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10 mL of 0.2 N HCl and 30 mL of 0.1 N HC...

`10 mL` of `0.2 N HCl` and `30 mL` of `0.1 N HCl` to gether exaclty neutralises `40 mL` of solution of `NaOH`, which is also exactly neutralised by a solution in water of `0.61 g` of an organic acid. What is the equivalent weight of the organic acid?

A

61

B

91.5

C

122

D

183

Text Solution

Verified by Experts

The correct Answer is:
C
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 xx 0.2 + 30 xx0.1 -= (0.61)/(E) xx 1000 , " "5 = (0.61 xx 1000)/(E) rArr E = (610)/(5) = 122`
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