Answer :
To determine which solution will exhibit the largest increase in boiling point compared to plain water, we need to utilize the concept of boiling point elevation. The boiling point elevation is determined by the formula:
[tex]\[ \Delta T_b = i \times K_b \times m \][/tex]
Where:
- [tex]\(\Delta T_b\)[/tex] is the boiling point elevation
- [tex]\(i\)[/tex] is the van't Hoff factor (the number of particles the solute dissociates into)
- [tex]\(K_b\)[/tex] is the ebullioscopic constant of the solvent (for water, [tex]\(K_b = 0.512 \, ^\circ \mathrm{C}/\mathrm{m}\)[/tex])
- [tex]\(m\)[/tex] is the molality of the solution
Let's analyze each solution step-by-step:
1. 0.5 M [tex]\(\mathrm{Al(NO_3)_3}\)[/tex]:
- The molality of the solution is 0.5 m.
- The van't Hoff factor ([tex]\(i\)[/tex]) for [tex]\(\mathrm{Al(NO_3)_3}\)[/tex] is 4 because it dissociates into four ions: 1 [tex]\(\mathrm{Al^{3+}}\)[/tex] and 3 [tex]\(\mathrm{NO_3^-}\)[/tex].
- Boiling point elevation: [tex]\( \Delta T_b = 4 \times 0.512 \times 0.5 = 1.024 \, ^\circ \mathrm{C} \)[/tex]
2. 0.5 M [tex]\(\mathrm{KOH}\)[/tex]:
- The molality of the solution is 0.5 m.
- The van't Hoff factor ([tex]\(i\)[/tex]) for [tex]\(\mathrm{KOH}\)[/tex] is 2 because it dissociates into two ions: 1 [tex]\(\mathrm{K^+}\)[/tex] and 1 [tex]\(\mathrm{OH^-}\)[/tex].
- Boiling point elevation: [tex]\( \Delta T_b = 2 \times 0.512 \times 0.5 = 0.512 \, ^\circ \mathrm{C} \)[/tex]
3. 4.0 M [tex]\(\mathrm{CH_2O}\)[/tex] (Formaldehyde):
- The molality of the solution is 4.0 m.
- The van't Hoff factor ([tex]\(i\)[/tex]) for [tex]\(\mathrm{CH_2O}\)[/tex] is 1 because it does not dissociate into ions.
- Boiling point elevation: [tex]\( \Delta T_b = 1 \times 0.512 \times 4.0 = 2.048 \, ^\circ \mathrm{C} \)[/tex]
Now, let’s compare the boiling point elevations of the three solutions:
- 0.5 M [tex]\(\mathrm{Al(NO_3)_3}\)[/tex]: [tex]\(1.024 \, ^\circ \mathrm{C}\)[/tex]
- 0.5 M [tex]\(\mathrm{KOH}\)[/tex]: [tex]\(0.512 \, ^\circ \mathrm{C}\)[/tex]
- 4.0 M [tex]\(\mathrm{CH_2O}\)[/tex]: [tex]\(2.048 \, ^\circ \mathrm{C}\)[/tex]
From these calculations, the solution with the largest increase in boiling point is 4.0 M [tex]\(\mathrm{CH_2O}\)[/tex], which exhibits a boiling point elevation of [tex]\(2.048 \, ^\circ \mathrm{C}\)[/tex]. Therefore, 4.0 M [tex]\(\mathrm{CH_2O}\)[/tex] will exhibit the largest increase in boiling point compared to plain water.
[tex]\[ \Delta T_b = i \times K_b \times m \][/tex]
Where:
- [tex]\(\Delta T_b\)[/tex] is the boiling point elevation
- [tex]\(i\)[/tex] is the van't Hoff factor (the number of particles the solute dissociates into)
- [tex]\(K_b\)[/tex] is the ebullioscopic constant of the solvent (for water, [tex]\(K_b = 0.512 \, ^\circ \mathrm{C}/\mathrm{m}\)[/tex])
- [tex]\(m\)[/tex] is the molality of the solution
Let's analyze each solution step-by-step:
1. 0.5 M [tex]\(\mathrm{Al(NO_3)_3}\)[/tex]:
- The molality of the solution is 0.5 m.
- The van't Hoff factor ([tex]\(i\)[/tex]) for [tex]\(\mathrm{Al(NO_3)_3}\)[/tex] is 4 because it dissociates into four ions: 1 [tex]\(\mathrm{Al^{3+}}\)[/tex] and 3 [tex]\(\mathrm{NO_3^-}\)[/tex].
- Boiling point elevation: [tex]\( \Delta T_b = 4 \times 0.512 \times 0.5 = 1.024 \, ^\circ \mathrm{C} \)[/tex]
2. 0.5 M [tex]\(\mathrm{KOH}\)[/tex]:
- The molality of the solution is 0.5 m.
- The van't Hoff factor ([tex]\(i\)[/tex]) for [tex]\(\mathrm{KOH}\)[/tex] is 2 because it dissociates into two ions: 1 [tex]\(\mathrm{K^+}\)[/tex] and 1 [tex]\(\mathrm{OH^-}\)[/tex].
- Boiling point elevation: [tex]\( \Delta T_b = 2 \times 0.512 \times 0.5 = 0.512 \, ^\circ \mathrm{C} \)[/tex]
3. 4.0 M [tex]\(\mathrm{CH_2O}\)[/tex] (Formaldehyde):
- The molality of the solution is 4.0 m.
- The van't Hoff factor ([tex]\(i\)[/tex]) for [tex]\(\mathrm{CH_2O}\)[/tex] is 1 because it does not dissociate into ions.
- Boiling point elevation: [tex]\( \Delta T_b = 1 \times 0.512 \times 4.0 = 2.048 \, ^\circ \mathrm{C} \)[/tex]
Now, let’s compare the boiling point elevations of the three solutions:
- 0.5 M [tex]\(\mathrm{Al(NO_3)_3}\)[/tex]: [tex]\(1.024 \, ^\circ \mathrm{C}\)[/tex]
- 0.5 M [tex]\(\mathrm{KOH}\)[/tex]: [tex]\(0.512 \, ^\circ \mathrm{C}\)[/tex]
- 4.0 M [tex]\(\mathrm{CH_2O}\)[/tex]: [tex]\(2.048 \, ^\circ \mathrm{C}\)[/tex]
From these calculations, the solution with the largest increase in boiling point is 4.0 M [tex]\(\mathrm{CH_2O}\)[/tex], which exhibits a boiling point elevation of [tex]\(2.048 \, ^\circ \mathrm{C}\)[/tex]. Therefore, 4.0 M [tex]\(\mathrm{CH_2O}\)[/tex] will exhibit the largest increase in boiling point compared to plain water.
