Question

An ethylene glycol solution contains 19.3G of ethylene glycol (C2H6O2) in 82.4mL of water.

(a) Compute the freezing point of the solution. (Assume a density of 1.00 g/mL for water.)
(b)Compute the boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

Express your answer using five significant figures.

Answer 1

Answer:

$\boxed{\text{(a) -7.0188 ^{\circ} C; (b) 101.93 ^{\, \circ} C}}$

Explanation:

1. Calculate the molal concentration of ethylene glycol

$b = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}\\n = \text{19.3 g} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{0.3109 mol}\\\text{Mass} = \text{82.4 g} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{0.0824 kg}\\b = \dfrac{\text{0.3109 mol}}{\text{0.0824 kg}} = \text{3.774 mol/kg}$

2. Calculate the freezing point

The formula for the freezing point depression ΔTf by a nonelectrolyte is

$\Delta T_{f} = K_{f}b\\\Delta T_{f} = 1.86 \times 3.774 = 7.0188 \,^{\circ}\text{C}\\T_{f} = T_{f}^{^\circ} - \Delta T_{f} = 0.00 -7.0188 = \textbf{-7.0188 ^{\circ} C}$

3. Calculate the boiling point

The formula for the boiling point elevation ΔTb by a nonelectrolyte is

$\Delta T_{b} = K_{b}b\\\Delta T_{b} = 0.512 \times 3.774 = 1.932 \,^{\circ}\text{C}\\T_{b} = T_{b}^{^\circ} + \Delta T_{b} = 100.00 + 1.932 = \textbf{101.93 ^{\, \circ} C}$