What would the boiling point be of 300 ml with 40 g CuCl2 dissolved in it?
To determine the boiling point of a solution, we need to consider the nature of the solute and solvent. In this case, the solute is CuCl2 (copper chloride), and the solvent is water.
To calculate the boiling point elevation, we can use the equation:
ΔTb = Kb * m
Where:
ΔTb = boiling point elevation
Kb = molal boiling point elevation constant (dependent on the solvent)
m = molality of the solution (moles of solute divided by mass of the solvent in kg)
First, we need to calculate the molality (m) of the solution:
m = moles of solute / mass of the solvent in kg
The molar mass of CuCl2 is 63.55 g/mol (atomic mass of copper = 63.55 g/mol, and two chlorine atoms at 35.45 g/mol each).
moles of CuCl2 = mass of CuCl2 / molar mass of CuCl2
= 40 g / 63.55 g/mol
≈ 0.630 mol
Now, we need the mass of the solvent. Since we have 300 ml (or 300 g) of solution, and the mass of the solute is 40 g, the mass of the solvent is:
mass of solvent = total mass - mass of solute
= 300 g - 40 g
= 260 g
Converting the mass of the solvent to kg:
mass of solvent in kg = 260 g / 1000
= 0.26 kg
Substituting the values into the equation:
ΔTb = Kb * m
We need the Kb value for water. The Kb value for water is 0.512 ºC/m.
ΔTb = 0.512 ºC/m * (0.630 mol / 0.26 kg)
Calculating:
ΔTb = 0.512 ºC/m * 2.423 mol/kg
ΔTb ≈ 1.24 ºC
The boiling point elevation of the solution is approximately 1.24 ºC.
To find the boiling point of the solution, we add the boiling point elevation to the boiling point of the pure solvent. The boiling point of water is approximately 100 ºC.
Boiling point of the solution = boiling point of the solvent + boiling point elevation
= 100 ºC + 1.24 ºC
≈ 101.24 ºC
Therefore, the boiling point of the 300 ml solution with 40 g of CuCl2 dissolved in it would be approximately 101.24 ºC.