Determination of the molar mass of ethylene glycol Weight of water: 99.95 g Weight of added ethylene glycol: 55.7 g Boiling Point of Solution: 104.6 °C Molar mass of ethylene glycol: Mmol
The Correct Answer and Explanation is:
Given:
- Mass of water (solvent), msolvent=99.95 g=0.09995 kgm_{\text{solvent}} = 99.95 \, \text{g} = 0.09995 \, \text{kg}msolvent=99.95g=0.09995kg
- Mass of ethylene glycol (solute), msolute=55.7 gm_{\text{solute}} = 55.7 \, \text{g}msolute=55.7g
- Boiling point of solution = 104.6 °C
- Boiling point of pure water = 100.0 °C
- Boiling point elevation, ΔTb=104.6−100.0=4.6∘C\Delta T_b = 104.6 – 100.0 = 4.6^\circ \text{C}ΔTb=104.6−100.0=4.6∘C
- Boiling point elevation constant of water, Kb=0.512 °C\cdotpkg/molK_b = 0.512 \, \text{°C·kg/mol}Kb=0.512°C\cdotpkg/mol
Step 1: Use the boiling point elevation formula
ΔTb=i⋅Kb⋅m\Delta T_b = i \cdot K_b \cdot mΔTb=i⋅Kb⋅m
Where:
- ΔTb\Delta T_bΔTb = boiling point elevation
- iii = van’t Hoff factor (for ethylene glycol, a non-electrolyte, i=1i = 1i=1)
- KbK_bKb = ebullioscopic constant
- mmm = molality of the solution (mol of solute per kg of solvent)
4.6=1⋅0.512⋅(mol solute0.09995)4.6 = 1 \cdot 0.512 \cdot \left( \frac{\text{mol solute}}{0.09995} \right)4.6=1⋅0.512⋅(0.09995mol solute)mol solute0.09995=4.60.512≈8.984\frac{\text{mol solute}}{0.09995} = \frac{4.6}{0.512} \approx 8.9840.09995mol solute=0.5124.6≈8.984mol solute=8.984×0.09995≈0.897 mol\text{mol solute} = 8.984 \times 0.09995 \approx 0.897 \, \text{mol}mol solute=8.984×0.09995≈0.897mol
Step 2: Find molar mass
Mmol=mass of solutemol solute=55.70.897≈62.1 g/molM_{\text{mol}} = \frac{\text{mass of solute}}{\text{mol solute}} = \frac{55.7}{0.897} \approx \boxed{62.1 \, \text{g/mol}}Mmol=mol solutemass of solute=0.89755.7≈62.1g/mol
Explanation
To determine the molar mass of a solute using boiling point elevation, we rely on colligative properties—those that depend solely on the number of solute particles, not their identity. Ethylene glycol, a common antifreeze, is a non-electrolyte and does not dissociate in solution, so it has a van’t Hoff factor (iii) of 1.
The boiling point of a solution is higher than that of the pure solvent. This increase in boiling point (ΔTb\Delta T_bΔTb) is directly proportional to the molality of the solution (moles of solute per kilogram of solvent). The relationship is described by the equation:ΔTb=i⋅Kb⋅m\Delta T_b = i \cdot K_b \cdot mΔTb=i⋅Kb⋅m
In our case, the boiling point of the solution is 104.6 °C, which is 4.6 °C higher than the boiling point of pure water. Given that the boiling point elevation constant for water (KbK_bKb) is 0.512 °C·kg/mol and the solvent is water weighing 99.95 g (or 0.09995 kg), we substitute into the equation and solve for the molality. This gives us the number of moles of ethylene glycol dissolved.
With the moles calculated (approximately 0.897 mol) and the known mass of the solute (55.7 g), we calculate the molar mass using the formula:Molar mass=massmoles\text{Molar mass} = \frac{\text{mass}}{\text{moles}}Molar mass=molesmass
This yields a molar mass of approximately 62.1 g/mol, which closely matches the known molar mass of ethylene glycol (C₂H₆O₂), confirming the accuracy of the calculation.
