Question

Assuming ideal solution behavior, what is the boiling point of a solution of 115.0 g of nonvolatile sucrose, C₁₂H₂2O₁₁, in 350.0 g of water?

Answer 1

The boiling point of the solution is calculated using the molal boiling point elevation constant and the molality of the solution, resulting in a boiling point of approximately 100.49°C.

Step-by-step explanation:

To calculate the boiling point of a solution of sucrose in water, we will use the molal boiling point elevation constant (Kb) for water, which is given as 0.51°C/m. The steps include:

1. Calculate the number of moles of sucrose using its molar mass.
2. Calculate the molality of the solution.
3. Use the molality and Kb to find the boiling point elevation, ΔTb.
4. Add ΔTb to the normal boiling point of water to get the solution's boiling point.

First, the molar mass of sucrose (C12H22O11) = 12(12.01 g/mol) + 22(1.008 g/mol) + 11(16.00 g/mol) = 342.3 g/mol. Calculating moles of sucrose: 115.0 g / 342.3 g/mol = 0.336 moles.

Next, molality (m) = moles of solute / kg of solvent = 0.336 moles / 0.350 kg = 0.960 m.

Then, calculate the boiling point elevation: ΔTb = Kb × m = 0.51°C/m × 0.960 m = 0.4896°C.

Therefore, boiling point of the solution = normal boiling point of water + ΔTb = 100.00°C + 0.4896°C = 100.49°C.