Answer :
Final answer:
The boiling point of the solution with 21.4 g of ethylene glycol in 97.6 mL of water is approximately 101.80 °C. This is calculated using the molality of the solution and the boiling point elevation constant for water.
Explanation:
To compute the boiling point of the solution containing 21.4 g of ethylene glycol (C₂H₆O₂) in 97.6 mL of water, we need to first determine the molality of the solution. Because we are asked to assume the density of water is 1.00 g/mL, the mass of water used to dissolve the ethylene glycol can be found to be 97.6 g.
Using the molar mass of ethylene glycol, which is 62.07 g/mol, the moles of ethylene glycol are calculated as:
Mo₂H₄ (OH)₂ = 21.4 g C₂H₆O₂ / 62.07 g/mol = approx. 0.344 moles of ethylene glycol.
Te molality (m) of the solution, which is the number of moles of solute per kilogram of solvent, is then:
m = moles of solute / mass of solvent in kg = 0.344 moles / 0.0976 kg = 3.52 mol/kg
To calculate the boiling point elevation, the van't Hoff factor (i), which is 1 for ethylene glycol since it does not dissociate in solution, and the boiling point elevation constant (ΔT₂) for water, which is 0.512 °C/m, are needed. Thus the boiling point elevation is calculated as:
ΔT₂ = i * K₂ * m = 1 * 0.512 °C/m * 3.52 m = 1.80 °C
The boiling point of the solution can be estimated by adding the boiling point elevation to the normal boiling point of pure water, which is 100 °C. Therefore, the boiling point of this ethylene glycol solution is approximately 101.80 °C.