Solutions Problems Answers

Solutions Problems Answers

1. Salt and water make a homogeneous mixture, a solution; where salt and sand make a non-uniform heterogeneous mixture.

2. KNO3 is the solute and water is the solvent because it is an aqueous solution (something dissolved in water).

3. Ammonium chloride is a soluble salt. It dissolves readily in water, which is a physical change, so I would have to set-up a distillation apparatus and evaporate the water from the solution. Once all the water has evaporated (and been collected from the condenser) I would be left with the ammonium chloride in the beaker.

4. NaNO3 is a solute that when dissolved in water that solution conducts electricity, hence the light coming on, so it is considered to be an electrolyte.

Solution A – supersaturated / Solution B – Unsaturated / Solution C – Saturated
Many crystals formed because by adding a few crystals to the supersaturated solution it changed the dynamic of the solution causing all the extra crystals that had been dissolved to come out of solution at once, hence a BIG crystal forming. / Because crystals could still dissolve in the solution it had yet to reach its maximum amount of solubility. / Because when crystals were added they did not dissolve but sank to the bottom because maximum amounts of solubility had already been reached.

6. It is a physical change because it is easily reversible.

7. The magnitude of the two first steps of forming a solution versus the magnitude of the last step (3rd) for forming a solution effects if it will be an exo or endothermic process. Step 1 and 2 are the breaking up of the solute and also the solvents interactions and are always endothermic, while step 3 is the forming of the solution (making interactions between solute and solvent) which is always exothermic.

8. The intermolecular attractions between solute and solvent would be the most important reason a solution occurs. If it is too difficult to break up the solute with the solvent then there will be no solution made.

9. X-Y would be polar because it would have different electronegativities. X-X would be non polar because they would have no electronegative difference.

10. The diagram should show K+ ions being surrounded by the oxygen atom of the water molecule and the Br- ions should be surrounded by the hydrogen atoms of the water molecule. There must be an interaction (not bond) between the ion of the salt and the oppositely charged polar end of the molecule.

11. Show the opposite charged ends of each molecule interacting with each other. The slightly positive end of one molecule interacting with the slightly negative end of the other molecule.

12. Non polar molecules dissolve primarily due to London dispersion forces. Although it seems as though there are no intermolecular attractions between non-polar molecules, there are very small temporary dipoles that can occur and therefore cause a certain amount of slight + and – within the molecules. Often non-polar interactions discuss dissolving as a blending of two non-polar molecules than a breaking up of solute.

13. The forces from the non-polar benzene (London dispersion) would not be strong enough to overcome the ionic bonding within KCl, so it would not dissolve.

14. Yes, because HCl is a dipole and it will interact readily with the polarity of water.

15. (a)

(b) predicted solubility is 130 g

(d) a supersaturated solution could form, looking like a saturated solution (completely uniform, but we would know that more than the max of solute had completely dissolved) OR a saturated solution where the excess solute couldn’t stay in solution forms and the excess falls to the bottom of the test tube.

(e) 171.5 g at 80 oC in 100 g of water, so in 250g of water, simply multiply by 2.5. Answer = 428.75 g

(f) 21 oC

(g) 98.0 g of NaClO3 in solution of 100mL let’s say…

Moles = 98.0 g / 106.44g/mol = 0.9207mol

Molarity = 0.9207 mol / 0.100L = 9.207 M

16. NaNO3 at 20 oC has a max solubility of 88g (from textbook pg. 254), so its molar solubility would be 88 g/ 85g/mol = 1.04 mol

Molar concentration = 1.04 mol/ 0.100 L = 10.35 M

17. (a) Mg(NO3)2(aq) + 2 KOH(aq) à Mg(OH)2(s) +2 KNO3(aq)

Mg2+ + 2NO3- +2K+ + 2OH- à Mg(OH)2(s) + 2K+ + 2NO3-

Mg2+ + 2OH- à Mg(OH)2(s)

(b) 2 NH4NO3(aq) + K2CO3(aq) à (NH4)2CO3 (aq) + 2 KNO3(aq)

2NH4+ + 2NO3- + 2K+ + CO32- à 2NH4+ + CO32- + 2K+ + 2NO3-

No net ionic equation because they all cancel out. There is no reaction.

18. Percentage by mass = 35 g / 135g of solution x 100% = 25.9 % of this solution is because of the mass of sugar.

19. 0.20 mol/L x 0.560 L =0.112 mol NH4Cl

0.112 mol x 53.5 g/mol = 5.99 g

20. 32.0 g/ 40.0 g/mol = 0.80 mol NaOH

Molarity = 0.80 mol/ 0.750 L = 1.07 mol/L