Answer:
[tex]\large \boxed{\text{102.57 $\, ^{\circ}$C}}[/tex]
Explanation:
Data:
m₂ = 482.6 g sucrose
m₁ = 0.2804 kg water
Calculations:
The formula for the boiling point elevation ΔTb is
[tex]\Delta T_{\text{b}} = iK_{\text{b}}b[/tex]
i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For sucrose, i = 1.
1. Moles of sucrose
[tex]n = \text{482.6 g} \times \dfrac{\text{1 mol}}{\text{342.30 g}} = \text{1.410 mol}[/tex]
2. Molal concentration of sucrose
[tex]b = \dfrac{\text{1.410 mol}}{\text{0.2804 kg}} = \text{5.029 mol/kg}[/tex]
3. Increase in boiling point
[tex]\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 5.029 \cdot mol \cdot kg^{-1} = 2.574 \, ^{\circ}\text{C}[/tex]
4. Boiling point
[tex]\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 2.574 \, ^{\circ}\text{C} = \mathbf{102.57 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{102.57 \, ^{\circ}C}}$}[/tex]
