Answer:0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point-C
0.13 m FeCl3---- Lowest boiling point-D
Explanation:
Using the boilng point elevation formula
ΔTb=m* kb *i
where m= molality
kb= elevated boiling point constant( here kb values will be same for all soluton)
i= vant hoff factor = number of ions present in a solution
Using the number of ions and molarity present in a solution as a collagative property, since kb is constant, we can determine which of the species has the highest boiling point.
1.) 0.13 m FeCl3= Fe³⁻ + Cl⁻
i=4
ΔTb=m* kb* i= molarity x number of ionsx Kb= 0.13 x 4= 0.52kb
2) 0.19 m Mg(CH3COO)2 = Mg²⁺ + CH₃COO⁻
i= 3
ΔTb=m* kb* i= molarity x number of ions= 0.19 x 3= 0.57kb
3. 0.30 m KI = K⁺ + I⁻
i= 2
ΔTb=m *kb *= imolarity x number of ions xKb= 0.30x 2= 0.60kb
4. 0.53 m Glucose(nonelectrolyte) =
i= 1 for nonelectroytes
ΔTb=m* kb* i = molarity x number of ionsx Kb= 0.53 x 1= 0.53Kb
therefore,
0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point
0.13 m FeCl3---- Lowest boiling point
