Ten millilitre of a gaseous hydrocarbon was burnt completely in 80 ml of `O_2` at STP. The volume of the remaining gas is 70 ml. The volume became 50 ml, on treatment with `NaOH`. The formula of the hydrocarbon is:

Options

A. `C_2H_6`
B. `C_2H_4`
C. `C_3H_8`
D. `C_3H_6`

Answer

B

Explanation

`C_xH_y+(x+(y)/(4))O_2rarrxCO_2+(y)/(2)+(y)/(2)H_2O`
`1ml(x+(y)/(4))ml x ml _`
`10ml 10(x+(y)/(4))ml 10x ml _`
Volume of `CO_2=(70-50)=20ml`
`10x=20` therefore `x=2`
Volume of `CO_2+`volume of `O_2` (left)`=70ml`
volume of `O_2` (left)`=70-20=50ml`
Volume of `O_2` (used) `=80-50=30ml`
`10(x+(y)/(4))=30`
Solve for `y`, putting `x=2,y=4`
hence the formula is`C_2H_4`.