Question

Equal masses of $H _{2}, O _{2}$ and methane have been taken in a container of volume $V$ at temperature $27^{\circ} C$ in identical conditions. The ratio of the volumes of gases $H _{2} ; O _{2} ; CH _{4}$ would be

• A. 8 : 16 : 1
• B. 16 : 8 : 1
• C. 16 : 1 : 2
• D. 8 : 1 : 2

Answer 1

Answer: (C)

According to Avogadro's hypothesis,
Volume of a gas $( V ) \propto$ number of moles $(n)$
Therefore, the ratio of the volumes of gases can be
determined in terms of their moles.
$\therefore$ The ratio of volumes of $H _{2}: O _{2}:$ methane $\left( CH _{4}\right)$ is given by
$V_{H_{2}}: V_{O_{2}}: V_{C_{4}}=n_{H_{2}}: n_{O_{2}}: n_{ CH _{4}}$
$\Rightarrow V_{H_{2}}: V_{O_{2}}: V_{C_{4}}:=\frac{m_{H_{2}}}{M_{H_{2}}}: \frac{m_{O_{2}}}{M_{O_{2}}}: \frac{m_{C H_{4}}}{M_{C H_{4}}}$
Given, $m_{H_{2}}=m_{O_{2}}=m_{C_{4}}=m$
$\left[\because n=\frac{\text { mass }}{\text { molarmass }}\right]$
Thus, $V_{H_{2}}: V_{O_{2}}: V_{CH_{4}}:=\frac{m}{2}: \frac{m}{32}: \frac{m}{16}$
$=16: 1: 2$