1.

Methanoic acid, the first member of carboxylic acid series, when warmed with concentrated sulphuric acid decompose in the following way and evolve carbon monoxide. The driving force for this reaction lies in the fact that the`HC-=O^(+)` ion is very unstable acid and thus easily loses `H^(+)`. Formic acid on heating with conc. `H_2 SO_4` givesA. `CO_(2)+H_(2)`B. `CO+H_(2)O`C. `CO`D. `H_(2)O`

Answer» Correct Answer - B
`HCO OH overset(H_(2)SO_(4))underset(Delta)toCO+H_(2)C`


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