Question

A solution contains 482.6 g

Answer 1

Answer:

$\large \boxed{\text{102.57 \, ^{\circ} C}}$

Explanation:

Data:

m₂ = 482.6 g sucrose

m₁ = 0.2804 kg water

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{\text{b}} = iK_{\text{b}}b$

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

$n = \text{482.6 g} \times \dfrac{\text{1 mol}}{\text{342.30 g}} = \text{1.410 mol}$

2. Molal concentration of  sucrose

$b = \dfrac{\text{1.410 mol}}{\text{0.2804 kg}} = \text{5.029 mol/kg}$

3. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 5.029 \cdot mol \cdot kg^{-1} = 2.574 \, ^{\circ}\text{C}$

4. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 2.574 \, ^{\circ}\text{C} = \mathbf{102.57 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is \large \boxed{\mathbf{102.57 \, ^{\circ}C}} }$