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Answer :
To predict the boiling point of a 4.5 m aqueous solution of MgCl2, we can use the boiling-point-elevation formula: ΔTb = Kb × m × i, where ΔTb is the boiling point elevation, Kb is the boiling-point-elevation constant for water (0.51 °C/m), m is the molality of the solution (4.5 m), and i is the van't Hoff factor (number of ions produced per formula unit of solute).
The closest answer is 102.3 °C.
To predict the boiling point of the solution, we need to use the boiling-point-elevation constant (Kb) and the molality of the solution (4.5 m). The formula we use is:
ΔTb = Kb x m
where ΔTb is the boiling point elevation, Kb is the boiling-point-elevation constant, and m is the molality of the solution.
Substituting the values we have, we get:
ΔTb = 0.51 °C/m x 4.5 m
ΔTb = 2.295 °C
This means that the boiling point of the solution will be 2.295 °C higher than the boiling point of pure water, which is 100 °C at standard pressure.
Therefore, the boiling point of the solution can be calculated as:
Boiling point = 100 °C + ΔTb
Boiling point = 100 °C + 2.295 °C
Boiling point = 102.295 °C
Therefore, the answer is d. 102.3 °C.
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