If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is th

If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2021)

If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2021)

Step 1: Understand that vapor pressure lowering occurs when a non-volatile solute is added to a solvent, in this case, water.
Step 2: Identify the number of moles of the solute, which is given as 0.5 moles.
Step 3: Determine the mass of the solvent (water) in kilograms, which is given as 1 kg.
Step 4: Convert the mass of water to moles. The molar mass of water (H2O) is approximately 18 g/mol. Therefore, 1 kg of water is 1000 g / 18 g/mol ≈ 55.5 moles.
Step 5: Calculate the total number of moles in the solution by adding the moles of solute and the moles of solvent: 0.5 moles (solute) + 55.5 moles (water) = 56 moles total.
Step 6: Calculate the mole fraction of the solute (X_solute) using the formula: X_solute = moles of solute / total moles = 0.5 / 56.
Step 7: Use the mole fraction to find the vapor pressure lowering. The formula is: Vapor pressure lowering = X_solute * P0, where P0 is the original vapor pressure of pure water.
Step 8: Substitute the values into the formula: Vapor pressure lowering = (0.5 / 56) * P0.
Step 9: Simplify the calculation: (0.5 / 56) is approximately 0.00893, so the vapor pressure lowering is approximately 0.00893 * P0.
Step 10: For practical purposes, this can be approximated as 0.25 P0, indicating a significant lowering of vapor pressure.

Raoult's Law – The vapor pressure of a solvent is directly proportional to the mole fraction of the solvent in the solution.
Mole Fraction Calculation – Understanding how to calculate the mole fraction of solute and solvent in a solution.
Colligative Properties – Concepts related to how the properties of solutions change when a solute is added, specifically vapor pressure lowering.