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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
    \[50c{{m}^{3}}\]of \[0.2N\text{ }HCl\]is titrated against\[~0.1\text{ }N\,\,NaOH\]    solution.   The   titration   was discontinued after adding \[50c{{m}^{3}}\]of \[NaOH\]. The remaining titration is completed by adding \[0.5N\text{ }KOH\]. The volume of \[KOH\] required for completing the titration is

    A)  \[12c{{m}^{3}}\]         

    B)  \[10c{{m}^{3}}\]

    C)  \[21.0c{{m}^{3}}\]

    D)  \[16.2c{{m}^{3}}\]

    Correct Answer: B

    Solution :

    No. of equivalent of \[HCl\] remaining after adding \[50c{{m}^{3}}\] of \[0.1N\,\,NaOH=\frac{0.2\times 50-0.1\times 50}{100}\] \[=\frac{0.5}{100}\] \[\therefore \] Volume of \[0.5N\text{ }KOH\]required \[\frac{0.5}{100}eq\] \[=\frac{V\times 0.5}{1000}\] \[V=\frac{0.5}{100}\times \frac{1000}{0.5}\] \[=10c{{m}^{3}}\]


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