An aqueous sugar solution is formed by adding 50.1 g sucrose (C12H22O11) to 200.0 g water. What is the boiling point (in Celsius) of this sugar solution (this is NOT the ΔTb).
Kb H2O = 0.51 Celsius/m
moles sucrose = grams/molar mass
molality = moles/kg solvent
Solve for molality = m
delta T = Kb*m
Solve for delta T, then add that to the normal boiling point.
To find the boiling point of the sugar solution, we need to use the formula:
ΔTb = Kb * m
Where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point constant of water, and
m is the molality of the solution.
To calculate the molality (m) of the solution, we need to determine the moles of solute (sucrose) and the mass of the solvent (water).
1. Calculate the moles of sucrose (C12H22O11):
- Given mass of sucrose = 50.1 g
- Calculate the molar mass of C12H22O11:
12.01 g/mol (C) + 1.01 g/mol (H) + 16.00 g/mol (O)
= 342.34 g/mol
- Moles of sucrose = mass / molar mass
= 50.1 g / 342.34 g/mol
≈ 0.1462 mol
2. Calculate the molality (m) of the solution:
- Given mass of water = 200.0 g
- Moles of water = mass / molar mass of water
= 200.0 g / 18.01 g/mol
≈ 11.105 mol
- Molality (m) = moles of solute / mass of solvent (in kg)
= 0.1462 mol / 0.200 kg
= 0.731 mol/kg
3. Calculate the boiling point elevation (ΔTb):
- Given Kb (molal boiling point constant of water) = 0.51 Celsius/m
- ΔTb = Kb * m
= 0.51 Celsius/m * 0.731 mol/kg
≈ 0.373 Celsius
4. Finally, add the boiling point elevation to the boiling point of pure water (100 degrees Celsius) to get the boiling point of the sugar solution:
- Boiling point = 100 degrees Celsius + ΔTb
= 100 degrees Celsius + 0.373 Celsius
≈ 100.373 degrees Celsius.
Therefore, the boiling point of the sugar solution is approximately 100.373 degrees Celsius.