## Chemistry: The Molecular Science (5th Edition)

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. $S^{\circ}(KOH,s)=78.9\,JK^{-1}mol^{-1}$ $S^{\circ}(KOH,aq)=91.6\,JK^{-1}mol^{-1}$