Question

17. A solution is prepared by dissolving 5.00 g of aspirin, C9H3O4, in 215 g of chloroform, has a boiling point that is elevated by AT = 0.47 °C over that of pure chloroform. What is the value of the boiling point elevation constant for chloroform? 18. What osmotic pressure in atmospheres would you expect for each of the following solutions? a) 5.00 g of NaCl in 350.0 mL of aqueous solution at 50 °C

Answer 1

® Ceren data - WA = mass of chlow form = 215g - Wo= mess of aspiranthony sing Mpz molar Mess of aspirin, cqho Oy = 9x12 + 8 x1 +4816 = 108 + 8 + 64 1 = 180 g/mol ATG = Elevation in boiling point = 0.476 kb = ? = Molar elevation constant for chloroform. using , Elevation in boiling point expression At low Works - MBX WA 0.476 = low xky x 5g loog x R15g Mb = 180x215 X 0.47 °C by maps? loro x 5 kg = 3.64 °C kg mot 1oox 215x0.47

- 18. Criven date Wo= Mess of Nall (solute ) = 5g My Moler trees of Nall = 58.5 gl mite ve volume of solution - 35000m2 = 0.350 Lite T = Temperature = 50°ca (50 1273)2 = 323k R= gas constant = 0.0821 atm Lkt mot Osmotic pressure a ? using omos Osmotic pressure expression. T = nRT - WB ORXT 59 x 0.0821 X 323k 58.59 not atm Lktrots 0.350L 1/ 5 X 0.0821 X323 at 58.5X0.350 IT 26.48 atm ,