If boiling point of urea solution is 100.18 °C and kb for water is 0.512 K kg mol', molality of solution is? (Boiling point of water = 100°C) A 0.25 mol kg' B 0.6 mol kg C 0.45 mol kg D 0.35 mol kg

If boiling point of urea solution is 100.18 °C and kb for water is 0.

1.	If boiling point of urea solution is 100.18 °C and kb for water is 0.512 K kg mol', molality of solution is? (Boiling point of water = 100°C) A 0.25 mol kg' B 0.6 mol kg C 0.45 mol kg D 0.35 mol kg
Answer» Explanation:Kb = 0.512 K kg mol-1 KF = 1.86 K kg mol-1 ΔTb = 100.18° – 100° = 0.18 ΔT = Kb x molality molality = 0.18/0.512 Now, ΔTf = Kf x molality = 1.86 x 0.18 0.512 0.180.512 = 0.6539 FREEZING POINT of solutions = 0 – 0.6539 = -0.6539Read more on Sarthaks.com - https://www.sarthaks.com/830773/solution-of-urea-in-water-has-boiling-point-of-100-18c-calculate-the-freezing-point-of-the