Q5 / 155 · WBJEE 2017 · Chemistry

$0.126 \mathrm{g}$ of an acid is needed to completely neutralise $20 \mathrm{mL} 0.1$ (N) NaOH solution. The equivalent weight of the acid is

  • A
    53
  • B
    40
  • C
    45
  • Correct answer
    D
    63

Solution

$\because$ Number of milli equivalents of NaOH $\left(n^{\prime}\right)=N V=0.1 \times 20=2$ $\therefore$ Number of equivalent $=0.002$ For Neutralisation Number of equivalent of base (NaOH) Number of equivalent of acid $0.002=\frac{W}{E}=\frac{0.126}{E}$ $\therefore \quad E=\frac{0.126}{0.002}=63$

Hence, the correct answer is option: 'D'.

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