Problem 76 - Chemistry The Central Science

Question

What is the freezing point of an aqueous solution that boils at \(105.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The freezing point of the aqueous solution that boils at \(105.0^{\circ}\mathrm{C}\) is approximately \(-18.17^{\circ}\mathrm{C}\).

Step 1: Determine the boiling point elevation of the solution.

The boiling point elevation, ΔTb, can be found by subtracting the normal boiling point of the solvent (water) from the boiling point of the solution: ΔTb = Tb - Tb,water In this case, the boiling point of water is \(100^{\circ}\mathrm{C}\). So: ΔTb = \(105.0^{\circ}\mathrm{C}\) - \(100^{\circ}\mathrm{C}\) = \(5^{\circ}\mathrm{C}\).

Step 2: Find the molality of the solute in the solution.

The boiling point elevation formula is given by: ΔTb = Kb * molality Where Kb is the ebullioscopic constant, which for water is 0.512 \(^{\circ}\mathrm{C/m}\). Rearranging the equation and plugging in the values, we get: molality = ΔTb / Kb = \(\frac{5^{\circ}\mathrm{C}}{0.512^{\circ}\mathrm{C/m}}\) = 9.7656 m

Step 3: Determine the freezing point depression of the solution.

The freezing point depression formula is given by: ΔTf = Kf * molality Where Kf is the cryoscopic constant, which for water is 1.86 \(^{\circ}\mathrm{C/m}\). Plugging in the values, we get: ΔTf = 1.86 \(^{\circ}\mathrm{C/m}\) * 9.7656 m = 18.1656 \(^{\circ}\mathrm{C}\)

Step 4: Find the freezing point of the aqueous solution.

The freezing point of the solution can be found by subtracting the freezing point depression from the normal freezing point of the solvent (water). In this case, the freezing point of water is \(0^{\circ}\mathrm{C}\). So: Freezing point of the solution = Tf,water - ΔTf Freezing point of the solution = \(0^{\circ}\mathrm{C}\) - 18.1656 \(^{\circ}\mathrm{C}\) = \(-18.1656^{\circ}\mathrm{C}\) The freezing point of the aqueous solution that boils at \(105.0^{\circ}\mathrm{C}\) is approximately \(-18.17^{\circ}\mathrm{C}\).