When 0.78 g of Al(OH)₃ is dehydrated to get Al₂O₃ which is reacted with 6 milliequivalents of HCl. The equivalents of AlCl₃ formed are: 

$\mathrm{Al}{\left(\mathrm{OH}\right)}_3\stackrel{\mathrm{Dehydration}}{\to }{\mathrm{Al}}_2{\mathrm{O}}_3\stackrel{6 \mathrm{milliequivalents}}{\to }{\mathrm{AlCl}}_3$

Calculate the milliequivalents of Al(OH)₃ is:

$\begin{array}{ccc}\mathrm{Milliequivalent} \mathrm{of} \mathrm{Al}{\left(\mathrm{OH}\right)}_3& =& \frac{\mathrm{Weight}}{{\displaystyle \frac{\mathrm{Molecular} \mathrm{weight}}{\mathrm{valency}}}}\times 1000\\ & =& \frac{0.78}{{\displaystyle \frac{78}{3}}}\times 1000\\ & =& 3\end{array}$

Milliequivalents of Al₂O₃ is also 3 because milliequivalents of Al(OH)₃ and Al₂O₃ is same. Milliequivalents of Al₂O₃ and AlCl₃ is also same because HCl is present in excess. Therefore, milliequivalents of AlCl₃ is also 3. Equivalents of AlCl₃ is:

$\begin{array}{ccc}\mathrm{Equivalents} \mathrm{of} {\mathrm{AlCl}}_3& =& 3\times \frac{1}{1000}\\ & =& 3\times {10}^{-3}\end{array}$