Sea water contains 29.25% NaCl and 19% MgCl2 by weight of — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Total weight of solute in solution = 29.25 + 19 = 48.25 gm Total weight of solute in solution = 100 – 48.25 = 51.75 gm So Tb = i × Kb × m = (i × m) × Kb mole of NaCl = . 25 . = 0.5 Mole of MgCl2 = 95 19 = 0.2 Tb = . 75 . 1000 19 75 . 1000 . 25 . 2             So Tb Boling point of solution Tb = 100 + Tb = 116