Question

Which one of the following solutions would yield the highest boiling point?

A) 0.10m NH3 B) 0.10m KCl C) 0.10m K2SO4 D) 0.10m sucrose

C is the answer

Answer 1

A) 0.10m NH3

i= 1

m= 0.1m

kb = 0.512C⁰/m

Tb = i*Kb*m

          = 1*0.512*0.1 = 0.0512C⁰

boliling point = 100+ 0.0512 = 100.0512C⁰

B) 0.10m KCl

KCl -----------> K⁺ + Cl^-

i = 2

Tb = i*Kb*m

         = 2*0.512*0.1

        = 0.1024C⁰

boliling point = 100+ 0.1024 = 100.1024C⁰

C) 0.10m K2SO4

   K2SO4 ----------> 2K⁺ + SO4^2-

    i = 3

Tb = i*Kb*m

          = 3*0.512*0.1 = 0.1536C⁰

boiling point = 100+ 0.1536 = 100.1536C⁰   Highest bolinling point

D) 0.10m sucrose

   i = 1 for non electrolyte

Tb = i*Kb*m

          = 1*0.512*0.1 = 0.0512C⁰

boiling point = 100+ 0.0512 = 100.0512C⁰