RAJASTHAN ­ PET Rajasthan PET Solved Paper-2004

  • question_answer
    The vapour density of\[PC{{l}_{5}}\]at\[300{}^\circ C\]is 60. Its dissociation percentage will be at this temperature

    A)  73              

    B)  77

    C)  83              

    D)  87

    Correct Answer: A

    Solution :

     Normal molecular weight of \[PC{{l}_{5}}=2\times \] vapour density \[=2\times 60\] \[=120\] Observed molecular weight of \[PC{{l}_{5}}=31+(35.5)\times 5\] \[=31+177.5\] \[=208.5\] van-t Hoff factor\[(i)=\frac{208.5}{120}\] \[=1.7375=1.73\] Degree of dissociation\[=\frac{i-1}{n-1}\] \[=\frac{1.73-1}{2-1}\] \[=0.73\] % dissociation \[=0.73\times 100=73%\]


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