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