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What is the normal boiling point of 1.21 m solution of CaI2 that has a density of 1.92 g/ml? ( kb for H2O =0.512 °c .kg/mole)
Accepted Answer
To determine the normal boiling point of a 1.21 M CaI2 solution, we need to calculate the boiling point elevation. Here's how:
1. Calculate the molality (m) of the solution:
- Assume 1 L of solution.
- Mass of solution = volume x density = 1 L x 1.92 g/mL = 1920 g
- Mass of solute (CaI2) = moles x molar mass = 1.21 mol x 293.89 g/mol = 355.3 g
- Mass of solvent (H2O) = mass of solution - mass of solute = 1920 g - 355.3 g = 1564.7 g = 1.5647 kg
- Molality (m) = moles of solute / mass of solvent (in kg) = 1.21 mol / 1.5647 kg = 0.774 mol/kg
2. Calculate the boiling point elevation (ΔTb):
- ΔTb = Kb x m = 0.512 °C.kg/mol x 0.774 mol/kg = 0.396 °C
3. Determine the normal boiling point:
- The normal boiling point of water is 100 °C.
- The boiling point of the solution = normal boiling point + ΔTb = 100 °C + 0.396 °C = 100.396 °C
Therefore, the normal boiling point of the 1.21 M CaI2 solution is approximately 100.396 °C.
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