JEE Main & Advanced AIEEE Solved Paper-2011

  • question_answer
    The degree of dissociation \[(\alpha )\] of a weak electrolyte, \[{{A}_{x}}{{B}_{y}}\] is related to van't Hoff factor (i) by the expression:   AIEEE  Solved  Paper-2011

    A) \[\alpha =\frac{x+y+1}{i-1}\]         

    B) \[\alpha =\frac{i-1}{\left( x+y-1 \right)}\]

    C) \[\alpha =\frac{i-1}{x+y+1}\]                         

    D) \[\alpha =\frac{x+y-1}{i-1}\]

    Correct Answer: B

    Solution :

                 Van?t Hoff factor (i) \[=\frac{Observed\,colligative\,property}{Normal\,colligative\,property}\] \[{{A}_{x}}{{B}_{y}}\xrightarrow{{}}x{{\underset{x\alpha }{\mathop{A}}\,}^{+y}}+yB_{y\alpha }^{^{-x}}\] Total moles \[=1-\alpha +x\alpha +y\alpha \]                    \[=1+\alpha (x+y-1)\]    \[i=\frac{1+\alpha (x+y-1)}{1}\] \[\Rightarrow \,\,\,\,\alpha =\frac{i-1}{x+y-1}\]


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