A) \[2:3\]
B) \[3:2\]
C) \[4:1\]
D) \[1:4\]
Correct Answer: C
Solution :
[c] \[\underset{y}{\mathop{{{H}_{2}}{{C}_{2}}{{O}_{4}}}}\,\xrightarrow[\Delta ]{{{H}_{2}}S{{O}_{4}}}{{H}_{2}}O+\underset{y}{\mathop{CO}}\,+\underset{y}{\mathop{C{{O}_{2}}}}\,\] \[\underset{x}{\mathop{HCOOH}}\,\xrightarrow[\Delta ]{{{H}_{2}}S{{O}_{4}}}{{H}_{2}}O+\underset{x}{\mathop{CO}}\,\] Let x and moles of \[HCOOH\]and \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\]be present in the original mixture. Total moles of CO formed\[~=x+y\] moles of \[C{{O}_{2}}\] formed =y Total moles of gases \[=x+y+y=x+2y\] Since \[KOH\]solution absorbs \[C{{O}_{2}},\] and volume reduces by \[1/6,\] moles of \[C{{O}_{2}}=\frac{1}{6}(x+2y)\] \[y=\frac{1}{6}(x+2y)\] \[\therefore \,\,\,\,\,\frac{x}{y}=4\] \[\therefore \,\,\,\,\,x:y::4:1\]You need to login to perform this action.
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