  #### At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion.  After combustion, the gases occupy 330 mL.  Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is : Option: 1  C4H8   Option: 2  C4H10 Option: 3  C3H6 Option: 4  C3H8 Volume of N in air = 375 × 0.8 = 300 ml

Volume of O2 in air = 375 × 0.2 = 75 ml

15ml

0                         0                            15x                 -

After combustion total volume

330 = 300 + 15x

x = 2

Volume of O2 used

y = 12

So hydrocarbon is = C2H12

None of the options matches it therefore it is a BONUS.

----------------------------------------------------------------------

Alternatively  Solution

15ml

0                         0                            15x                 -

Volume of O2 used

If further information (i.e., 330 ml) is neglected, option (C3H8 ) only satisfy the above equation.

#### Ritika Jonwal 