What is the vapor pressure of a solution containing 0.1 mol of a non-volatile solute in 1 kg of water? (Vapor pressure of pure water = 23.76 mmHg)

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Q: What is the vapor pressure of a solution containing 0.1 mol of a non-volatile solute in 1 kg of water? (Vapor pressure of pure water = 23.76 mmHg)

Solution: Vapor pressure = P°_solvent * (n_solvent / (n_solvent + n_solute)) = 23.76 * (55.5 / (55.5 + 0.1)) = 22.88 mmHg

Steps: 9

Step 1: Identify the vapor pressure of pure water, which is given as 23.76 mmHg.
Step 2: Determine the number of moles of the solvent (water). Since 1 kg of water is approximately 55.5 moles (using the molar mass of water, which is about 18 g/mol).
Step 3: Identify the number of moles of the solute, which is given as 0.1 mol.
Step 4: Use the formula for the vapor pressure of the solution: Vapor pressure = P°_solvent * (n_solvent / (n_solvent + n_solute)).
Step 5: Substitute the values into the formula: Vapor pressure = 23.76 mmHg * (55.5 / (55.5 + 0.1)).
Step 6: Calculate the denominator: 55.5 + 0.1 = 55.6.
Step 7: Calculate the fraction: 55.5 / 55.6.
Step 8: Multiply the result by 23.76 mmHg to find the vapor pressure of the solution.
Step 9: The final result is approximately 22.88 mmHg.