a solvent, entropy increases. So, KOH(aq) is expected to have greater entropy than KOH(s).
We see from table 16.1 that our prediction is correct.
$S^{\circ}(KOH,s)=78.9\,JK^{-1}mol^{-1}$
$S^{\circ}(KOH,aq)=91.6\,JK^{-1}mol^{-1}$
