JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Volumetric Analysis

  • question_answer
    0.45 g of an acid (mol wt. = 90) required 20 ml of 0.5 N \[KOH\] for complete neutralization. Basicity of acid is [CPMT 1979]

    A) 1

    B) 2

    C) 3

    D) 4

    Correct Answer: B

    Solution :

    Normality = N = \[\frac{{{W}_{B}}\ \times \ 1000}{Eq.\,wt\times V}\] \[\therefore \ \text{Eq}\text{.}\ \text{Wt}\ =\ \frac{\text{0}\text{.45}\ \times \ \text{1000}}{0.5\ \times \ 20}\ =\ 45\] \[\therefore \ \text{Basicity}\ =\ \frac{\text{Molec}\text{.}\ \text{Wt}}{\text{Eq}\text{. Wt}}\ =\ \frac{90}{45}\ =\ 2\]


You need to login to perform this action.
You will be redirected in 3 sec spinner