See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(A) 0 2 2 NH OH N  x 3 2 0 x 1 n factor 1 1 1       (B) n = factor of 2 4 3 Fe (SO ) 2  Eq. weight = M/2 (C) Mili eq. = milimoles × n = factor = 3 × 2 = 6.