BCECE Medical BCECE Medical Solved Papers-2010

  • question_answer
    Volume of 0.6 M \[NaOH\] required to neutralize \[30\,c{{m}^{3}}\] of 0.4 M HCl is

    A)  \[30\,\,c{{m}^{3}}\]

    B)  \[45\,\,c{{m}^{3}}\]

    C)  \[20\,\,c{{m}^{3}}\]

    D)  \[50\,\,c{{m}^{3}}\]

    Correct Answer: C

    Solution :

                    \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] or            \[{{V}_{1}}\times 0.6=30\times 0.4\] \[\therefore \] \[V=\frac{30\times 0.4}{0.6}\] \[=20\,\,c{{m}^{3}}\]


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