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The boiling point of an aqueous solution is 102.32 °C. What is the freezing point? Constants...

The boiling point of an aqueous solution is 102.32 °C. What is the freezing point? Constants can be found here.

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Expert Solution

Boiling point elivation Tb = Kb * m

Boiling poit of water = 100C0

boliling elivation   T = 102.32-100 = 2.32C0

Ebulliscopic constant Kb = 0.512 C0/m

   m = T/Kb = 2.32/0.512 =4.53m

Elivation of freezing point   Tf = m*Kf

= 4.53*1.86 = 8.42C0

normally freezing point of water 100C0 but depresed freezing point = -8.42C0

                             

queen_honey_blossom answered 2 years ago
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