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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    In a \[0.2\] molal aqueous solution of a weak acid\[H\,\,X\], the degree of ionization is\[0.3\]. Taking \[{{k}_{f}}\] for water as\[1.85\], the freezing point of the solution will be nearest to

    A) \[-{{0.480}^{o}}C\]                         

    B)        \[-{{0.360}^{o}}C\]

    C)        \[-{{0.260}^{o}}C\]                         

    D)        \[+{{0.480}^{o}}C\]

    Correct Answer: A

    Solution :

    \[\Delta {{T}_{f}}=molarity\times {{k}_{f}}\times i\]             \[HX{{H}^{+}}+{{X}^{-1}}\] \[=0.2\times 1.85\times 1.3\]         \[\alpha =0.3\] \[={{0.481}^{o}}\]                                  \[i=1+\alpha =1.3\] \[\therefore \]\[f.p.={{0.481}^{o}}C\]


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