Using Colligative Properties to Find the Example: The Tf of camphor is 179.80°C and it’s Solute Molar Mass kf is 39.7°C/m. When 200.0 mg of a compound (X) • The molar mass of the solute can be obtained by are added to 100.0 g of camphor, it’s freezing point drops to 179.29°C. What is the molar mass of X? measuring one of the colligative properties of the solution, the mass of the solute and the mass of → Calc. the m of X: ∆Tf (179.80 -179.29)°C mol the solvent (or the volume of the solution) ∆T f = k f m → m = = = 0.013 k f 39.7°C/m kg – ∆P and ∆Tb are not very sensitive to changes in the solute concentration (rarely used for molar mass → Calc. the number of moles of X: mol( X ) mol determinations) m = → 0.013 × 0.1000 kg = 0.0013 mol kg( camphor ) kg – ∆Tf is more sensitive especially when solvents with large kf constants are used → Calc. the molar mass of X: – Π is most sensitive and can be used for substances grams( X ) 0.2000 g MW = → = 160 g/mol with low molar solubility such as large biomolecules mol( X ) 0.0013 mol Example: A 3.0 g polymer sample is dissolved in Volatile Nonelectrolyte Solutions enough benzene to produce 150. mL of solution. If – No dissociation; both solvent and solute have vapor the solution’s osmotic pressure is 0.0119 atm at pressures (mixtures of volatile organic compounds) 25°C, what is the average MW of the polymer? • Raoult’s law can be applied to the vapor Calc. M of the solution: pressures of both the solute and the solvent Π 0.0119 atm mol M = = = 4.87 ×10−4 RT 0.08206 L.atm/mol. K × 298 K L Psolv = XsolvP°solv Psolute = XsoluteP°solute → Calc. the number of moles of the polymer: ⇒The presence of each component lowers the mol( pol ) mol vapor pressure of the other component M = → 4.87×10−4 × 0.150 L = 7.30×10−5 mol L( so ln) L – The total pressure over the solution is the sum of → Calc. the molar mass of the polymer: the partial pressures of the solute and the solvent grams( polym ) 3.0 g MW = → = 41000 g/mol Ptot = Psolv + Psolute = XsolvP°solv + XsoluteP°solute mol( polym ) 7.30×10−5 mol 1 1.0 3.0 – The mole fractions of the solute and solvent in the X = = 0.25 X = = 0.75 vapor phase above the solution can be calculated b 1.0 + 3.0 t 1.0 + 3.0 using Dalton’s law o Pb = X b Pb = 0.25× 95.1 torr = 24 torr vap vap Psolv = X solv Ptot Psolute = X solute Ptot o Pt = X t Pt = 0.75× 28.4 torr = 21 torr P P vap solv vap solute Ptot = Pb + Pt = 24 + 21 = 45 torr X solv = X solute = Ptot Ptot vap Pb 24 vap Pt 21 X b = = = 0.53 X t = = = 0.47 Example: The equilibrium vapor pressures of Ptot 45 Ptot 45 pure benzene and toluene are 95.1 and 28.4 torr, ⇒The mole fractions of the solute and solvent in the respectively at 25°C. Calculate the total vapor phase are different from those in the liquid pressure, partial pressures and mole fractions of phase → The vapor phase is enriched with the benzene and toluene over a solution of 1.0 mol of more volatile component → Condensation of the benzene in 3.0 mol of toluene at 25°C. vapors leads to a solution which is enriched with the more volatile component → distillation Electrolyte Solutions – For ideal solutions (very dilute solutions), the ideal value of i represents the number of moles of – The solute dissociates to ions particles to which a mole of the solute dissociates – The number of particles in the solution is greater than the one implied by the solute concentration – For real solutions, i is smaller than the ideal value • The formulas for the colligative properties – The deviation is due to clustering of cations and must be modified anions (ionic atmosphere) which reduces the “effective” concentration of particles → (i×Conc.) – van’t Hoff factor (i) – accounts for the dissociation of the solute Example: Select the solution with the higher – i is determined experimentally as a ratio of the osmotic pressure: 0.1M NaCl or 0.08M CaCl2 measured colligative property versus the one → Calculate the effective concentrations: expected for a nonelectrolyte → 0.1M NaCl → i×M = 2×0.1 = 0.2 0.08M CaCl i M = 3 0.08 = 0.24 (higher) ∆P= i(XsoluteP°solv) ∆Tb = i(kb m) → 2 → × × 0.08M CaCl has higher ( = i MRT) Π = i(MRT) ∆Tf = i(kf m) ⇒ 2 Π Π × 2.