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Answer :
The boiling point of the solution is 100.6213 ∘C.
To calculate the freezing and boiling points of the solution, we need to use the formula:
ΔTf = Kf * molality
ΔTb = Kb * molality
where:
ΔTf is the change in freezing point
Kf is the freezing point depression constant for the solvent
molality is the molal concentration of the solute in the solvent
ΔTb is the change in boiling point
Kb is the boiling point elevation constant for the solvent
First, let's find the molality of the solution.
Molar mass of the compound = 62 grams
Mass of the compound = 186 grams
Number of moles of the compound = mass / molar mass = 186 / 62 = 3 moles
Mass of water = 2.505 kg = 2505 grams
Molality (m) = moles of solute / mass of solvent in kg
= 3 / 2.505 = 1.197 mol/kg
Now we can calculate the freezing point depression and boiling point elevation.
ΔTf = Kf * molality
= 1.86 ∘C/m * 1.197 mol/kg
= 2.2282 ∘C
ΔTb = Kb * molality
= 0.52 ∘C/m * 1.197 mol/kg
= 0.6213 ∘C
To find the freezing point of the solution, we subtract the change in freezing point from the freezing point of the pure solvent (water).
Freezing point of water = 0 ∘C
Freezing point of the solution = Freezing point of water - ΔTf
= 0 ∘C - 2.2282 ∘C
= -2.2282 ∘C
So, the freezing point of the solution is -2.2282 ∘C.
To find the boiling point of the solution, we add the change in boiling point to the boiling point of the pure solvent (water).
Boiling point of water = 100 ∘C
Boiling point of the solution = Boiling point of water + ΔTb
= 100 ∘C + 0.6213 ∘C
= 100.6213 ∘C
So, the boiling point of the solution is 100.6213 ∘C.
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