Which aqueous solution has a change in its boiling point temperature of 0.026 0C (kb for water is 0.512 0C/m)?
0.0050 m CO2
0.050 m NaCl
0.050 m C6H12O6
0.020 m NH4Cl
0.010 m Al(NO3)3
To determine the change in boiling point temperature, we can use the formula:
ΔT = i * Kb * m
Where ΔT is the change in boiling point temperature, i is the van't Hoff factor (the number of particles that the solute dissociates into when dissolved), Kb is the molal boiling point constant, and m is the molality of the solution.
Let's calculate the change in boiling point temperature for each solution and determine which one has a change of 0.026 0C:
1) For CO2, i = 1 (since it does not dissociate in water), Kb = 0.512 0C/m, and m = 0.0050 m.
ΔT = (1)(0.512 0C/m)(0.0050 m) = 0.00256 0C
2) For NaCl, i = 2 (since it dissociates into Na+ and Cl- ions), Kb = 0.512 0C/m, and m = 0.050 m.
ΔT = (2)(0.512 0C/m)(0.050 m) = 0.0512 0C
3) For C6H12O6, i = 1 (since it does not dissociate in water), Kb = 0.512 0C/m, and m = 0.050 m.
ΔT = (1)(0.512 0C/m)(0.050 m) = 0.0256 0C
4) For NH4Cl, i = 2 (since it dissociates into NH4+ and Cl- ions), Kb = 0.512 0C/m, and m = 0.020 m.
ΔT = (2)(0.512 0C/m)(0.020 m) = 0.020 0C
5) For Al(NO3)3, i = 4 (since it dissociates into Al3+ and 3 NO3- ions), Kb = 0.512 0C/m, and m = 0.010 m.
ΔT = (4)(0.512 0C/m)(0.010 m) = 0.020 0C
From the calculations, the aqueous solution with a change in boiling point temperature of 0.026 0C is 0.050 m NaCl.