10 ml of a gaseous hydrocarbon was burnt completely in 80 ml of O2 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

Question

10 ml of a gaseous hydrocarbon was burnt completely in 80 ml of O2 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) C2H4 (B) CH4 (C) C2H2 (D) C2H6

Tardigrade Admin · Accepted Answer

Suppose the molecular formula of hydrocarbon

= C_{x} H_{y}

C_{x} H_{y} (g) + (x + \frac{y}{4}) O_{2} (g) \to x C O_{2} (g) + \frac{y}{2} H_{2} O (ℓ)

Volume of hydrocarbon

= 10 m l

Decreases in volume

= 10 + 80 - 70 = 20 m l

Volume of

C O_{2} = 70 - 50 = 20 m l

x = \frac{20}{10} = 2

V_{O_{2}} (l e f t) = 50 m l

V_{O_{2}} (

Consumed

) = 80 - 50 = 30 m l

V_{O_{2}} (

Consumed

) = V_{H.C.} \times (x + \frac{y}{4})

30 = 10 \times (2 + \frac{y}{4})

y = (3 - 2) \times 4 = 4

Molecular formula is

C_{2} H_{4}

.

Q. 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

$C_{2} H_{4}$

$C H_{4}$

$C_{2} H_{2}$

$C_{2} H_{6}$

Q. 10 ml of a gaseous hydrocarbon was burnt completely in 80 ml of O2​ 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

C2​H4​

CH4​

C2​H2​

C2​H6​

Q. 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

$C_{2} H_{4}$

$C H_{4}$

$C_{2} H_{2}$

$C_{2} H_{6}$