See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

1 4 7 16 a a a ........ a 147      1 1 2 3 4 16 2a 15d 49 a a a a .......... a 392 N 98; M 49              Using   16 1 i 16 1 2 16 1 2 16 a 49 A.M G.M a a .......a a a ........a 16 2               S = 49 and W = 2