Question

10 ml of a gaseous hydrocarbon was burnt completely in 80 ml of $O_{2}$ at NTP. The remaining gas occupied 70 ml at NTP. This volume become 50 ml on treatment with KOH solution. The molecular formula of the hydrocarbon is

• A. $C_{2}H_{4}$
• B. $CH_{4}$
• C. $C_{2}H_{2}$
• D. $C_{2}H_{6}$

Answer 1

Answer: (A)

Suppose the molecular formula of hydrocarbon $= C _{ x } H _{ y }$
$ C _{ x } H _{ y }( g )+\left( x +\frac{ y }{4}\right) O _2( g ) \rightarrow xCO _2( g )+\frac{ y }{2} H _2 O (\ell) $
Volume of hydrocarbon $=10 ml$
Decreases in volume $=10+80-70=20 ml$
Volume of $CO _2=70-50=20 ml$
$ x=\frac{20}{10}=2 $
$V _{ O _2}( left )=50 ml$
$V _{ O _2}($ Consumed $)=80-50=30 ml$
$V _{ O _2}($ Consumed $)= V _{\text {H.C. }} \times\left( x +\frac{ y }{4}\right)$
$30=10 \times\left(2+\frac{ y }{4}\right)$
$ y=(3-2) \times 4=4 $
Molecular formula is $C _2 H _4$.