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 $${H_2}:{O_2}:C{H_4}$$ would be
A.
8 : 16 : 1
B.
16 : 8 : 1
C.
16 : 1: 2
D.
8 : 1 : 2
Answer :
16 : 1: 2
Solution :
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 $${H_2}:{O_2}:{\text{methane}}\left( {C{H_4}} \right)$$ is
given by
$$\eqalign{
& {V_{{H_2}}}:{V_{{O_2}}}:{V_{C{H_4}}} = {n_{{H_2}}}:{n_{{O_2}}}:{n_{C{H_4}}} \cr
& \Rightarrow {V_{{H_2}}}:{V_{{O_2}}}:{V_{C{H_4}}}:\, = \frac{{{m_{{H_2}}}}}{{{M_{{H_2}}}}}:\frac{{{m_{{O_2}}}}}{{{M_{{O_2}}}}}:\frac{{{m_{C{H_4}}}}}{{{M_{C{H_4}}}}} \cr} $$
$${\text{Given,}}$$ $${m_{{H_2}}} = {m_{{O_2}}} = {m_{C{H_4}}} = m$$ $$\left[ {\because \,n = \frac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}} \right]$$
$${\text{Thus}},$$ $${V_{{H_2}}}:{V_{{O_2}}}:{V_{C{H_4}}} = \frac{m}{2}:\frac{m}{{32}}:\frac{m}{{16}}$$ $$ = 16:1:2$$
Releted MCQ Question on Physical Chemistry >> States of Matter Solid, Liquid and Gas
Releted Question 1
Equal weights of methane and oxygen are mixed in an empty container at $${25^ \circ }C.$$ The fraction of the total pressure exerted by oxygen is