16 " mL of " a gaseous compound `C_(n)H_(3n)O_(m)` was mixed with 60 " mL of " `O_2` and sparked. The gas mixture on cooling occupied 44 mL. After treatment with NaOH solution, the volume of gas remaining was 12 mL. Deduce the formula of the compound.

Explanation

`C_(n)H_(3n)O_(m)+[n+(3n)/(4)-(m)/(2)]O_2tonCO_2+[(3n)/(2)]H_2O(l)`
After the reaction: `60-16[n+(3n)/(4)-(m)/(2)]`
Volume of `CO_2=16n=` volume absorbed by NaOH
`16n=44-12=32`
`n=2`
`V_(O_2)(l eft)=12mL`
`V_(O_2)(used)=60-12=48mL`
`therefore16[n+(3n)/(4)-(m)/(2)]=48becausem=1`
Formula of compound`=C_2H_6O`