If the vapor pressure of pure water is 23.8 mmHg at 25°C, what is the vapor pressure of a solution containing 1 mole of NaCl in 1 kg of water?

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Q: If the vapor pressure of pure water is 23.8 mmHg at 25°C, what is the vapor pressure of a solution containing 1 mole of NaCl in 1 kg of water?

Solution: Using Raoult's law, the vapor pressure lowering is calculated considering the van 't Hoff factor for NaCl.

Steps: 10

Step 1: Understand that vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature.
Step 2: Know that pure water has a vapor pressure of 23.8 mmHg at 25°C.
Step 3: Recognize that when a solute (like NaCl) is added to a solvent (like water), the vapor pressure of the solution decreases.
Step 4: Use Raoult's law, which states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
Step 5: Calculate the number of moles of water in 1 kg. The molar mass of water is about 18 g/mol, so 1 kg (1000 g) of water is approximately 55.56 moles.
Step 6: Determine the van 't Hoff factor (i) for NaCl, which is 2 because NaCl dissociates into two ions: Na+ and Cl-.
Step 7: Calculate the total number of moles in the solution: 1 mole of NaCl + 55.56 moles of water = 56.56 moles total.
Step 8: Calculate the mole fraction of water in the solution: mole fraction of water = moles of water / total moles = 55.56 / 56.56.
Step 9: Calculate the vapor pressure of the solution using Raoult's law: vapor pressure of solution = vapor pressure of pure water * mole fraction of water.
Step 10: Substitute the values to find the new vapor pressure of the solution.