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NEET Chemistry Analytical Chemistry Question Bank Self Evaluation Test - Analytical Chemistry

  • question_answer
    0.45 g of acid (molecular weight 90) is neutralized by 20 ml of 0.5N caustic potash. The basicity of acid is

    A) 1                                 

    B) 2    

    C) 3                                 

    D) 4

    Correct Answer: B

    Solution :

    [b] Eq. of acid = Eq of base, \[\therefore \frac{0.45}{E.wt}=\frac{20\times 0.5}{1000},=E.wt=45\] \[Basicity=\frac{M.wt}{E.wt}=\frac{90}{45}=2\]


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