The freezing point of aqueous solution containing 5% by mass urea, 10% by mass KCl and 10% by mass of glucose, is1) ans (264.67 K).plz give me the solution part Share with your friends Share 7 Geetha answered this ∆Tf = i × Kf × m = i × Kf × W2M2 × W1 × 1000∆Tf = ∆Tf (urea) + ∆Tf (KCl) + ∆Tf (glucose) = 1 × 1.82 × 5× 100060 × 100 + 2 × 1.82 × 10× 100074.5 × 100 + 1 × 1.82 × 10× 1000180 × 100 = 1.52 + 4.89 + 1.01 = 7.42 oCFreezing point = 273 - 7.42 = 265.58 oC 11 View Full Answer
