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asked ago in Chemistry by (23.4k points)

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

Sarthaks Test For Class 7-12

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answered ago by (24.7k points)
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Best answer

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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