Question

What is the freezing point of vinegar, which is an aqueous solution of \(5.00 \%\) acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2},\) by mass \((d=\) \(\left.1.006 \mathrm{~g} / \mathrm{cm}^{3}\right) ?\)

Short answer

Answer: The freezing point of vinegar is approximately -1.63°C.

Step-by-step solution

Calculate the molality of the solution

First, we have to calculate the molality (mol solute/kg solvent) of the acetic acid solution. We are given a \(5.00\%\) aqueous solution, meaning there are \(5.00\) grams of \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) in \(100.0\) grams of the solution. 1. Calculate the mass of water: \(100.0 - 5.00 = 95.0\) grams 2. Convert mass of water to kilograms: \(\frac{95.0}{1000} = 0.0950\,\mathrm{kg}\) 3. Calculate moles of \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\): \(\frac{5.00 \,\mathrm{g}}{60.05\frac{\mathrm{g}}{\mathrm{mol}}} = 0.0833\,\mathrm{mol}\) Now, we can calculate the molality: \(m = \frac{0.0833\,\mathrm{mol}}{0.0950\,\mathrm{kg}} = 0.876\,\mathrm{mol\, kg^{-1}}\)

Calculate the change in freezing point

Now that we have the molality of the acetic acid solution, we can use the freezing-point depression constant (Kf) for water to calculate the change in the freezing point. The Kf value for water is \(1.86\, \mathrm{K\, kg\, mol^{-1}}\). The formula for calculating the change in freezing point is: \(\Delta T_f = -K_f \times m \times i\) where \(\Delta T_f\) is the change in freezing point, \(K_f\) is the freezing-point depression constant, \(m\) is the molality of the solution, and \(i\) is the van't Hoff factor (number of particles the solute dissociates into). Since acetic acid is a weak acid and does not completely dissociate in water, we will assume its van't Hoff factor is \(1\): \(\Delta T_f = -1.86\, \mathrm{K\, kg\, mol^{-1}} \times 0.876\,\mathrm{mol\, kg^{-1}} \times 1 = -1.63\,\mathrm{K}\)

Find the freezing point of vinegar

The freezing point of vinegar can be found by subtracting the change in freezing point from the normal freezing point of water (\(0^\circ\mathrm{C}\)): \(T_{f(vinegar)} = T_{f(water)} + \Delta T_f = 0^\circ\mathrm{C} - 1.63^\circ\mathrm{C} = -1.63^\circ\mathrm{C}\) So, the freezing point of vinegar is approximately \(-1.63^\circ\mathrm{C}\).