Honors Chemistry
1.
a. Solute = Iron III Chloride or FeCl3, solvent = water
b. Bonds made between Iron III ions and water molecules, chloride ions and water molecules, bonds broken between Iron II ions and chloride ions, bonds broken between water molecules
c. If solid is present after extensive stirring, solution would be saturated.
d. Add water or increase temperature to make solution unsaturated.
e. Aqueous Iron II and aqueous nitrate ions
f. Each ion is surrounded by, and bonded to, several water molecules.
g. Molar mass = 241.859 g/mol, moles of solute = 0.0023816 mol, Molarity = 0.0191 M
h. 0.0191 M Fe+3 ion, 0.0572 M Cl- ion
2.
a. Saturated with nitrogen gas, oxygen gas, carbon dioxide gas and argon gas. Plenty of time for these gases to dissolve into the water and establish an equilibrium. Amount of each dissolved gas would be determined by the partial press ure of each as well as the temperature.
b. Solutes would include each of the above gases, solvent would be water
c. Lower temperature or increase pressure of each gas above the liquid would make the solution temporarily unsaturated, but given enough time the solution would return to being saturated.
d. Increase in temperature would lower solubility, so some gases would leave the solution, perhaps with bubbles forming. After some time the solution would return to being saturated, but with less of each of the solutes.
3.
a. q = mcΔT = (125 g)(4.18 J/goC)(3.6oC) = 1881J
b. 1881J/.576 g = 3266 J/g or –3266 J/g or -3.67 kJ/g
c. (–3.27 kJ/g)(241.859 g/mol) = 790 kJ/mol
d. Since the overall reaction was exothermic, more bonds or stronger bonds (or both) were made than were broken.
4. Boiling point would increase. The presence of the solute would interfere with evaporation inside a bubble, but not condensation. Since there would be less evaporation at 100oC that condensation, vapor pressure would drop and be lower that 1 atmosphere, so water would not boil at 100oC. Water temperature would then be allowed to increase above 100oC (if being heated). When temperature rose, evaporation would increase, condensation would decrease, vapor pressure would increase and the solution would boil, but at a higher temperature.
5. The melting point would drop. The presence of solute slows the rate of solidification, but does not affect melting. Since solidification is the heat releasing process, the temperature would be allowed to drop (if being cooled). At the new lower temperature solidification rate would increase, melting rate would decrease, and the solution would come to a new thermal equilibrium, but at a lower temperature.
6. V = n/M = 0.0445 mol/0.786 M = 0.0566 l = 56.6 ml
7. Molar mass = 261.338 g/mol (Ba(NO3)2), so we would need 11.6 grams of the solute. We would add this to maybe 50 ml of water, dissolve completely, then add additional water until we had the 56.6 ml of solution needed.
8. Molar mass = 132.139 g/mol ((NH4)2SO4), moles of solute = MV = (0.0667 M)(.023 l) = 0.0015341 moles, or .203 g
9. Molar mass = 262.855 g/mol (Mg3(PO4)2), moles of solute = .36142 mol, M = .36142 mol/.450 l = 0.803 M
10. 2 Al(s) + 6 HNO3(aq) à 3 H2(g) + 2 Al(NO3)3(aq) 0.70417 moles of Al require 2.1125 moles of HNO3 or 1.056 liters or 1056 ml
11. 2 Na(s) + 2 H2O(l) à H2(g) + 2 NaOH(aq) 0.10874 mol of sodium would produce 0.10874 moles of NaOH, dissolved in enough water to make 100 ml would give a concentration of approx. 1.0 M (sodium would take up some space).
12. AgNO3(aq) + NaCl(aq) à AgCl(s) + NaNO3(aq) 0.0344 moles of AgNO3 reacting with 0.0413 moles of NaCl would produce 0.0344 moles of AgCl (molar mass of 143.323 g/mol) or 4.93 g
13. 0.0344 moles of NaCl would also be made, dissolved in 0.160 liters of solution gives 0.215 M NaCl solution
14. 2 H3PO4(aq) + 3 (NH4)2S(aq) à 3 H2S(g) + 2 (NH4)3PO4(aq) 0.025 moles of H3PO4 react with 0.033 moles of (NH4)2S would produce 0.033 moles of H2S since the (NH4)2S would be the limiting reactant. V=nRT/P = (0.033 moles)(8.31)(448K)/200 kPa) = .614 liters or 614 milliliters