Equal masses of H_2, O_2 and methane have been taken in a contain…

MCQ

Equal masses of $H_2, O_2$ and methane have been taken in a container of volume $V$ at temperature $27\, ^o C$ in identical conditions. The ratio of the volumes of gases $H_2 : O_2 :$ methane would be

A
$8 : 16 : 1$
B
$16 : 8 : 1$
✓
$16 : 1 : 2$
D
$8 : 1 : 2$

✓

Answer

Correct option: C.

$16 : 1 : 2$

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. The ratio of volumes of $\mathrm{H}_{2}: \mathrm{O}_{2}:$ methane $\left(\mathrm{CH}_{4}\right)$ is given by

$v_{H_{2}}: v_{O_{2}}: v_{C H_{2}}=n_{H_{12}}: n_{O_{2}}: n_{C H_{4}}$

$\Rightarrow v_{H_{2}}: v_{O_{2}}: v_{C H_{4}}=\frac{m_{H_{2}}}{M_{H_{2}}}: \frac{m_{O_{2}}}{M_{O_{2}}}: \frac{m_{C H_{4}}}{M_{C H_{4}}}$

But $m_{H_{2}}=m_{O_{2}}=m_{C H_{4}}=m\left[\therefore n=\frac{\text { mass }}{\text { molar massd }}\right]$

Thus, $\quad v_{H_{2}}: v_{O_{2}}: v_{C H_{4}}=\frac{m}{2}=\frac{m}{1}=\frac{m}{16}=16: 1: 2$