Question

When 2.0 moles of KCl is dissolved in 1.0 L of water, the boiling point of the solution will be ___ than the boiling point of pure water.

Answer 1

When a solute, such as KCl, is dissolved in a solvent like water, it causes the boiling point of the solution to increase compared to the boiling point of the pure solvent. This phenomenon is known as boiling point elevation.

To calculate the boiling point elevation, we can use the formula:

ΔTb = i * K * molality

Where: ΔTb = Boiling point elevation i = Van't Hoff factor (the number of particles the solute dissociates into in the solution) K = Boiling point elevation constant for the solvent molality = moles of solute / mass of solvent (in kg)

For KCl dissolved in water, the Van't Hoff factor (i) is 2 because KCl dissociates into two ions (K+ and Cl-) in the water.

The boiling point elevation constant for water is approximately 0.512 °C/m.

Now, let's calculate the boiling point elevation:

molality = moles of solute / mass of solvent (in kg) molality = 2.0 moles / 1.0 kg = 2.0 mol/kg

ΔTb = 2 * 0.512 °C/m * 2.0 mol/kg ΔTb = 2.048 °C

The boiling point of the solution will be 2.048°C higher than the boiling point of pure water.