an aqueous solution of calcium chloride boils at 101.3 degrees Celsius. How many kilograms of calcium chloride were dissolved in 1000.0 g of the solvent??

Answer 1

We need to determine the boiling point elevation due to the presence of solutes in the solution, and then use this to find the molality of the solution, which will then allow us to find the mass of solute.

The boiling point elevation formula is:
ΔTb = Kb * molality, where ΔTb is the boiling point elevation, Kb is the ebullioscopic constant of the solvent, and molality (m) is the molality of the solution.

For water, Kb = 0.512 K kg/mol.

We are given that an aqueous solution of calcium chloride boils at 101.3°C. The boiling point of pure water is 100°C. Therefore, the boiling point elevation ΔTb due to the presence of solutes is (101.3 - 100) °C = 1.3°C.

Now we can find the molality of the solution:
1.3°C = 0.512 K kg/mol * molality
molality = 1.3 / 0.512 = 2.5390625 mol/kg

Now we need to find the mass of the calcium chloride solute. Calcium chloride has a molar mass of approximately 110.98 g/mol.

mass of calcium chloride = molality * 1000.0 g solvent * Molar mass CaCl2
mass of calcium chloride = 2.5390625 mol/kg * 1.0 kg * 110.98 g/mol = 281.72 g

Therefore, 281.72 g (or 0.2817 kg) of calcium chloride were dissolved in 1000.0 g of the solvent.