Assuming 100% dissociation, calculate the freezing point ( 𝑇f ) and boiling point ( 𝑇b ) of 2.25 𝑚 AgNO3(aq) .

Answer 1

To calculate the freezing and boiling points of a solution, we need to use the following equations:

Δ𝑇f = 𝑘𝑓 × 𝑚 Δ𝑇b = 𝑘𝑏 × 𝑚

where Δ𝑇f is the change in freezing point, Δ𝑇b is the change in boiling point, 𝑘𝑓 and 𝑘𝑏 are the cryoscopic and ebullioscopic constants of the solvent, and 𝑚 is the molality of the solute in the solution.

For water, the cryoscopic constant is 1.86 °C/m and the ebullioscopic constant is 0.51 °C/m.

To calculate 𝑚, we need to first calculate the number of moles of AgNO3 in the solution. Since we have a 2.25 𝑚 solution, we can assume that we have 2.25 moles of AgNO3 in 1 kg of water. Therefore, the molality of the solution is:

𝑚 = moles of solute / mass of solvent (in kg) 𝑚 = 2.25 mol / 1 kg = 2.25 m

Using these values, we can now calculate the freezing point depression and boiling point elevation of the solution.

Freezing point depression:

Δ𝑇f = 𝑘𝑓 × 𝑚 Δ𝑇f = 1.86 °C/m × 2.25 m = 4.185 °C

The freezing point of pure water is 0°C, so the freezing point of the solution will be:

𝑇f = 0°C - 4.185°C = -4.185°C

Boiling point elevation:

Δ𝑇b = 𝑘𝑏 × 𝑚 Δ𝑇b = 0.51 °C/m × 2.25 m = 1.1475 °C

The boiling point of pure water is 100°C, so the boiling point of the solution will be:

𝑇b = 100°C + 1.1475°C = 101.1475°C

Therefore, the freezing point of a 2.25 𝑚 AgNO3(aq) solution is -4.185°C and the boiling point is 101.1475°C, assuming 100% dissociation.