What mass of sucrose should be added to 75 gram of water to raise the boiling point about 100.35°C.What mass of sucrose is required to produce the solution of this molality?
To calculate the mass of sucrose needed to raise the boiling point of water by a certain amount and to determine the required molality, we need to use the concept of molality and the formula for the boiling point elevation.
1. Boiling Point Elevation Formula:
The formula to calculate the boiling point elevation (∆Tb) is:
∆Tb = Kb * m
where:
∆Tb is the boiling point elevation (in degrees Celsius),
Kb is the molal boiling point elevation constant (unique for each solvent),
m is the molality of the solution (moles of solute per kilogram of solvent).
2. Solving for Molality:
To find the molality (m), we rearrange the formula as follows:
m = ∆Tb / Kb
3. Finding the Molal Boiling Point Elevation Constant:
For sucrose in water, the molal boiling point elevation constant (Kb) is approximately 0.512 °C/m.
4. Solving for Molality:
Substituting the values into the formula, we have:
m = 100.35 °C / 0.512 °C/m
m ≈ 196.08 m
So, the molality of the solution is approximately 196.08 m.
5. Calculating the Mass of Sucrose:
To calculate the mass of sucrose needed, we need to use the formula:
m = (moles of solute) / (kg of solvent)
First, we need to convert the mass of water into kilograms:
75 grams of water = 75 / 1000 = 0.075 kg of water
Next, we can calculate the moles of sucrose needed using the molality formula:
moles of solute = molality * kg of solvent
moles of solute = 196.08 * 0.075
moles of solute ≈ 14.71 mol
Finally, we need to convert the moles of sucrose to grams:
mass of sucrose = moles of solute * molar mass of sucrose
The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol.
mass of sucrose = 14.71 * 342.3
mass of sucrose ≈ 5034.84 g
Therefore, the mass of sucrose required to produce the desired solution with a molality of approximately 196.08 m is approximately 5034.84 grams.