What mass of sucrose, C12H22O11, should be added to 75.0 g H2O to raise the boiling point to 100.35 degrees C?
Molal boiling point of water:
0.512 Kb, degrees C m^-1
Answer: 18 g sucrose
delta T = Kb*m
Solve for m
m = mols/kg solvent\
Solve for mol
mols = grams/molar mass
Solve for grams.;
To calculate the mass of sucrose needed to raise the boiling point of water, we can use the equation:
ΔTb = Kb * m
where ΔTb is the change in boiling point, Kb is the molal boiling point constant for water, and m is the molality of the solution.
First, we need to calculate the change in boiling point (ΔTb). The question states that the boiling point needs to be raised to 100.35 degrees C, which means ΔTb = 100.35 - 100 = 0.35 degrees C.
Next, we need to calculate the molality (m) of the solution. Molality is defined as the moles of solute per kilogram of solvent. In this case, the solute is sucrose and the solvent is water.
To calculate the molality, we need to convert the given mass of water (75.0 g) to kilograms:
mass of water = 75.0 g = 0.075 kg
Now, we need to calculate the moles of sucrose. To do this, we need the molar mass of sucrose, which is:
molar mass of sucrose = 12 * 12.01 g/mol + 22 * 1.01 g/mol + 11 * 16.00 g/mol = 342.34 g/mol
Next, we can calculate the moles of sucrose using the formula:
moles of sucrose = mass of sucrose / molar mass of sucrose
Since we want to solve for the mass of sucrose, we can rearrange the formula as:
mass of sucrose = moles of sucrose * molar mass of sucrose
Finally, we can substitute the values into the formula:
mass of sucrose = moles of sucrose * molar mass of sucrose = m * (ΔTb / Kb) * molar mass of sucrose
mass of sucrose = (0.075 kg) * (0.35 degrees C / 0.512 degrees C m^-1) * (342.34 g/mol)
Simplifying the equation gives:
mass of sucrose ≈ 18 g
Therefore, approximately 18 grams of sucrose should be added to 75.0 grams of water to raise the boiling point to 100.35 degrees C.