Question
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?
Answers
GPT 3.5
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.
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.
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