$20 cm ^{3}$ of a gaseous hydrocarbon and $150 cm ^{3}$ of oxygen were exploded in a closed vessel at room temperature. After cooling, $110 cm ^{3}$ of gases remained. After absorption by concentrated sodium hydroxide solution the volume left was $50 cm ^{3}$, the molecular formula of the hydrocarbon is ____________

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$20 c m^{3}$ of a gaseous hydrocarbon and $150 c m^{3}$ of oxygen were exploded in ...

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20 c m^{3}

of a gaseous hydrocarbon and

150 c m^{3}

of oxygen were exploded in a closed vessel at room temperature. After cooling,

110 c m^{3}

of gases remained. After absorption by concentrated sodium hydroxide solution the volume left was

50 c m^{3}

, the molecular formula of the hydrocarbon is ____________

A.

C_{3} H_{6}

C_{4} H_{10}

C_{3} H_{8}

C_{4} H_{8}

Explanation

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

Volume of hydrocarbon,

C_{x} H_{y} = 20 c m^{3}

Total volume of

O_{2}

supplied

= 150 c m^{3}

fter cooling, steam will not be part of the residual gas again as it would have condensed.
Then, volume of

C O_{2}

and unused

O_{2} = 110 c m^{3}

. After absorption by Conc.

N a O H, C O_{2}

is removed. This means that the

50 c m^{7}

left after passing the residual gas through Conc.

N a O H

is the volume of unused

O_{2}

\begin{aligned} V_{toat} & = V_{used} O_{2} + V_{un-ued} \\ V_{uiedo} & = 150 c m^{3} - 50 c m^{3} \\ = 1000 c m^{3} \\ V_{C O_{2}} & + V_{unused} = 110 c m^{3} \\ ∴ V_{C_{7}} & = 110 c m^{3} - 50 c m^{3} = 60 c m^{3} \end{aligned}

rom Avogadro's law,

\frac{n_{1}}{v_{1}} = \frac{n_{2}}{v_{2}}

. Then,

\begin{aligned} \frac{n_{C_{,} H_{H}}}{V_{C_{,} H_{r}}} & = \frac{n_{C O_{I}}}{v_{C O_{2}}} \\ \frac{1}{20} & = \frac{x}{60} \\ 20 x & = 60 \\ x & = \frac{60}{20} = 3 \end{aligned}

lso,

\begin{aligned} \frac{n_{C_{,} H_{1}}}{V_{C_{,} H,}} & = \frac{n_{C O_{2}}}{v_{C O_{2}}} \\ \frac{1}{20} & = \frac{(x + \frac{y}{4})}{100} \\ 100 & = 20 (x + \frac{y}{4}) \\ 100 & = 20 x + 5 y \\ Put x & as 3 \\ 100 & = 20 (3) + 5 y \\ 100 & = 60 + 5 y \\ 5 y & = 100 - 60 = 40 \\ y & = \frac{40}{5} = 8 \end{aligned}

Thus,

C_{x} H_{y} = C_{3} H_{8}

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