$$ _{x} H _{y}+\left(x+\frac{y}{4}\right) O _{2} \rightarrow x CO _{2}+\frac{y}{2} H _{2} O
$$
Volume of hydrocarbon, \(C _{x} H _{y}=20 cm ^{3}\)
Total volume of \(O _{2}\) supplied \(=150 cm ^{3}\)fter cooling, steam will not be part of the residual gas again as it would have condensed.
Then, volume of \(CO _{2}\) and unused \(O _{2}=110 cm ^{3}\). After absorption by Conc. \(NaOH , CO _{2}\) is removed. This means that the \(50 cm ^{7}\) left after passing the residual gas through Conc. \(NaOH\) is the volume of unused \(O _{2}\).
$$
\begin{aligned}
V_{\text {toat }} &=V_{\text {used }} O_{2}+V_{\text {un-ued }} \\
V_{\text {uiedo }} &=150 cm ^{3}-50 cm ^{3} \\
&=1000 cm ^{3} \\
V_{ CO _{2}} &+V_{\text {unused }}=110 cm ^{3} \\
\therefore V_{C_{7}} &=110 cm ^{3}-50 cm ^{3}=60 cm ^{3}
\end{aligned}
$$rom Avogadro's law, \(\frac{n_{1}}{v_{1}}=\frac{n_{2}}{v_{2}}\). Then,
$$
\begin{aligned}
\frac{n_{ C _{,} H _{ H }}}{V_{ C _{,} H _{ r }}} &=\frac{n_{ CO _{ I }}}{v_{ CO _{2}}} \\
\frac{1}{20} &=\frac{x}{60} \\
20 x &=60 \\
x &=\frac{60}{20}=3
\end{aligned}
$$lso,
$$
\begin{aligned}
\frac{n_{ C _{,} H _{1}}}{V_{ C _{,} H ,}} &=\frac{n_{ CO _{2}}}{v_{ CO _{2}}} \\
\frac{1}{20} &=\frac{\left(x+\frac{y}{4}\right)}{100} \\
100 &=20\left(x+\frac{y}{4}\right) \\
100 &=20 x+5 y \\
\text { Put } x & \text { as } 3 \\
100 &=20(3)+5 y \\
100 &=60+5 y \\
5 y &=100-60=40 \\
y &=\frac{40}{5}=8
\end{aligned}
$$
Thus, \(C _{x} H _{y}= C _{3} H _{8}\)