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Grade 11Physical Chemistry

Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

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12 Years agoGrade 11
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1 Answer

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ApprovedApproved Tutor Answer1 Year ago

To determine how much sucrose needs to be added to 500 g of water to raise its boiling point to 100°C, we can use the concept of boiling point elevation. This phenomenon occurs when a non-volatile solute, like sucrose, is dissolved in a solvent, such as water. The boiling point elevation can be calculated using the formula:

Boiling Point Elevation Formula

The formula for boiling point elevation is:

ΔT_b = i * K_b * m

  • ΔT_b = change in boiling point (°C)
  • i = van 't Hoff factor (for sucrose, i = 1 since it does not dissociate)
  • K_b = ebullioscopic constant of the solvent (for water, K_b = 0.512 °C kg/mol)
  • m = molality of the solution (mol of solute/kg of solvent)

Step-by-Step Calculation

1. **Determine the change in boiling point (ΔT_b)**: Since we want the boiling point to reach 100°C and it currently boils at 99.63°C, the change is:

ΔT_b = 100°C - 99.63°C = 0.37°C

2. **Substitute known values into the formula**: We know that for sucrose, the van 't Hoff factor (i) is 1, and the ebullioscopic constant (K_b) for water is 0.512 °C kg/mol. Plugging these values into the equation gives:

0.37°C = 1 * 0.512 °C kg/mol * m

3. **Solve for molality (m)**:

m = 0.37°C / 0.512 °C kg/mol ≈ 0.722 mol/kg

4. **Calculate the number of moles of sucrose needed**: Since molality (m) is defined as moles of solute per kilogram of solvent, we can find the number of moles needed for 500 g (0.5 kg) of water:

moles of sucrose = m * kg of solvent = 0.722 mol/kg * 0.5 kg = 0.361 mol

5. **Convert moles of sucrose to grams**: The molar mass of sucrose (C12H22O11) is approximately 342 g/mol. Therefore, the mass of sucrose needed is:

mass of sucrose = moles * molar mass = 0.361 mol * 342 g/mol ≈ 123.5 g

Final Result

To achieve a boiling point of 100°C in 500 g of water, you would need to add approximately 123.5 grams of sucrose.

This calculation illustrates how colligative properties, such as boiling point elevation, depend on the number of solute particles in a solution rather than their identity. By understanding these principles, we can manipulate boiling points for various applications in chemistry and cooking.