A definite amount of gaseous hydrocarbon having `("carbon atoms less than" 5)` was burnt with sufficient amount of `O_(2)`. The volume of all reactants was `600mL`. After the explosion the volume of the product `[CO_(2)(g)` and `H_(2)O (g)]` was found to be `700 mL` under the similar conditions. The molecular formula of the compound 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:-

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

Use equation 2H_(2)O(l) to 2H_(2)(g) to answer the following What volume of O_(2) will be produced if the volume of H_(2) produced is 2500 cm^(3) under similar conditions?

10 " mL of " a gaseous organic compound containing C, H and O only was mixed with 100 " mL of " O_(2) and exploded under condition which allowed the H_(2)O formed to condense. The volume of the gas after explosion was 90 mL. On treatment with KOH solution, a further contraction of 20 mL in volume was observed. The vapour density of the compound is 23. All volume measurements were made under the same condition. Q.The molecular formula of the compound is

10 " mL of " a gaseous organic compound containing C, H and O only was mixed with 100 " mL of " O_(2) and exploded under condition which allowed the H_(2)O formed to condense. The volume of the gas after explosion was 90 mL. On treatment with KOH solution, a further contraction of 20 mL in volume was observed. The vapour density of the compound is 23. All volume measurements were made under the same condition. Q.The molecular formula of the compound is