Science, Tech, Math › Science Boiling Point Elevation Example Problem Calculate Boiling Point Elevation Temperature Share Flipboard Email Print The boiling point temperature can be elevated by the addition of a solute to the water. David Murray and Jules Selmes / 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 December 09, 2019 This example problem demonstrates how to calculate boiling point elevation caused by adding salt to water. When salt is added to water, the sodium chloride separates into sodium ions and chloride ions. The premise of boiling point elevation is that the added particles raise the temperature needed to bring water to its boiling point. The extra particles interfere with the interactions between solvent molecules (water, in this case). Boiling Point Elevation Problem 31.65 g of sodium chloride is added to 220.0 mL of water at 34 °C. How will this affect the boiling point of the water? Assume the sodium chloride completely dissociates in the water. Given:density of water at 35 °C = 0.994 g/mLKb water = 0.51 °C kg/mol Solution To find the temperature change elevation of a solvent by a solute, use the equation:ΔT = iKbmwhere:ΔT = Change in temperature in °Ci = van't Hoff factorKb = molal boiling point elevation constant in °C kg/molm = molality of the solute in mol solute/kg solvent Step 1. Calculate the Molality of the NaCl molality (m) of NaCl = moles of NaCl/kg water From the periodic table: atomic mass Na = 22.99atomic mass Cl = 35.45moles of NaCl = 31.65 g x 1 mol/(22.99 + 35.45)moles of NaCl = 31.65 g x 1 mol/58.44 gmoles of NaCl = 0.542 molkg water = density x volumekg water = 0.994 g/mL x 220 mL x 1 kg/1000 gkg water = 0.219 kgmNaCl = moles of NaCl/kg watermNaCl = 0.542 mol/0.219 kgmNaCl = 2.477 mol/kg Step 2. Determine the Van 't Hoff Factor The van't Hoff factor, "i," is a constant associated with the amount of dissociation of the solute in the solvent. For substances which do not dissociate in water, such as sugar, i = 1. For solutes that completely dissociate into two ions, i = 2. For this example, NaCl completely dissociates into the two ions, Na+ and Cl-. Therefore, here, i = 2. Step 3. Find ΔT ΔT = iKbmΔT = 2 x 0.51 °C kg/mol x 2.477 mol/kgΔT = 2.53 °C Answer Adding 31.65 g of NaCl to 220.0 mL of water will raise the boiling point by 2.53 °C. Boiling point elevation is a colligative property of matter. That is, it depends on the number of particles in a solution and not their chemical identity. Another important colligative property is freezing point depression.