See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

    sol mole of I2 reacted =5×10–2 × 2 = 10–1 mole 2 2 4 PbO 2H O Pb(OH)   4 2 6 Pb(OH) 2NaOH Na [Pb(OH) ]   Na2PbO2 mole of NaOH required = 10–1× 2 3 1 1 V 10 2 10 10      3 2 V 2000 mili litre 10   = 2 litre