See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Diborane forms adduct at low temperature 2 6 3 2 6 3 B H 2NH B H .2NH   2 6 3 B H .2NH on heating to 200oC gives borazole o 2 6 3 3 3 6 2 200 C 3B H .2NH 2B N H 12H    Moles of addition product = moles of diborane 24 28  Moles of borazole 2 24 3 28   Weight of borazole 2 24 81