See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

  3 4 2 3 2 4 2 Fe C O Fe 2CO oxidation in acidic medium       n-factor = (3 + 2 4) – (2 + 2  3) = 3 No. of equivalents of ferrous oxalate = 0.1  3 = 0.3   2 2 4 2 2 2 CaC O .H O Ca 2CO H O oxidation acidic medium     n-factor = (2 + 2  4) – (2 + 2  3) = 2 no. of equivalents of CaC2O4.H2O =