What is the boiling point of a solution of 20.0g carbon tetra chloride (CCl4) dissolved in 250.0g benzene (C6H6)? The boiling point of benzene is 80.10C, and it’s boiling point elevation constant is 2.53C/m. Assume solute is a nonelectrolyte. What is the morality of this solution?
To find the boiling point elevation of the solution, we first need to calculate the molality of the solution:
1. Calculate the molality (m) of the solution:
molality (m) = moles of solute / kg of solvent
a. Calculate the moles of CCl4:
molar mass of CCl4 = 12.01 g/mol (C) + 4(35.45 g/mol) (Cl) = 153.82 g/mol
moles of CCl4 = 20.0 g / 153.82 g/mol = 0.130 mol
b. Calculate the kg of benzene:
kg of benzene = 250.0 g / 1000 = 0.250 kg
c. Calculate the molality:
molality = 0.130 mol / 0.250 kg = 0.52 m
2. Calculate the boiling point elevation (ΔTb) of the solution:
ΔTb = iKb * m
Given:
Kb = 2.53 ̊C/m
i = 1 (since CCl4 is a non-electrolyte)
ΔTb = 1 * 2.53 ̊C/m * 0.52 m = 1.31 ̊C
3. The boiling point of the solution:
Boiling point of the solution = Boiling point of benzene + ΔTb
Boiling point of the solution = 80.1 ̊C + 1.31 ̊C = 81.41 ̊C
Therefore, the boiling point of the solution is 81.41 ̊C.
4. To find the morality of this solution, we need to calculate the moles of benzene present:
molar mass of C6H6 = 12.01 (C) + 6(1.01)(H) = 78.11 g/mol
moles of benzene = 250.0 g / 78.11 g/mol = 3.20 mol
Since benzene does not dissociate in solution, the morality of this solution is the same as the molality, which is 0.52 m.