`5mL` of a gaseous hydrocarbon reacts with 30mL of `O_2`. The resultant gas, on cooling is found to measure 25mL of which 10mL are absorbed by NaOH and the remainder by pyrogallol. Determine molecular formula of hydrocarbon. All measurements are made at constant pressure and temperature.

Options

A. `C _(2) H _(6)`
B. `C_(3) H _(6)`
C. `C _(2) H _(4)`
D. `CH_(3)`

Answer

C

Explanation

Let formula of hydrocarbon be `C _(6) H _(6)`
`therefore underset("Volume taken")(C _(a) H _(b) + )[a + (b)/(4) ]O _(2) underset(5 mL 30 mL )(toaCO _(2) ) + (b)/(2) H _(2) O (l)0-`
Volume left `0 30 - 5 [ a + (b)/(4) ] 5a -`
Volume absorbed by `NaOH = CO _(2)` formed = 10 mL
Volume absorbed by pyrogallol `= O _(2)` left unreacted = 15 mL
Volume of `O _(2)` used `= 30 - 15 = 15 mL = > 5a = 10 or a =2`
`5 (a + (b)/(4)) = 15 or 5 (2 + (b)/(4)) = 15 implies b= 4`
`therefore` Hydrocarbon is `C _(2) H _(4)`