At 25 $\degree$C, the vapor pressure of pure water is 25.756 mmHg.What is the vapor pressure of water in a solution that contains 2.922 g of sodium chloride for every 100.0 g of water?

Question

Quizplus · Accepted Answer

DTo find the vapor pressure of water in a solution, we use Raoult's law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. Sodium chloride (NaCl) dissociates into two ions (Na+ and Cl-) in solution, effectively doubling the number of particles compared to the initial moles of NaCl added.First, calculate the moles of NaCl:Molar mass of NaCl = 58.44 g/molMoles of NaCl = 2.922 g / 58.44 g/mol = 0.05 molesSince NaCl dissociates into 2 ions, the total moles of particles = 0.05 moles * 2 = 0.1 molesNext, calculate the moles of water:Molar mass of water = 18.015 g/molMoles of water = 100.0 g / 18.015 g/mol ≈ 5.55 molesTotal moles in solution = moles of water + moles of particles from NaCl = 5.55 + 0.1 = 5.65 molesMole fraction of water = moles of water / total moles = 5.55 / 5.65 ≈ 0.9823Vapor pressure of water in the solution = vapor pressure of pure water * mole fraction of water = 25.756 mmHg * 0.9823 ≈ 25.3 mmHgThe answers closest to this calculation are C) 25.52 mmHg and D) 25.35 mmHg, which account for the slight decrease in vapor pressure due to the presence of NaCl in the solution.

At 25 °\degree° C, the Vapor Pressure of Pure Water Is 25

At 25 $\degree$ C, the Vapor Pressure of Pure Water Is 25