Question

What is the freezing point of an aqueous solution that boils at 105.4 ∘c?

Answer 1

The freezing point of an aqueous solution boiling at 105.4 ℃ can be determined by back-calculating the molality from the observed boiling point elevation and then using this molality to calculate the freezing point depression. Assuming the solute does not ionize (van't Hoff factor of 1), the freezing point of the solution is -19.694 ℃.

Step-by-step explanation:

To determine the freezing point of an aqueous solution that boils at 105.4 ℃, we must first understand the process of boiling point elevation and freezing point depression. These are colligative properties, meaning they depend on the number of particles in the solution, not their identity. The key detail provided is that the solution boils at 105.4 ℃, indicating the presence of a solute that raises the boiling point above that of pure water, which is 100 ℃ at 1 atm pressure.

First, calculate the boiling point elevation (ΔTb) using the equation ΔTb = i * Kb * m, where 'i' is the van't Hoff factor, 'Kb' is the molal boiling point elevation constant of water (0.51 ℃/m), and 'm' is the molality of the solution. Since we are not given explicit information about the solute or its van't Hoff factor, we'll assume the solute is non-electrolyte (i.e., it does not dissociate into ions) and thus has an 'i' value of 1. This assumption could lead to inaccuracies if the solute were, in fact, an electrolyte.

To find the molality, we use the observed boiling point elevation to back-calculate. The observed elevation is 105.4 ℃ - 100 ℃ = 5.4 ℃. Thus, m = ΔTb / (i * Kb) = 5.4 / (1 * 0.51) = 10.588 mol/kg.

Knowing the molality, we can now calculate the freezing point depression (ΔTf) using the equation ΔTf = i * Kf * m, where 'Kf' is the molal freezing point depression constant of water, which is 1.86 ℃/m for non-electrolytes. Assuming the same molality and that 'i' is 1, we can find ΔTf = 1 * 1.86 * 10.588 = 19.694 ℃ of depression. The freezing point is then calculated by subtracting this value from the normal freezing point of water, 0 ℃: 0 ℃ - 19.694 ℃ = -19.694 ℃.

Answer 2

Answer : -19.6

Explanation : We can calculate this using the combine equation of boiling point and depression in freezing point and obtain a combined equation which is

∆T(f) = ∆T(b) x {K(f) / K(b)} ;

Where ∆T(f) is depression in freezing point,

∆T(b) is elevation in boiling point,

K(f) is cryoscopic constant for water it is 1.86

K(b) is ebullioscopic constant for water it is 0.512

So we can solve the equation,

∆T(f) = (105.4 -100) X
`(1.86)/(0.512)` = 19.6

So when we have the freezing point depression as 19.6,

we know water freezes at 0° C so the freezing point will be 0 - 19.6 = -19.6 °C