Arrange the following aqueous solutions in order of increasing boiling point temperature (lowest to highest temperature, with 1 being the lowest and 4 being the highest): 0.20 m glucose, 0.30 m BaCl2, 0.40 m NaCl, 0.50 m Na2SO4.

0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.

Explanation:

Step 1: Data given

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation =  Shows how much the boiling point increases

⇒i = the van't Hoff factor: Says in how many particles the compound will dissociate

⇒ Since all are aqueous solutions Kb for all solutions is the same (0.512 °C/m)

⇒m = the molality

Step 2:

0.20 m glucose

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for glucose = 1

⇒ Kb = 0.512 °C/m

⇒m = 0.20 m

ΔT = 1*0.512 * 0.20

ΔT = 0.1024 °C

0.30 m BaCl2

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for BaCl2 = Ba^2+ + 2Cl- : i = 3

⇒ Kb = 0.512 °C/m

⇒m = 0.30 m

ΔT = 3*0.512 * 0.30

ΔT = 0.4608 °C

0.40 m NaCl

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for NaCl = Na+ + Cl- : i = 2

⇒ Kb = 0.512 °C/m

⇒m = 0.40 m

ΔT = 2*0.512 * 0.40

ΔT = 0.4096 °C

0.50 m Na2SO4.

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for Na2SO4 = 2Na+ + SO4^2- : i =3

⇒ Kb = 0.512 °C/m

⇒m = 0.50 m

ΔT = 3*0.512 * 0.50

ΔT = 0.768 °C

0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.

i have no clue hope you find the answer

the correct answer is:

a. formation of precipitation