Science, Tech, Math › Science Raoult's Law Example Problem - Volatile Mixture Calculating Vapor Pressure of Volatile Solutions Share Flipboard Email Print rclassenlayouts / 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 February 05, 2019 This example problem demonstrates how to use Raoult's Law to calculate the vapor pressure of two volatile solutions mixed together. Raoult's Law Example What is the expected vapor pressure when 58.9 g of hexane (C6H14) is mixed with 44.0 g of benzene (C6H6) at 60.0 °C?Given:Vapor pressure of pure hexane at 60 °C is 573 torr.Vapor pressure of pure benzene at 60 °C is 391 torr. Solution 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 by the vapor pressure equation:Psolution = ΧsolventP0solventwherePsolution is the vapor pressure of the solutionΧsolvent is mole fraction of the solventP0solvent is the vapor pressure of the pure solventWhen two or more volatile solutions are mixed, each pressure component of the mixed solution is added together to find the total vapor pressure.PTotal = Psolution A + Psolution B + ...Step 1 - Determine the number of moles of each solution in order to be able to calculate the mole fraction of the components.From the periodic table, the atomic masses of the carbon and hydrogen atoms in hexane and benzene are:C = 12 g/molH = 1 g/mol Use the molecular weights to find the number of moles of each component:molar weight of hexane = 6(12) + 14(1) g/molmolar weight of hexane = 72 + 14 g/molmolar weight of hexane = 86 g/molnhexane = 58.9 g x 1 mol/86 gnhexane = 0.685 molmolar weight of benzene = 6(12) + 6(1) g/molmolar weight of benzene = 72 + 6 g/molmolar weight of benzene = 78 g/molnbenzene = 44.0 g x 1 mol/78 gnbenzene = 0.564 molStep 2 - Find mole fraction of each solution. It doesn't matter which component you use to perform the calculation. In fact, a good way to check your work is to do the calculation for both hexane and benzene and then make sure they add up to 1.Χhexane = nhexane/(nhexane + nbenzene)Χhexane = 0.685/(0.685 + 0.564)Χhexane = 0.685/1.249Χhexane = 0.548Since there are only two solutions present and the total mole fraction is equal to one:Χbenzene = 1 - ΧhexaneΧbenzene = 1 - 0.548Χbenzene = 0.452Step 3 - Find the total vapor pressure by plugging the values into the equation:PTotal = ΧhexaneP0hexane + ΧbenzeneP0benzenePTotal = 0.548 x 573 torr + 0.452 x 391 torrPTotal = 314 + 177 torrPTotal = 491 torr Answer: The vapor pressure of this solution of hexane and benzene at 60 °C is 491 torr.