The boiling point of water at 750 mm Hg is 99.63°C. How much sucrose should be added to 500 g of water to raise its boiling point to 100°C?

To raise the boiling point of water from 99.63°C to 100°C at 750 mm Hg, you will need to add a solute (in this case, sucrose) to the water. The change in boiling point is determined by the elevation of the boiling point due to the presence of the solute. You can use the formula for elevation of boiling point:

ΔT_b = K_b * m

Where:

ΔT_b is the change in boiling point.
K_b is the ebullioscopic constant (a property of the solvent).
m is the molality of the solution.
First, you need to calculate the change in boiling point (ΔT_b) required to raise the boiling point from 99.63°C to 100°C:

ΔT_b = 100°C - 99.63°C = 0.37°C

Now, you need to find the ebullioscopic constant (K_b) for water. The ebullioscopic constant for water is approximately 0.512°C·kg/mol.

Next, calculate the molality (m) of the solution. Molality is defined as the moles of solute per kilogram of solvent:

m = (moles of solute) / (mass of solvent in kg)

You want to find out how much sucrose to add to 500 g (0.5 kg) of water to achieve the desired boiling point elevation. First, you need to calculate the moles of water:

Moles of water = (mass of water) / (molar mass of water)
Moles of water = 0.5 kg / 18.015 g/mol ≈ 27.75 moles
Now, you can use the ΔT_b formula to find the moles of sucrose required:

0.37°C = 0.512°C·kg/mol * m

m = 0.37°C / 0.512°C·kg/mol ≈ 0.723 kg/mol

Now, you need to convert the molality (m) into moles of sucrose (C12H22O11) per kilogram of water. The molar mass of sucrose is approximately 342.3 g/mol.

Moles of sucrose per kilogram of water = (0.723 kg/mol) * (1000 g/kg) / 342.3 g/mol ≈ 2.12 mol/kg
So, you need to add approximately 2.12 moles of sucrose to 500 g of water to raise its boiling point from 99.63°C to 100°C at 750 mm Hg.