Question

10 ml of 0.2 N HCl and 30 ml of 0.1 N HCl together exactly neutralise 40 ml of a solution of NaOH, which is also exactly neutralized by a solution in water of 0.61 g of an organic acid. The equivalent weight of the organic acid is :

Answer 1

Answer:

122

Explanation:

10 ml of 0.2 N HCl + 30 ml of 0.1 N HCl=40 ml of NaOH (0.61 g organic acid)

10 ml of 0.2 N HCl + 30 ml of 0.1 N HCl=40 ml of NaOH (0.61 g organic acid)meq. of HCl= meq of NaOH=meq of organic acid

10 ml of 0.2 N HCl + 30 ml of 0.1 N HCl=40 ml of NaOH (0.61 g organic acid)meq. of HCl= meq of NaOH=meq of organic acid10×0.2+30×0.1=0.61/E×1000

×10005=0.61×1000/E

E=160/5=122

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Answer 2

Answer:

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