This preview shows page 1. Sign up to view the full content.
Unformatted text preview: g solute/ g solute + g solvent x 100% = percent by mass ppt (per thousand) = ratio x 1000 ppm (per million) = ratio x 1,000,000 ppb (per billion) = ratio x 10 9 mole fraction Xsolute = moles solute/moles of all components 16 g ethanol, 200 g of water, final volume 214.0 mL 16 g / 46.01 g/mol = 0.35 mol ethanol 200/18.01 g/mol = 11.1 mol H2O Xsolute = .35/.35+11.1 = .03 Mole % = 3% M = .35 moles ethanol/.214 L solute = 1.63 M Molality = .35 mol ethanol/.2 kg = 1.75 molality Colligative properties of solution depend on number of particles, not on type *Ethanol 1 molal less massive than sugar because its a smaller molecule *Sugar 1 Molal NaCl 1 mol Na+1 mol Cl / 1 kg solvent 1 Molal CaCl2 1 mol Ca+2 mole Cl/ 1 kg solvent Affected by: Freezing point Boiling point Vapor pressure...
View Full Document
This note was uploaded on 01/23/2012 for the course CHEM 18a taught by Professor Dolnik during the Fall '06 term at Brandeis.
- Fall '06