JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Volumetric Analysis

  • question_answer
    The volume of \[\frac{N}{10}\] NaOH require to neutralise 100 ml of \[\frac{N}{25}\] HCl is [Pb. CET 2000]

    A) 30 ml

    B) 100 ml

    C) 40 ml

    D) 25 ml

    Correct Answer: C

    Solution :

    For complete neutralisation, milli equivalent of base = milli equivalent of acid \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\]Þ  \[\frac{1}{10}\times {{V}_{1}}=\frac{1}{25}\times 100\]; \[{{V}_{1}}=40\ ml.\]


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