`E` is.

A

`Na_(3) GeF_(6)`

B

`Na_(3) AlF_(6)`

C

`Al_(2) O_(3)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`underset((A))(Al_(2)O_(3).2H_(2)O) + 2NaOH + H_(2) O rarr underset("Sodium aluminate (B)")(2Na[Al(OH)_(4)])`
`2Na[Al(OH)_(4)] + underset((C))(2CO_(2)) rarr underset("Pure")(Al_(2)O_(3)).2H_(2)O + 2NaHCO_(3) + H_(2) O`
`Al_(2).O_(3). 2H_(2) O` is mixed with cryolite, `Na_(3) AlF_(6) (E)` to reduce the melting point of the mixture and increases conductivity. On electrolysis. `Al` is deposited at cathode `(F)`
At cathode : `At^(3+) + 3e^(Ө) rarr Al`
Al anode : `C + O^(2-) rarr CP + 2e^(Ө)`
`C + 2O^(2-) rarr CO_(2) + 4e^(Ө)`.
Hence,
`E` is `Na_(3) AlF_(6)`
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