`10mL` of gaseous hydrocarbon on combustion gives `40ml` of `CO_(2)(g)` and `50mL` of `H_(2)O` (vapour) The hydrocarbon is .

10 ml of gaseous hydrocarbon on combustion gives 40 ml of CO_2_(g) and 50 ml of H_2O (vap) . The hydrocarbon is :

10 ml of gaseous hydrocarbon on combution gives 20ml of CO_(2) and 30ml of H_(2)O(g) . The hydrocarbon is :-

10 ml of gaseous hydrocarbon on combution gives 20ml of CO_(2) and 30ml of H_(2)O(g) . The hydrocarbon is :-

10ml of a gaseous hydrocarbon on combustion gives 40ml of CO_(2) and 50ml of H_(2)O vapour under the same conditions. The hydrocarbon is

A gaseous hydrocarbon on complete combustion gives 3.38 g of CO_(2) and 0.690 g of H_(2)O and no other product. The empirical formula of the hydrocarbon is

500 ml of a gaseous hydrocarbon when burnt in excess of O_(2) gave 2.0litres of CO_(2) and 2.5 litres of water vapours under same conditions. molecular formula of the hydrocarbon is:-

500 ml of a gaseous hydrocarbon when burnt in excess of O_(2) gave 2.0litres of CO_(2) and 2.5 litres of water vapours under same conditions. molecular formula of the hydrocarbon is:-

30mL of a gaseous hydrocarbon requires 90mL of O_(2) for complete oxidation, 60mL of CO_(2) gas is formed in the process. The molecular formula of the hydrocarbon will be: