If the vapour density of a gaseous hydrocarbon X' is 35, determine its molecular formula if 0.60g of carbon is present in 0.70g of the hydrocarbon 'X' [C=12, H=1].

If the mass of 10^22 molecules of a hydrocarbon is about 1.2 g, then the gram molecular mass of hydrocarbon is :

Give the structures of the hydrocarbons corresponding to the molecular formula C_(8)H_(12)

A hydrocarbon contains 17.2% H. If the vapour density is 29, calculate its molecular formula.

A gaseous hydrocarbon gives upon combustion, 0.72 g of water and 3.08 g of CO_(2) . The empirical formula of the hydrocarbon is

A gaseous hydrocarbon has 85% carbon and vapour density of 28. The possible formula of the hydrocarbon will be

A hydrocarbon contain 80% C . The vapour density of compound in 30 . Molecular formula of compound is :-

2g of a non-volatile hydrocarbon solute dissolved in 100g of a hypothetical organic solvent (molar mass =50 ) was found to lower the vapour pressure from 75.00 to 74.50 mm of Hg at 20^(@)C . Given that the hydrocarbon contains 96% of C , what is the molecular formula of the hydrocarbon ?

0.1 moles of Hydrocaron on complete combustion produced 17.6 gms of CO_(2) . How many Carbon atoms are present in each molecule of the hydrocarbon.

A gaseous hydrocarbon consumed 5 times its volume of oxygen as for combustion. The volume of CO_(2) produced in the reaction is thrice the volume of hydrocarbon under the same conditions How many grams of water is produced by combustion of 0.1 mol of the givenn hydrocarbon?

A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula (C = 12, H = 1).