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BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    Volume of \[\text{0}\text{.6 M NaOH}\]required to neutralize \[\text{30}\,\text{c}{{\text{m}}^{\text{3}}}\]of \[\text{0}\text{.4}\,\text{M}\,\text{HCl}\]is

    A) \[30c{{m}^{3}}\]                              

    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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