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

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

Step-by-step explanation:

1. Calculate the molal concentration of ethylene glycol

`b = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\\n = \text{19.3 g} * \frac{\text{1 mol}}{\text{62.07 g}} = \text{0.3109 mol}\\\text{Mass} = \text{82.4 g} * \frac{\text{1 kg}}{\text{1000 g}} = \text{0.0824 kg}\\b = \frac{\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 * 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 * 3.774 = 1.932 \,^(\circ)\text{C}\\T_(b) = T_(b)^(^\circ) + \Delta T_(b) = 100.00 + 1.932 = \textbf{101.93 $^(\, \circ)$C}`