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generalnov14notes

# generalnov14notes - g solute g solute g solvent x 100 =...

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∆Hsoln = ∆Hsolute always endothermic +∆Hsolvent +∆Hmix = ∆Hhydration ∆Hsolute > ∆Hhydration endothermic ∆G = ∆H-T∆S *Gibb’s free energy – measure of spontaneity Unsaturated solution haven’t dissolved as much solute as you can Saturated all that can dissolve is dissolved – equilibrium Super-saturated heat to dissolve, dissolve as much as possible, then cool – generally unstable non equilibrium Gases are less soluble at higher T than lower Henry’s Law : Sgas = k u (P gas ) O2 – 1.3 x 10 -3 N2 – 1.1 x 10 -4 NH3 – 5.8 x 10 What pressure of CO2 is required to make a 0.12 M solution of CO2 (in water) at 25C? 0.12 M = 3.4x10 -2 M/p (P CO2 ) M – Molarity – moles solute/L solution – useful for measuring M - molality – moles solute/1000g of solvent – useful when we need fixed solute:solvent ratios Percent by mass – 5g/5g O2 + 5g = .50 = 50%
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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 it’s 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...
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