JEE Main & Advanced Sample Paper JEE Main Sample Paper-14

  • question_answer
    Direction (Q. Nos. 57): A given sample of \[{{N}_{2}}{{O}_{4}}\] in a closed shows 20% dissociation in \[N{{O}_{2}}\] at\[{{27}^{o}}C\]and 1 atm. The sample is now heated up to\[{{127}^{o}}C\]and the analysis of the mixture shows 60% dissociation at\[{{127}^{o}}C\].
    The molecular weight of mixture at \[{{27}^{o}}C\] is

    A)  76.66                                   

    B)  78.69             

    C)  66.52                                   

    D)  80.24

    Correct Answer: A

    Solution :

     \[0.8a\times 92+0.4a\times 46\] \[=1.2\times a\times M\] at 300 K \[M=76.66\,g\,\,mo{{l}^{-1}}\]


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