`AlCl_(2)` is soluble is axcess of `NaOH` forming sodium metaaluminate `Na[Al(OH)_(4)]`.
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Correct Answer – T
`AICI_(3) + NaOH rarr 3NaCI + AI(OH)_(3) darr`
`AI(OH)_(3) + NaOH rarr underset(“Soluble”)(Na[AI(OH)_(4)])`
Hence true.