See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Given,   4 6 1 10 K Fe CN N      4 6 1 1 10 4 K Fe CN M N M valence factor     For   1 4 6 3 3 3 1 , / 40 1 10 25 40 K Fe CN n V mol litre mol m mol m      N N n V nST ST V        4 2 25 8.314 300