JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Volumetric Analysis

  • question_answer
    Required amount of crystalline oxalic acid (eq. wt. = 63) to prepare \[{N}/{10}\;\] 250 ml oxalic acid solution is [MP PMT 1996]

    A) 0.158 g

    B) 1.575 g

    C) 15.75 g

    D) 6.3 g

    Correct Answer: B

    Solution :

    \[N\ =\ \frac{{{W}_{B}}\ \times \ 1000}{E\ \times \ V}\ \Rightarrow \ \frac{N}{10}\ =\ \frac{x\ \times \ 1000}{63\ \times \,250}\] \[\therefore \ x\ =\ 1.575grams\]


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