Molarity (M) of an aqueous — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

n n n sol sol sol M v d   500 1.25 625g    Mass of solute (x) = 0.2 × 0.5 × 159.5 = 15.95 nsolute = 0.1, Mass of solvent = Mass of solution – Mass of solute = 625 – 15.95 = 609.05 0.1 m 609.05  m 0.164 164 10    NO2