JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Volumetric Analysis

  • question_answer
    25 ml of a solution of\[N{{a}_{2}}C{{O}_{3}}\]having a specific gravity of 1.25 required 32.9 ml of a solution of HCl containing 109.5 grams of the acid per litre for complete neutralization. Calculate the volume of \[0.84N{{H}_{2}}S{{O}_{4}}\] that will be completely neutralized by 125 grams of the\[N{{a}_{2}}C{{O}_{3}}\]solution [UPSEAT 2001]

    A) 460 ml

    B) 540 ml

    C) 480 ml

    D) 470 ml

    Correct Answer: D

    Solution :

    \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[N\times 25=\frac{109.5\times 32.9}{36.5}\]Þ  \[N=\frac{109.5\times 32.9}{36.5\times 25}\] \[{{N}_{3}}{{V}_{3}}={{N}_{4}}{{V}_{4}}\] (\[{{V}_{3}}=\frac{m}{d}\], \[{{V}_{3}}=\frac{125}{1.25}\]) \[\frac{109.5\times 32.9}{36.5\times 25}\times 100=0.84\times V\]Þ  \[V=470\,ml\]


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