Answer:
0.100 m AlCl3 will have the highest boiling point
Explanation:
Step 1: Data given
The molal boiling point elevation constant for water is 0.51°C/m
Since those are all aqueous solutions, the have the same molal boiling point elevation constant
Step 2:
0.100 m C6H12O6
ΔT = i*Kb*m
⇒with ΔT is the boiling point elevation = TO BE DETERMINED
⇒with i = the van't Hoff factr = 1
⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m
⇒with m = the molality = 0.100m
ΔT = 1 * 0.51 * 0.100
ΔT = 0.051 °C
0.100 m NaCl
ΔT = i*Kb*m
⇒with ΔT is the boiling point elevation = TO BE DETERMINED
⇒with i = the van't Hoff factr = Na+ + Cl- = 2
⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m
⇒with m = the molality = 0.100m
ΔT =2 * 0.51 * 0.100
ΔT = 0.102 °C
0.100 m AlCl3
ΔT = i*Kb*m
⇒with ΔT is the boiling point elevation = TO BE DETERMINED
⇒with i = the van't Hoff factr = Al^3+ + 3Cl- = 4
⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m
⇒with m = the molality = 0.100m
ΔT =4 * 0.51 * 0.100
ΔT = 0.204 °C
0.100 m MgCl2
ΔT = i*Kb*m
⇒with ΔT is the boiling point elevation = TO BE DETERMINED
⇒with i = the van't Hoff factr = Mg^2+ +2Cl- = 3
⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m
⇒with m = the molality = 0.100m
ΔT =3 * 0.51 * 0.100
ΔT = 0.153 °C
0.100 m AlCl3 will have the highest boiling point
