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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 orgainc acid
`10 xx 0.2 + 30 xx 0.1 -= (0.61)/(e) xx 100`
`5 = (0.61 xx 1000)/(E)`
`E = (610)/(5) = 122`
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