Boiling point of water at 750mmHg is 96.63°C. How much sucrose is added to 500g of water such that it boils at 100°C? Molal elevation constant (Kb) of H₂O is 0.52 K·kg·mol⁻¹.

To determine how much sucrose needs to be added to 500g of water to raise its boiling point to 100°C, we can use the formula for boiling point elevation:

Boiling Point Elevation Formula

The formula is given by:

ΔT_b = K_b × m

Where:

• ΔT_b = change in boiling point
• K_b = molal elevation constant
• m = molality of the solution

Step 1: Calculate the Change in Boiling Point

Since the boiling point of pure water at 750 mmHg is 96.63°C, the change in boiling point needed to reach 100°C is:

ΔT_b = 100°C - 96.63°C = 3.37°C

Step 2: Calculate the Molality Required

Using the boiling point elevation constant:

3.37°C = 0.52 K·kg·mol⁻¹ × m

Rearranging gives:

m = 3.37°C / 0.52 K·kg·mol⁻¹ ≈ 6.47 mol/kg

Step 3: Convert Mass of Water to kg

Since we have 500g of water, we convert this to kilograms:

500g = 0.5 kg

Step 4: Calculate Moles of Sucrose Needed

Now, we can find the number of moles of sucrose needed:

m = moles of solute / kg of solvent

Rearranging gives:

moles of sucrose = m × kg of solvent = 6.47 mol/kg × 0.5 kg = 3.235 moles

Step 5: Calculate Mass of Sucrose

The molar mass of sucrose (C₁₂H₂₂O₁₁) is approximately 342 g/mol. Therefore, the mass of sucrose needed is:

mass = moles × molar mass = 3.235 moles × 342 g/mol ≈ 1106.07 g

Final Result

To raise the boiling point of 500g of water to 100°C, you would need to add approximately 1106.07 grams of sucrose.