JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Volumetric Analysis

  • question_answer
    The volume of 0.05 M \[{{H}_{2}}S{{O}_{4}}\] required to neutralise 80 ml of 0.13 N \[NaOH\] will be [CPMT 1989]

    A) 104 ml

    B) 52 ml

    C) 10.4 ml

    D) 26 ml

    Correct Answer: A

    Solution :

    \[{{N}_{1}}{{V}_{1}}\ =\ {{N}_{2}}{{V}_{2}}\] \[0.1\times {{V}_{1}}=0.13\times 80\Rightarrow {{V}_{1}}=104\,ml\]


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