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What is the expected boiling point of a solution with 128 g of KCl (a strong electrolyte) dissolved in 1.3 kg of water?

- Molar mass of KCl: 74.55 g/mol
- Boiling point elevation constant, [tex]K_b[/tex]: 0.51°C/m

Answer :

Final answer:

The expected boiling point of the KCl solution is calculated by first determining the molality of the solution, then calculating the change in boiling point (∆Tb) using the formula ∆Tb = i * Kb * m, and finally adding this to the normal boiling point of water. The expected boiling point is approximately 101.35 °C.

Explanation:

To calculate the expected boiling point of a KCl solution, you would first find the molality (m) using the formula: m = moles of solute/kg of solvent. First, convert the 128 g of KCl into moles by dividing by the molar mass of KCl (74.55 g/mol). The result is roughly 1.72 mol.

Then, plug the moles of KCl and the kg of water into the molality formula. The result is about 1.32 m.

Next, calculate the change in boiling point (∆Tb) using the formula: ∆Tb = i * Kb * m. 'i' is the van 't Hoff factor, which represents the number of particles the solute splits into in solution. As KCl is a strong electrolyte, it fully disassociates into 2 ions in solution, namely K+ and Cl-, so 'i' is 2. 'Kb' is the molal boiling point elevation constant for water, which is given as 0.51 °C/m.

Plugging these values into the formula gives: ∆Tb = 2 * 0.51 °C/m * 1.32m, which equals approximately 1.35 °C.

Finally, add this value to the normal boiling point of water (100 °C) to get the expected boiling point of the solution: 100 °C + 1.35 °C = 101.35 °C. Therefore, the expected boiling point of the KCl solution is approximately 101.35 °C.

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Final answer:

To find the expected boiling point of a KCl solution, the boiling point elevation is calculated using the molality, the van't Hoff factor, and the solvent's ebullioscopic constant, resulting in an expected boiling point of 101.35 °C for the solution.

Explanation:

To determine the expected boiling point of a solution consisting of KCl dissolved in water, we can use the boiling point elevation formula ΔTb = i · m · Kb, where ΔTb is the change in boiling point, i is the van't Hoff factor which indicates the number of particles into which a compound dissociates in solution, m is the molality of the solution, and Kb is the ebullioscopic constant of the solvent (in this case water, which is given as 0.51 °C/m).

First, calculate the molarity of KCl:

  • Number of moles of KCl = mass (g) / molar mass (g/mol) = 128 g / 74.55 g/mol = 1.716 mol.
  • Molality (m) = moles of solute / kg of solvent = 1.716 mol / 1.3 kg = 1.32 mol/kg.

KCl completely dissociates into K+ and Cl- ions, so the van't Hoff factor (i) for KCl is 2.

Now, calculate the boiling point elevation using the formula:

  • ΔTb = i · m · Kb = 2 · 1.32 mol/kg · 0.51 °C/m = 1.3452 °C.

Since the boiling point of pure water is 100 °C, the expected boiling point of the solution will be:

  • 100 °C + 1.3452 °C = 101.3452 °C (rounded to 101.35 °C).