What is the boiling point of a 3.40 m aqueous solution of a nonvolatile nonelectrolyte? (The boiling point elevation constant for water is 0.512^° C/m) A)1.74^° C B)98.3^° C C)100.^° C D)102^° C

Question

Quizplus · Accepted Answer

The boiling point elevation equation is: 
ΔTb = Kb * m
where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant (0.512 °C/m for water), and m is the molality of the solution (moles of solute per kg of solvent).
Since the solute is a nonvolatile nonelectrolyte, it does not dissociate into ions and does not affect the van't Hoff factor (i). Therefore, m = 3.40 mol/kg.
Plugging in the values: 
ΔTb = (0.512 °C/m) * 3.40 mol/kg = 1.74 °C
Since the boiling point elevation is added to the normal boiling point of water (100 °C), the boiling point of the solution is:
Boiling point = 100 °C + 1.74 °C = 101.74 °C
Rounding to the nearest tenth gives the answer: 
Boiling point = 102.0 °C, which is choice D.

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