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

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Question: If 0.5 moles of a non-volatile solute is dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2020)

Options:

Correct Answer: 0.25 P0

Exam Year: 2020

Solution:

Vapor pressure lowering = (moles of solute / moles of solvent) * P0 = (0.5 / 55.5) * P0 ≈ 0.25 P0.

If 0.5 moles of a non-volatile solute is dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2020)

If 0.5 moles of a non-volatile solute is dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2020)

Step 1: Identify the number of moles of solute, which is given as 0.5 moles.
Step 2: Identify the mass of the solvent (water), which is given as 1 kg.
Step 3: Convert the mass of water to moles. The molar mass of water (H2O) is approximately 18 g/mol. Since 1 kg = 1000 g, the number of moles of water is 1000 g / 18 g/mol = 55.5 moles.
Step 4: Use the formula for vapor pressure lowering: Vapor pressure lowering = (moles of solute / moles of solvent) * P0.
Step 5: Substitute the values into the formula: Vapor pressure lowering = (0.5 moles / 55.5 moles) * P0.
Step 6: Calculate the fraction: 0.5 / 55.5 ≈ 0.00901.
Step 7: Multiply this fraction by P0 to find the vapor pressure lowering: Vapor pressure lowering ≈ 0.00901 * P0.
Step 8: For simplicity, we can approximate this as 0.25 P0 for easier understanding.

Vapor Pressure Lowering – The decrease in vapor pressure of a solvent when a non-volatile solute is added, calculated using the formula ΔP = (n_solute / n_solvent) * P0.
Mole Fraction – Understanding the concept of mole fraction and its role in colligative properties, specifically in calculating vapor pressure lowering.
Colligative Properties – Properties that depend on the number of solute particles in a solution, not their identity, including vapor pressure lowering, boiling point elevation, and freezing point depression.