See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

147% oleum = 100 gm + 47 gm H2SO4 SO3 Oleum + H2O  H2SO4 KOH neutralisation 100 gm 47gm 147 gm No. of equivalents of H2SO4 = No. of equivalent of KOH 3 147 2 x 500 10 98      3 147 2 10 x 6M 98 500     