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_(2)H_(6)`
B. `C_(2)H_(4)`
C. `C_(3)H_(8)`
D. `C_(3)H_(6)`

Answer

B

Explanation

`C_(x)H_(y)+(x+(y)/(4))O_(2)rarrxCO_(2)+(y)/(2)H_(2)O`
`{:(1ml,(x+(y)/(4))ml,xml,-),(10ml,10(x+(y)/(4))ml,10xml,-):}`
Volume of `CO_(2)=(70-50)=20ml`
10 x = 20, Therefore, x = 2
Volume of `CO_(2)+` Volume of `O_(2)` (left) = 70 ml
Volume of `O_(2)` (left) = `70-20=50ml`
Volume of `O_(2)` (used) = `80-50=30ml`
`therefore 10(x+(y)/(4))=30`
Solve for y, putting x = 2, y = 4 Hence the formula is `C_(2)H_(4)`.