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JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Volumetric Analysis

  • question_answer
    0.1914g of an organic acid is dissolved in approx. 20 ml of water. 25 ml of 0.12 N NaOH required for the complete neutralization of the acid solution. The equivalent weight of the acid is [MP PET 2000]

    A) 65

    B) 64

    C) 63.80

    D) 62.50

    Correct Answer: C

    Solution :

    \[\text{Volume}=25\,\,ml=\frac{25}{1000}litre\] \[\text{Normality}=\frac{wt}{eq.wt\times \text{Volume}}\]Þ \[0.12=\frac{0.1914\times 1000}{E\times 25}\] \[eq.\,wt.=\frac{0.1914\times 1000}{0.12\times 25}=63.8\]


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