Question

56.8 grams of glucose is dissolved in acetic acid. The boiling point of the acetic acid raises to 126.6°C. What is the molal concentration of the solution?

Answer 1

Molal concentration of glucose = 2.64 m

Step-by-step explanation:

The excersise can be solved by the formula for colligative property, elevation of boiling point.

ΔT = Kb . m . i

As glucose (C₆H₁₂O₆) is an organic compound, i = 1. (No ions generated)

ΔT = Boiling T° for solution - Boiling T° of pure solvent

Kb = Ebulloscopic constant.

We need data from acetic acid.

Boiling T° of pure solvent: 118.1°C

Kb = 3.22 °C/m

We replace data:

126.6°C - 118.1°C = 3.22°C/m . m . 1

(126.6°C - 118.1°C) / 3.22 m/°C = m

Molality = 2.64 mol/kg