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0.45 gm of an acid of mol. wt. 90 was neutralised by 20 ml of 0.5 (N) caustic potash. The basicity of acid is-

(1) 1 

(2) 2

(3) 3 

(4) 4

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Correct Answer is: (2) 2

20 ml of 0.5 (N) caustic potash

= 20 × 0.5 × 10–3 equiv. of caustic potash

∴ 20 × 0.5 × 10–3 equiv. of acid = 0.45 gm

∴ 1 equiv. of acid = 0.45/10 x 10-3 = 45 gm

∴ x for acid = 90/45 = 2

Hence, basicity of acid = 2

followed by formation of carbocation

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