How much calcium chloride would you need to dissolve in 500 grams of water to create a solution whose boiling point is 100.40 C?
To determine the quantity of calcium chloride required to dissolve in water to reach a specific boiling point, we need to use the concept of boiling point elevation. The boiling point elevation is a colligative property, which means it depends solely on the number of particles dissolved in the solvent, rather than their specific identities.
To calculate the quantity of calcium chloride needed, we need to consider the molality of the solution, which is the number of moles of solute per kilogram of solvent. The formula to calculate the boiling point elevation is as follows:
ΔTb = Kb * m
Where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point elevation constant (for water, the value is 0.512 ˚C/m), and
m is the molality of the solution.
In this case, the boiling point elevation is 0.40 ˚C (since the new boiling point is 100.40 ˚C), Kb is 0.512 ˚C/m, and we need to solve for m.
m = ΔTb / Kb
m = 0.40 ˚C / 0.512 ˚C/m
m = 0.78125 m
Now, we need to convert this molality value to moles of calcium chloride per kilogram of water (solvent). The molar mass of calcium chloride (CaCl2) is approximately 111 g/mol.
moles of CaCl2 = m * molar mass
moles of CaCl2 = 0.78125 m * 111 g/mol
moles of CaCl2 = 86.71875 mol
Since we are dealing with 500 grams of water, which is equivalent to 0.5 kg, we can determine the amount of calcium chloride needed.
mass of CaCl2 = moles of CaCl2 * molar mass
mass of CaCl2 = 86.71875 mol * 111 g/mol
mass of CaCl2 = 9630.859 g
Therefore, you would need approximately 9630.859 grams of calcium chloride to dissolve in 500 grams of water to create a solution with a boiling point of 100.40 ˚C.