1.

Pick out the incorrect statement regarding H_2SO_4.

Answer»

When treated with `H_2SO_(4), HCOOH" form "CO and H_2O`
Glucose, when treated with conc. `H_(2)SO_(4)`, forms carbon
Conc. `H_(2)SO_(4)` cannot oxidize HBr and HI to form `Br_(2), and l_(2)`, respectively
Conc. `H_2SO_(4)`, reacts with `NaNO_(3)`

Solution :`2HBr+H_(2)SO_(4) to 2H_(2)O+SO_(2)+Br`
`8HI+H_(2)SO_(4) to 4H_(2)O+H_(2)S+4I_(2)`
Thus, `H_(2)SO_(4)` oxidises both HBr and HI.
(a). `HCOOH underset(-H_(2)O)overset(H_(2)SO_(4))to CO+H_(2)O`
`(b). C_(6)H_(12)O_(6) underset(-6H_(2)O)overset(H_(2)SO_(4))to 6C+6H_(2)O`
`(d) (a) NaNO_(3)+H_(2)SO_(4) overset("Warming")to NaHSO_(4)+HNO_(3)`


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