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.
5.
Solution A – supersaturated
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.
Solution B – Unsaturated
Because crystals could still
dissolve in the solution it
had yet to reach its
maximum amount of
solubility.
Solution C – Saturated
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 nonpolar 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.
Solubility (g/100g H2O)
Solubility of NaClO3
250
200
150
100
50
0
0
20
40
60
80
Temperature (oC)
15. (a)
(b) predicted solubility is 130 g
100
120
(c) maximum mass is 85 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)  Mg(OH)2(s) +2 KNO3(aq)
Mg2+ + 2NO3- +2K+ + 2OH-  Mg(OH)2(s) + 2K+ + 2NO3Mg2+ + 2OH-  Mg(OH)2(s)
(b) 2 NH4NO3(aq) + K2CO3(aq)  (NH4)2CO3 (aq) + 2 KNO3(aq)
2NH4+ + 2NO3- + 2K+ + CO32-  2NH4+ + CO32- + 2K+ + 2NO3No 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