Question

I need help with problem 16. Not problem 16 under chapter 13 set 5.

*9. 9. The vapor pressure of pure toluene, CyHx, is 137 torr at 60°C. What is the vapor pressure of toluene in a solution made by dissolving 25.0 grams of the nonelectrolyte glucose, C6H12O6, in 50.0 grams of toluene at 60°C? 10. The vapor pressure of water is 118 torr at 55°C. How many grams of urea, CO(NH2)2 must be added to 67.4 grams of water to lower the vapor pressure to 109 torr? 11. How much is the vapor pressure lowered when a 12.0 gram sample of nonvolatile fructose, CH,206, is dissolved in 40.0 grams of ethanol, C2H5OH, (also known as "grain" or "drinking" alcohol) at 78.4°C? For pure ethanol, the vapor pressure is equal to 760 torr at 78.4°C. 13. What are the freezing point and the boiling point of the solution made in Exercise 11? For ethanol, Kris 1.99°C/m, the normal freezing point is -114.6°C; Kh is 1.22°C/m and the normal boiling point is 78.4°C. 15. What will be the freezing point and the boiling point of a solution prepared by dissolving 25.0 grams of the nonvolatile, nonelectrolyte CO(NH2)2 in 75.0 grams of nitrobenzene, C6H5NO2? For nitrobenzene, Kb=5.24°C/m and Ke=7.00°C/m. The normal boiling point is 210.88°C and the normal freezing point is 5.70°C. 16. Camphor, CH60, with a fragrant and penetrating odor, is used in cosmetics as a preservative. Camphor has a normal freezing point of 178.8°C and a Kf of 37.3°C/m. When a 2.96 gram sample of a nonelectrolyte with an empirical formula of C2H4Cl is dissolved in 47.21 grams of camphor, the freezing point of the solution is 160.4°C. What are the molecular weight and the molecular formula of the solute? 17. Fatty acids, such as stearic acid, C18H3802, are nonpolar because of their long hydrocarbon chains. Because they are nonpolar, they have relatively low freezing points. How many grams of iodine, 12, must be dissolved in 22.3 grams of stearic acid to lower its freezing point to 58.5°C? The normal freezing point = 70.1°C and Kr=7.40°C/m. 18. The normal boiling point of toluene, CHs, a common industrial solvent, is 110.6 1°C. When a 6.95 gram sample of the nonelectrolyte glucose, C6H12O6, is dissolved in 22.4 grams of toluene, the boiling point of the solution is 114.87°C. What is the boiling point elevation constant, Kb, of toluene? 30. What are the molarity, the molality, and the mole fraction of solute in an aqueous solution that is 9.97% by mass K2SO4? The specific gravity of the solution is 1.023. 37. When a 41.9 gram sample of the nonelectrolyte urea, CO(NH2)2, is dissolved in 125.0 grams of formic acid, HCOOH, the solution has a freezing point of 3.25°C. The normal freezing point of formic acid is 8.40°C. What is the freezing point depression constant of formic acid? 39. What are the molality and the molarity of KBr in an aqueous solution that is 3 8.00% by mass KBr. The solution density is 1.3525 g/mL. ANSWERS TO EXERCISES 1. 6.00 m NaOH 2. X= 0,0975 3. 2.33 g CO(NH2)2, 2.26% 4, 224 mL H2O 5. 3.84 m KCN; X-0.0646 6. X = 0.06728; 4.003 m KI 7. 19.1 g C2H5OH 8. 8.008 M; 34.00 %, X-0.1677 9.109 torr 10. 18.6 g CO(NH2) 11.54.2 torr 12. 146 torr 13. T. -80.4°C; T.--117.9°C 14.1.4x102 g/mol 15. To - 240.0°C; Tr=-33.2°C 16.127 g/mol; CH3C12 17. 8.88 g 12 18.2.48°C/m 30. X = 0.011; 0.637 m K2S04:0.587 M KS04 37.0.922°C/m 39.5.150 m KBr; 4.319 M KB Chapter 13 Set 5 -16. Solutions that are 5.50% by mass glucose (C6H12O6, also known as "blood sugar") are often administered intravenously to provide patients nourishment. -a) How many grams of glucose are in 450.0 grams of a 5.50% solution of glucose? -(b) How many grams of water are contained in 325 grams of the solution? - (c) How many grams of solution must be administered to provide 26.5 grams of glucose? -(d) How many grams of water must be added to 6.78 grams of glucose to make a solution that 5.50% by mass

Answer 1

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16 Freezing point of Camphor = 178-8°c Freezing point of solution = 160-4°C .: ATE = (178.8 – 160-4) °c 18.4°C Alf = Kfom (where m= molality) 18 4 = 37.3 m m = 0.493 molal molality = moles of solute weight of solvent (Ig) weight in gram Molar mass x weight of solvent (kg) 2.96 x 1000 0.493 Molar masy X 47.21 . Molar mass = [127.2 g Imol. 127 n = Empirical mass = 63-5 Molar mass = 127 Molar mass Empirical mass 63.5 Molecular formula = Empirical Formula xn. = ( Hy Cl) x2 = cq Hq Cl] .