Science, Tech, Math › Science Raoult's Law Example Problem - Vapor Pressure and Strong Electrolyte Share Flipboard Email Print A strong electrolyte is one that dissociates completely in water, such as a strong acid or base or a salt. Jacobs Stock Photography Ltd / Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Todd Helmenstine Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. our editorial process Todd Helmenstine Updated January 30, 2019 This example problem demonstrates how to use Raoult's Law to calculate the change in vapor pressure by adding a strong electrolyte to a solvent. Raoult's Law relates the vapor pressure of a solution on the mole fraction of the solute added to a chemical solution. Vapor Pressure Problem What is the change in vapor pressure when 52.9 g of CuCl2 is added to 800 mL of H2O at 52.0 °C.The vapor pressure of pure H 2O at 52.0 °C is 102.1 torrThe density of H2O at 52.0 °C is 0.987 g/mL. Solution Using Raoult's Law Raoult's Law can be used to express the vapor pressure relationships of solutions containing both volatile and nonvolatile solvents. Raoult's Law is expressed byPsolution = ΧsolventP0solvent wherePsolution is the vapor pressure of the solutionΧsolvent is mole fraction of the solventP0solvent is the vapor pressure of the pure solvent Step 1 Determine the mole fraction of solutionCuCl2 is a strong electrolyte. It will completely dissociate into ions in water by the reaction:CuCl2(s) → Cu2+(aq) + 2 Cl-This means we will have 3 moles of solute added for every mole of CuCl2 added.From the periodic table:Cu = 63.55 g/molCl = 35.45 g/molmolar weight of CuCl2 = 63.55 + 2(35.45) g/molmolar weight of CuCl2 = 63.55 + 70.9 g/molmolar weight of CuCl2 = 134.45 g/molmoles of CuCl2 = 52.9 g x 1 mol/134.45 gmoles of CuCl2 = 0.39 molTotal moles of solute = 3 x (0.39 mol)Total moles of solute = 1.18 molmolar weightwater = 2(1)+16 g/molmolar weightwater = 18 g/moldensitywater = masswater/volumewatermasswater = densitywater x volumewatermasswater = 0.987 g/mL x 800 mLmasswater = 789.6 gmoleswater = 789.6 g x 1 mol/18 gmoleswater = 43.87 molΧsolution = nwater/(nwater + nsolute)Χsolution = 43.87/(43.87 + 1.18)Χsolution = 43.87/45.08Χsolution = 0.97 Step 2 Find the vapor pressure of the solutionPsolution = ΧsolventP0solventPsolution = 0.97 x 102.1 torrPsolution = 99.0 torr Step 3 Find the change in vapor pressureChange in pressure is Pfinal - POChange = 99.0 torr - 102.1 torrchange = -3.1 torr Answer The vapor pressure of the water is reduced by 3.1 torr with the addition of the CuCl2.