Question

You have a 1.8M sodium phosphate solution , whose density is 1.13 g/ml. Calculate its: a) mol fraction. b) mass %. c) Molality. d) Normal freezing point. e) Normal boiling point.

Answer 1

LI Date: Giveffe , molasity of sodium phosphaterie Naz poy 5:458:Milan density d' = 1:13 amet A Page No. Assteming 12 solution is present → volume of solution (V) = 1 = !000ml Nous to determine the mass ok solution - with knowo, dom s mie dxv=1413 gmex 1000 ml = 1130 g now determine the masses of Na, Pou l water in and an o ikkea, molarity of Nazloy : 1:8 M = 18 moles pes t of solution -Nowy mass of 1 molé lof Na, poy ily mol mass = 1649 maxs 1.8 moles of Nazply =164X18 3 295.24 0_Mass of Wales = Total mass of solution – mass of NQ, POy 11309 - 295629_5 834,99 (a) to determine the mole fraction : Firstly we have to calculate - the no. of moles of both Na, Poy & Ho o mass 295.2 g = 1.8 moles Na3PO4) mol-mass 164 g/mol - mass - H2O) 834.8g u6i3 moles - mol-mass 18 g/mol Now mole fraction of Na₃PO, ath (Na₂ POų) ? 2 1.8 2. Afraz pOy) in naslon) + n o 18+46:3 → Unge PO,] = 0.037 16 Mass % :- Mars 1. Naz. POR mass Xino = 295.2_x100 = 26.146. Total mass (0) Molality : Leaotality' onardo -7_no.of moles of N9zPQy __ wt of solvent in kg. le H₂O 7.8 mot --0.8348kq 1130 0 = 2.1 malig = 2.1 m