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JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    The freezing point of a 1.00 m aqueous solution of HF is found to be \[-1.91\text{ }{}^\circ C\]. The freezing point constant of water, K. is 1.86 K  kg \[\text{mo}{{\text{l}}^{-1}}\] The percentage dissociation of HF at this concentration is.   JEE Main Online Paper (Held On 26-May-2012)  

    A) 30%                       

    B) 10%

    C) 5.2%                                     

    D) 2.7%

    Correct Answer: D

    Solution :

    \[\Delta {{T}_{f}}={{K}_{f}}\times m\times i\] \[i=\frac{\Delta T}{{{K}_{f}}\times m}=\frac{1.91}{1.86\times 1}=1.02\] For\[\underset{(1-\alpha )}{\mathop{HI}}\,\underset{\alpha }{\mathop{{{H}^{+}}}}\,+\underset{\alpha }{\mathop{{{I}^{-}}}}\,\] \[1-\alpha +\alpha =i=1.027\] \[1+\alpha =1.02\] \[\alpha =0.02\]or2.7%


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