Which one of the following aqueous solutions will exhibit highest boil...
Explanation:
Boiling point elevation is a colligative property that depends on the concentration of solute particles in a solution. The greater the concentration of solute particles, the higher the boiling point of the solution.
The formula for boiling point elevation is:
ΔTb = Kbm
where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant, and m is the molality of the solution (moles of solute per kilogram of solvent).
Now, let's compare the given solutions to determine which one will exhibit the highest boiling point:
a) 0.01 M KNO3
- KNO3 disassociates into 2 ions in solution: K+ and NO3-
- molality, m = 0.01 mol / 1 kg = 0.01 m
- Kb for water = 0.512 °C/m
- ΔTb = Kb x m = 0.512 x 0.01 = 0.00512 °C
b) 0.015 M urea
- Urea does not disassociate in solution
- molality, m = 0.015 mol / 1 kg = 0.015 m
- ΔTb = Kb x m = 0.512 x 0.015 = 0.00768 °C
c) 0.01 M Na2SO4
- Na2SO4 disassociates into 3 ions in solution: 2 Na+ and SO4 2-
- molality, m = 0.01 x 3 / 1 = 0.03 m (taking into account the number of ions)
- Kb for water = 0.512 °C/m
- ΔTb = Kb x m = 0.512 x 0.03 = 0.01536 °C
d) 0.015 M glucose
- Glucose does not disassociate in solution
- molality, m = 0.015 mol / 1 kg = 0.015 m
- ΔTb = Kb x m = 0.512 x 0.015 = 0.00768 °C
Therefore, the solution with the highest boiling point is 0.01 M Na2SO4, with a boiling point elevation of 0.01536 °C.
Which one of the following aqueous solutions will exhibit highest boil...
The correct answer is option C
Greater the number of solute particles in the solution, higher the elevation in boiling point. Among the given alternatives, Na2SO4 will exhibit the highest boiling point since it will dissociate into 3 ions when dissolved in the solution.
△ = i × kb×m
i × m of Na2SO4 is highest, hence its boiling point will also be highest.
Na2SO4 i × m=3×0.01=0.03
KNO3 i × m=2×0.01=0.02
Urea i×m=1×0.015=0.015
Glucose i×m=1×0.015=0.015