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AMU Medical AMU Solved Paper-2013

  • question_answer
    The degree of ionisation of HF in 0.100 M aqueous solution is (freezing point of the solution \[=-{{0.197}^{o}}C\] and \[{{K}_{f}}\] for water \[={{186}^{o}}C)\]

    A)  6%                        

    B)  12%  

    C)  3%                        

    D)  9%

    Correct Answer: A

    Solution :

                    \[\Delta {{T}_{f}}=i\,{{k}_{1}}\,m\] \[0.197=i\times 1.86\times 0.1\]                 \[i=\frac{0.197}{1.86\times 0.1}=1.059\approx 1.06\] In case of dissociation, \[\alpha =\frac{i-1}{n-1}=\frac{1.06-1}{2-1}\]                                                 = 0.06 = 6%


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