1.0 g of AlCl3 is added to 100.0 ml of water. What would be the resulting pH?
Not being given any other information assume the reaction goes to completion and that it is:

$$\ce{AlCl3 + 6H2O -&gt;[excess H2O] Al(H2O)5(OH)^{2+} + H+ +3Cl-}$$

molecular mass $\ce{AlCl3 = \pu{133.34 g/mol}}$. So:

$$\ce{\pu{mol AlCl_{3}} = mol [Al(H2O)_5OH^{2+}] = mol [H+]} = \frac{\pu{1 g}}{\pu{133.34 g/mol}} = \pu{7.496\cdot 10^{-3} mol}$$

$$\pu{Molarity} = \dfrac{\pu{moles}}{\pu{volume}} = \dfrac{\pu{7.496\cdot 10^{-2} mol}}{\pu{0.100 L}} = \pu{0.07496 M}$$

pH = - log(0.07496) = 1.12



Now the truth is a bit more complicated. You can assume the reaction happens in two steps. The first step will go to completion. 

$$\ce{AlCl3 + 6H2O -&gt;[excess H2O] Al(H2O)6^{3+}  +3Cl-}$$

The second step won't go to completion since $\ce{Al(H2O)6^{3+}}$ is a weak acid. There will be an equilibrium favoring the left side. But with no value for $K_\mathrm{eq}$ you are stymied. 

$$\ce{Al(H2O)6^{3+} -&gt;[excess H2O] Al(H2O)5(OH)^{2+} + H+}$$

$$K_\mathrm{eq} =\dfrac{\ce{[H+][Al(H2O)5(OH)^{2+}]}}{\ce{[Al(H2O)6^{3+}]}}$$