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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Self Evaluation Test - Equilibrium

  • question_answer
    A gaseous compound of molecular mass 82.1 dissociates on heating to 400 K as \[{{X}_{2}}{{Y}_{4}}(g)\rightleftharpoons {{X}_{2}}(g)+2{{Y}_{2}}(g)\] The density of the equilibrium mixture at a pressure of 1 atm and temperature of 400K is\[2.0g{{L}^{-1}}\]. The percentage dissociation of the compound is  

    A) 12.5%              

    B) 48.5%

    C) 90.1%              

    D) 25.0%

    Correct Answer: A

    Solution :

    [a] \[D=\frac{PM}{RT}=\frac{1\times 82.1}{0.0821\times 400}=2.5g{{L}^{-1}}\] \[d=2.0g{{L}^{-1}}\](given) \[\alpha =\frac{D-d}{d(n-1)}=\frac{2.5-2}{2(3-1)}=0.125=12.5%\] %


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