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 | 16th May, 2017, 10:16: PM
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
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 | 17th May, 2017, 11:44: AM
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