InterviewSolution
Saved Bookmarks
| 1. |
`AlCl_(2)` is soluble is axcess of `NaOH` forming sodium metaaluminate `Na[Al(OH)_(4)]`. |
|
Answer» 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. |
|