1 / 5

(B) Adding a nonvolatile solute to a solvent decreases

Phase diagram. P. pure solvent. solid. solution. liquid. Note that FP and BP . 1 atm. gas. T. (B) Adding a nonvolatile solute to a solvent decreases the solution’s freezing point (FP) and increases its boiling point (BP) . (not water’s). NFP. NBP.

signa
Download Presentation

(B) Adding a nonvolatile solute to a solvent decreases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Phase diagram P pure solvent solid solution liquid Note that FP and BP . 1 atm gas T (B) Adding a nonvolatile solute to a solvent decreases the solution’s freezing point (FP) and increases its boiling point (BP). (not water’s) NFP NBP

  2. The freezing point depression and boiling point elevation are given by: DTx = Kx m i DTx = FP depression or BP elevation Kx = Kf (molal FP depression constant) or Kb (molal BP elevation constant) -- they depend on the solvent -- for water: Kf= 1.86oC/m Kb = 0.52oC/m m = molality of solute

  3. i = van’t Hoff factor (accounts for # of particles in solution) In aq. soln., assume that… 1 -- i = __ for nonelectrolytes 2 -- i = __ for KBr, NaCl, etc. Jacobus Henricus van’t Hoff (1852 – 1911) 3 -- i = __ for CaCl2, etc. In reality, the van’t Hoff factor isn’t always an integer. Use the guidelines unless given information to the contrary.

  4. Kf = 1.86oC/m Kb = 0.52oC/m 0.690 m Find the FP and BP of a soln. containing 360. g barium chloride and 2.50 kg of water. DTx = Kx m i 360 g BaCl2 = 1.725 mol BaCl2 i = 3 DTf = Kf m i = 1.86(0.690)(3) = 3.85oC FP = –3.85oC DTb = Kb m i = 0.52(0.690)(3) = 1.08oC BP = 101.08oC

  5. 3.1oC 40.0 1 (organic = nonelec.) mol unknown Solve for m = 0.0775 kg camphor 1 mol unk. = 0.186 g unk. X g unk. Camphor, C10H16O, has an NFP of 179.8oC and a Kf of 40.0oC/m. When 0.186 g of a nonelectrolytic substance is dissolved in 22.01 g of camphor, the mixture’s new freezing point is 176.7oC. Find the unknown’s molar mass. DTf = Kf m i (Multiply by 0.02201 kg camphor to get…) 1.71 x 10–3 mol unk. X = 109 g/mol

More Related