What is the boiling point of a solution that contains 3 moles of KBr in 2000 g of water? (Kb = 0.512 C/m)

The answer to your question is   Tb = 100.768°C

Explanation:

Data

Tb = ?

moles of KBr = 3

mass of water = 2000 g

Kb = 0.512 C/m

Process

1.- Calculate the molality of the solution

molality = moles of solute/mass of solvent (kg)

-Substitution

molality = 3/2

-Result

molality = 1.5

2.- Calculate ΔT

    ΔTb = Kbm

-Substitution

    ΔTb = (0.512)(1.5)

-Result

    ΔTb = 0.768

3.- Calculate the boiling point, the boiliing point of pure water is 100°C

   ΔTb = Tb - Tb(solvent)

-Solve for Tb

   Tb = Tb(solvent) + ΔTb

-Substitution

   Tb = 100 + 0.768

-Result

  Tb = 100.768°C

output?

explanation: