Answer to Question #302006 in Chemistry for mcc

Question

Assuming Raoult’s law applies, calculate the vapor pressure of a solution of 45 g of glucose (C₆H₁₂O₆) in 95 g of water at 25 degrees Celsius

Answer 1

Solution:

Сalculate the moles of each component of the solution:

The molar mass of glucose (C₆H₁₂O₆) is 180 g/mol

The molar mass of water (H₂O) is 18.015 g/mol

Therefore,

Moles of C₆H₁₂O₆ = (45 g C₆H₁₂O₆) × (1 mol C₆H₁₂O₆ / 180 g C₆H₁₂O₆) = 0.25 mol C₆H₁₂O₆

Moles of H₂O = (95 g H₂O) × (1 mol H₂O / 18.015 g H₂O) = 5.27 mol H₂O

Сalculate the mole fraction of each component of the solution:

Mole fraction (χ) = Moles / Total moles

Total moles of solution = Moles of C₆H₁₂O₆ + Moles of H₂O = 0.25 mol + 5.27 mol = 5.52 mol

Therefore,

χ_C6H12O6 = Moles of C₆H₁₂O₆ / Total moles of solution = (0.25 mol) / (5.52 mol) = 0.0453

χ_H2O = Moles of H₂O / Total moles of solution = (5.27 mol) / (5.52 mol) = 0.9547

Raoult’s law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present:

P_solution = χ_solvent × P_solvent

The vapor pressure of water at room temperature (25°C) is 23.8 mmHg.

Therefore,

P_solution = 0.9547 × 23.8 mmHg = 22.72 mmHg

P_solution = 22.72 mmHg

Answer: The vapor pressure of a solution is 22.72 mmHg