A)
\[{{X}_{{{H}_{2}}}}\]
\[{{X}_{{{O}_{2}}}}\]
0.50
0.50
B)
\[{{X}_{{{H}_{2}}}}\]
\[{{X}_{{{O}_{2}}}}\]
0.22
0.78
C)
\[{{X}_{{{H}_{2}}}}\]
\[{{X}_{{{O}_{2}}}}\]
0.78
0.22
D)
\[{{X}_{{{H}_{2}}}}\]
\[{{X}_{{{O}_{2}}}}\]
0.35
0.65
Correct Answer: C
Solution :
[c] Initially \[{{H}_{2}}\]and \[{{O}_{2}}\]both are present \[pV=nRT\] \[{{n}_{{{H}_{2}}}}+{{n}_{{{o}_{2}}}}={{n}_{(total)}}=\frac{pV}{RT}=\frac{1\times V}{RT}=\frac{V}{RT}\] After reaction is complete, only \[{{H}_{2}}\](unreacted) is left Moles of unreacted \[{{H}_{2}}=\frac{pV}{RT}=\frac{0.35V}{RT}\] Thus, \[n{{'}_{{{H}_{2}}}}+n{{'}_{{{O}_{2}}}}+\]unreacted \[{{H}_{2}}=\frac{V}{RT}\] \[n{{'}_{{{H}_{2}}}}+n{{'}_{{{O}_{2}}}}=\frac{V}{RT}=\frac{0.35V}{RT}\Rightarrow n{{'}_{{{H}_{2}}}}+n{{'}_{{{O}_{2}}}}=\frac{0.65V}{RT}\] Also, \[2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l)\] Moles of \[{{H}_{2}}=2\times \]moles of \[{{O}_{2}}\] \[{{n}_{{{H}_{2}}}}=2n{{}_{{{O}_{2}}}}\] \[\therefore 3{{n}_{{{O}_{2}}}}=\frac{0.65V}{RT};{{n}_{{{O}_{2}}}}=\frac{0.65V}{3RT}\] \[\therefore {{\chi }_{{{O}_{2}}}}=\frac{{{n}_{{{O}_{2}}}}}{{{n}_{total}}}=\frac{0.65V/3RT}{V/RT}=0.22\] \[{{\chi }_{{{H}_{2}}}}=1-0.22=0.78\]
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