MCQ
$0.16\, g$ of dibasic acid required $25 \,ml$ of decinormal $NaOH$ solution for complete neutralisation. The molecular weight of the acid will be
- A$32$
- B$64$
- ✓$128$
- D$256$
✓
Answer
Correct option: C.
$128$
c
(c) Dibasic acid $NaOH$; ${N_1}{V_1} = {N_2}{V_2}$
(c) Dibasic acid $NaOH$; ${N_1}{V_1} = {N_2}{V_2}$
$\frac{W}{E} \times 1000 = \frac{1}{{10}} \times 25$; $\frac{{0.16}}{E} \times 1000 = \frac{{25}}{{10}}$
$M = 2 \times E = 2 \times 64 = 128$
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