Calculate the boiling-point elevation for 2.0 kg of water containing 300. g of the salt CaCl2 (Kb = 0.512 C/m)
dT = i*Kb*m i = 3 for CaCl2 since it gives three particles upon ionization Kb = 0.512 m = mol/kg solvent mols CaCl2 = 300 g/molar mass CaCl2 = ? m = ?mols CaCl2/2
ANSWER: 1.32 degrees
To calculate the boiling-point elevation, you'll need to know the molality of the solution and the molal boiling-point elevation constant.
First, let's find the molality (m). Molality is defined as the number of moles of solute (in this case, CaCl2) per kilogram of solvent (water).
Given:
Mass of water (solvent) = 2.0 kg
Mass of CaCl2 (solute) = 300. g
Convert the mass of CaCl2 to moles by dividing by its molar mass:
Molar mass of CaCl2 = 40.08 g/mol + (2 * 35.45 g/mol) = 110.98 g/mol
Number of moles of CaCl2 = Mass of CaCl2 / Molar mass of CaCl2
Next, calculate the molality, which is the moles of solute divided by the mass of the solvent (in kg):
Molality (m) = Moles of CaCl2 / Mass of water (in kg)
Now, let's calculate the molality:
Moles of CaCl2 = 300 g / 110.98 g/mol
Mass of water (in kg) = 2.0 kg
Molality (m) = (300 g / 110.98 g/mol) / 2.0 kg
Next, you need the molal boiling-point elevation constant (Kb). The value given in the question is 0.512 °C/m.
Finally, you can calculate the boiling-point elevation (∆Tb) using the formula:
∆Tb = Kb * m
∆Tb = 0.512 °C/m * Molality
Substitute the calculated molality into the equation to find the boiling-point elevation.
dT = i*Kb*m
i = 3 for CaCl2 since it gives three particles upon ionization
Kb = 0.512
m = mol/kg solvent
mols CaCl2 = 300 g/molar mass CaCl2 = ?
m = ?mols CaCl2/2
Substitute m, i, Kb and solve for the elevation.
Post your work if you get stuck.