Question

Exercise 7.0.968 g of an acid are present in 300 ml of a solution. 10 ml of this solution required exactly 20 ml of 0.05 N KOH solution. Calculate no. of neutralisable protons andequivalent weight of the acid (Given mol, wt. of acid is 98)​

Answer 1

Answer:

20 ml of 0.05 N KOH solution. Calculate no. of neutralisable protons andequivalent weight of the acid