See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Equivalents of Na2CO3 used = (100 – 60) × 10–3 × 1 50 Equivalent of Na2CO3 used = equivalent of Ca2+ or Mg2+ present in 100 ml of hard water = equivalent of Ca2+ or Mg2+ present in 100 ml of hard water. The amount of CaCO3 in 103 g of hard water = 3 40 10 10 1 50 50     = 400 × 10–3 g The amount of CaCO3 in 106 g of hard water = 400 g. Permanent hardness of water = 400 ppm. AITS-FT-I-PCM(Sol.)-JEE(Main)/2023 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 14 Mathematics PART – C SECTION – A