Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.13 m FeCl3 A. Highest boiling point 2. 0.19 m Mg(CH3COO)2 B. Second highest boiling point 3. 0.30 m KI C. Third highest boiling point 4. 0.53 m Glucose(nonelectrolyte) D. Lowest boiling point An error has been detected in your answer. Check for typos,

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

fe 55.85 g/mol and h2o 18.02 g/mol.

i do not get it   at all