Question

     2 Na₃PO₄(aq) + 3 MgCl₂(aq) ® 6 NaCl(aq) + Mg₃(PO₄)₂(s)

1. How many grams of Mg3(PO4)2 can be produced when 82.0 mL of 3.5 M MgCl2 reacts with excess sodium phosphate?

2. A solution is made up of 86.3g ethylene glycol (C₂H₆O₆) in 143.2g of water.

    Calculate the freezing point of this solution (∆T_f = m X K_f, For water K_f = 1.86 ^oC/m)

3. Calculate the boiling point of the ethylene glycol/water solution from question #2   (∆T_b = m X K_b, For water K_b = 0.512 ^oC/m)

Answer 1

(1)

2 Na3PO4(aq) + 3 MgCl2(aq) -------> 6 NaCl(aq) + Mg3(PO4)2(s)

From above chemical equation it is clear that one mole or 262.86g of Mg3(PO4)2 are produced by 3 moles or 3 X 95.2g of MgCl2.

Moles of MgCl2 added = molarity X volume in liters

Moles of MgCl2 added = 3.5 X 0.082 = 0.287 moles

3 moles of MgCl2 produce one mole of Mg3(PO4)2.

0.287 moles of MgCl2 will produce (1 / 3 ) X 0.287 = 0.09566 moles

Mass of Mg3(PO4)2. produced = no. of moles produced X mal mass

Mass of Mg3(PO4)2. produced = 0.09566 X 262.86 = 25.15 g

(2) ∆Tf = m X Kf

∆Tf = T_solvent - T_solution

m = molality = moles of solute / mass of solvent in kg

moles of ethylene glycol = 86.3 / 62.07 = 1.4

(NOTE: YOU HAVE WRITTEN WRONG FORMULLA . CORRCT FORMULLA OF ETHYLENE GLYCOL IS C2H6O2)

m =1.4 / 0.1432 = 9.7, 143.2 G = 0.1432kg

∆Tf = T_solvent - T_solution = 9.7 X 1.86

0 - T_solution = 18.05

becaus freezing point of wapure water = 0^oC

T_solution = - 18.05 ^oC

(3) ∆Tb = T_solutiont - T_solvent =  m X Kb

T_solutiont = 100 + 9.7 X 0.512 = 104.96^oC