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CBSE Class 11-science Answered

0.1914 g of an organic acid is dissolved in 20 ml of water,25 ml of 0.12 N NaOH is required for complete neutralisation of the acid solution. The equivalent weight of the acid is
Asked by Anil | 16 May, 2017, 10:16: PM
answered-by-expert Expert Answer

Moles of acid = 0.1914/M

Moles of NaOH = 25 x 10-3 x 0.12 = 3 x 10-3

Now, mole of acid = mole of NaOH

Or 0.1914/M = 3 x 10-3 or M = 0.1914/3x 10-3

                          = 63.8 gmol-1

Answered by Vaibhav Chavan | 17 May, 2017, 11:44: AM
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