a student dissolves 1.0mole of sucrose in 1000 grams of water at 1.0atm. Compared to the boiling point of pure water , the boiling point of the solution is?
dT - Kb*m
m = mols/kg solvent
Look up Kb for water, Substitute m and solve for dT. Add dT to 100 to find the new boiling point of the sugar solution.
To determine the boiling point of the solution, we need to consider the boiling point elevation caused by the presence of the solute (sucrose).
The boiling point elevation (∆Tb) can be calculated using the formula:
ΔTb = Kb * m
Where:
ΔTb = boiling point elevation
Kb = molal boiling point elevation constant
m = molality of the solute
The molality (m) of the solution can be calculated by dividing the moles of solute (sucrose) by the mass of the solvent (water) in kilograms.
m = moles of solute / mass of solvent (in kg)
Given:
Moles of sucrose (n) = 1.0 mole
Mass of water (mH2O) = 1000 grams
To convert grams to kilograms:
mass of water (mH2O) = 1000 grams / 1000 = 1 kg
Now, calculate the molality (m):
m = n / mH2O
m = 1.0 mole / 1 kg
m = 1.0 m
Next, we need to find the molal boiling point elevation constant (Kb) for water. The Kb value for water is approximately 0.512 °C/m.
Now, we can calculate the boiling point elevation (∆Tb):
ΔTb = Kb * m
ΔTb = 0.512 °C/m * 1.0 m
ΔTb = 0.512 °C
Therefore, the boiling point of the solution will be elevated by 0.512 °C compared to the boiling point of pure water at 1.0 atm.
To determine the boiling point of the solution, we need to use the concept of boiling point elevation.
The boiling point of a solution differs from that of a pure solvent because the solute particles (in this case, sucrose) interfere with the solvent particles (water) and make it more difficult for them to escape the liquid phase and enter the gas phase.
The boiling point elevation is given by the equation:
ΔTb = Kb × m
where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point constant specific to the solvent (water),
m is the molality of the solution.
In this particular case, we need to calculate the boiling point elevation of the sucrose solution in comparison to pure water.
Let's assume the molality (m) of the solution is "x" mol/kg. Since we dissolved 1.0 mole of sucrose in 1000 grams of water, the molality can be calculated as follows:
m = moles of solute / mass of solvent (in kg)
m = 1.0 mol / 1.0 kg
Thus, the molality of the solution is 1.0 mol/kg.
The molal boiling point constant (Kb) for water is approximately 0.512 °C/m.
Now let's calculate the boiling point elevation (ΔTb):
ΔTb = Kb × m
ΔTb = 0.512 °C/m × 1.0 mol/kg
Therefore, the boiling point elevation of the solution is 0.512 °C.
To find the boiling point of the solution, we need to add the boiling point elevation to the boiling point of pure water.
The boiling point of pure water at 1.0 atm is 100 °C.
Thus, the boiling point of the solution would be:
Boiling point of solution = Boiling point of pure water + ΔTb
Boiling point of solution = 100 °C + 0.512 °C
Hence, the boiling point of the solution would be approximately 100.512 °C.