Khi hòa tan tinh thể$A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.n{{H}_{2}}O$ vào nước thu được dung dịch $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$
$A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+BaC{{l}_{2}}\to 3BaS{{O}_{4}}\downarrow +2AlC{{l}_{3}}$
${{n}_{BaS{{O}_{4}}}}=\frac{0,699}{233}=0,003(mol)$ $\Rightarrow {{n}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}}=\frac{1}{3}{{n}_{BaS{{O}_{4}}}}=\frac{1}{3}\times 0,003=0,001(mol)$
$\Rightarrow {{n}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}(\text{dd}A)}}=2\times 0,001=0,002(mol)={{n}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.n{{H}_{2}}O}}$
${{M}_{tt}}=\frac{6,66}{0,002}=3330$
${{M}_{tt}}=2\times 27+3\times 96+n\times 18=3330$
342 + $n\times 18$ = 3330 $\Rightarrow $ n = 166
Vậy CT tinh thể là: $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.166{{H}_{2}}O$