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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2000

  • question_answer
    At \[{{90}^{o}}C\]pure water has \[[{{H}_{3}}{{O}^{+}}]={{10}^{-6}}\] \[mole\text{ }litr{{e}^{-1}},\]the value of \[{{K}_{\omega }}\] at \[{{90}^{o}}C\]is:

    A)  \[{{10}^{-14}}\]               

    B)         \[{{10}^{-12}}\]               

    C)         \[{{10}^{-8}}\]                 

    D)         \[{{10}^{-6}}\]

    Correct Answer: D

    Solution :

    \[{{K}_{\omega }}=[{{H}^{+}}]\,[O{{H}^{-}}]=[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]\] \[={{10}^{-6}}\times {{10}^{-6}}\] \[={{10}^{-12}}\]                 \[\because \] \[[{{H}_{3}}{{O}^{+}}]={{10}^{-6}}mole/litre=[O{{H}^{-}}]\]


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