Question

The boiling point of pure ethanol (C₂H₅OH, MM = 46.07 g/mol) is 78.4 °C. A solution is made using 100.0 g of ethanol (the solvent) and 19.1 g of glucose (MM = 180.16 g/mol). What is the boiling point of this solution, in °C? (Kb of ethanol is 1.22 °C/m)

Answer 1

The boiling point of the ethanol solution with glucose added is 79.6932 °C, calculated by finding the molality of glucose, using the ebullioscopic constant for ethanol, and adding the boiling point elevation to the pure ethanol boiling point.

Step-by-step explanation:

To find the boiling point elevation of an ethanol solution with a nonvolatile solute like glucose, we need to use the formula ΔT = i * Kb * m, where ΔT is the change in boiling point, i is the van't Hoff factor (i = 1 for non-electrolytes like glucose), Kb is the ebullioscopic constant of the solvent (ethanol in this case), and m is the molality of the solution.

First, we calculate the molality (m) of the glucose solution:

1. Find the moles of glucose: Moles = mass (g) / molar mass (g/mol). So, moles of glucose = 19.1 g / 180.16 g/mol = 0.106 mol.
2. Find the mass of ethanol in kilograms: 100.0 g ethanol = 0.100 kg.
3. Calculate the molality (m): m = moles of solute / mass of solvent (kg). So m = 0.106 mol / 0.100 kg = 1.06 m.

Next, we use the ebullioscopic constant for ethanol (Kb = 1.22 °C/m) to find the boiling point elevation:

ΔT = i * Kb * m = 1 * 1.22 °C/m * 1.06 m = 1.2932 °C.

Finally, we add the boiling point elevation to the boiling point of pure ethanol to get the boiling point of the solution:

Boiling point of solution = boiling point of pure ethanol + ΔT = 78.4 °C + 1.2932 °C = 79.6932 °C.

Therefore, the boiling point of the ethanol solution is 79.6932 °C.