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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 42

  • question_answer
    A mixture of \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\] and \[HCOOH\]is heated with cone. \[{{H}_{2}}S{{O}_{4}}\]. The gas produced is collected, and on treatment with \[KOH\]solution, the volume of the gas decreases by \[\frac{1}{6}\]. The molar ratio of the two acids in the original mixture is

    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 \[C{{O}_{2}}=\frac{1}{6}(x+2y)\] \[y=\frac{1}{6}(x+2y)\] \[\therefore \,\,\,\,\,\frac{x}{y}=4\] \[\therefore \,\,\,\,\,x:y::4:1\]


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