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JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    The vapour density of \[PC{{l}_{5}}\] at \[250{}^\circ C\] is found to be 57.9, percentage dissociation at this temperature is

    A)  80%                                      

    B)  22%

    C)  40%                                      

    D)  60%

    Correct Answer: A

    Solution :

    \[\alpha =\frac{{{V}_{\infty }}-{{V}_{0}}}{(n-1){{V}_{0}}}\] \[=\frac{104.25-57.9}{(2-1)\times 57.9}=0.80\] \[{{V}_{\infty }}=\frac{208.5}{2}\](\[\because {{V}_{0}}\times 2=\] molar mass) Hence, degree of dissociation = 80%


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