WS 6.1 Honors key
1.
Homogeneous mixture, parts indistinguishable. Solid-liquid= salt water, gas-liquid=O2 in water, gas-gas= CO2 in air.
2.
The ionic crystal is solvated by the polar water molecules through ion-dipole bonding. The cation is surrounded by the
negative side of the water molecule and the anion is surrounded by the positive side of the water molecule.
3.
The sucrose molecule has multiple sites (O-H) of H-boding where polar water molecules can attach and solvate the molecule.
4.
Solubility is determined by the extent to which the interacting particles (solvent-solute) have similar types of intermolecular
bonding. If the solute-solvent IMF’s are stronger than the solvent-solvent and solute-solute interactions than the substance will
be soluble.
5.
a) ∆𝐻𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = ∆𝐻𝑠𝑜𝑙𝑣𝑒𝑛𝑡 + ∆𝐻𝑠𝑜𝑙𝑢𝑡𝑒 + ∆𝐻𝑚𝑖𝑥. = 250 𝑘𝐽/𝑚𝑜𝑙 + 120𝑘𝐽/𝑚𝑜𝑙 + (−400𝑘𝐽/𝑚𝑜𝑙) = −𝟑𝟎𝒌𝑱/𝒎𝒐𝒍
b) since the dissolving process is exothermic, the energy released will transfer into the water and increase the kinetic energy
of the water molecules and therefore, increase the temperature.
6.
a) Methanol. NaCl is ionic, so it dissolves through ion-dipole forces. Both methanol and 1-propanol have a single polar OH
group, but the hydrocarbon portion interacts only weakly with the ions and 1-propanol has a longer hydrocarbon portion
than methanol.
b) Water. Ethylene glycol molecules have two OH groups, so they interact with each other through H-bonding. H bonds
formed with water can replace these H-bonds between molecules better than dipole-induced dipole forces with hexane can.
c) Ethanol. Diethyl ether molecules interact through dipole-dipole and dispersion forces. They can form H bonds to water or
to ethanol. But ethanol can also interact with the diethyl ether effectively through dispersion forces because it has a
hydrocarbon chain.
7.
No. The dipole-dipole forces attracting sugar molecules is stronger than the weak LDF for the nonpolar polish remover.
8.
No. There are no, or very few attachment points for the polar water molecule to solvate the oil.
9.
HCl is a polar molecule that can easily attract to the polar water molecule. Propane uses LDF’s to attract, polar water
molecule has no points of attachment for solvation.
10.
a)
b)
c)
The answer would be that the CH3OH molecule induces a dipole on the CCl4 molecule, allowing them to attract.
both molecules would attract through their dipole-dipole IMF’s.
The NaCl is a polar molecule (ionic bonding) and when it dissolves in water the charged ends of the water molecule
will attract to the opposite charges of Na+ and Cl- this is known as an ion-dipole type of IMF.
11. Gas solubility decreases with increasing temperature, so less dissolved oxygen for aquatic critters to breath.
12. Gas solubility increases with increasing pressure according to Henry’s Law S gas=K Pgas
13.
Solute
CO2(g)
NaCl(s)
increasing temp
decrease
increase
decreasing temp
increase
decrease
increasing pressure
increase
no effect
14. Sg = (3.1 x 10-2 M/atm)(4.0 atm) = 0.12 mol/L
15. Sg = kPg= (6.8 x 10-4 M/L-atm)(0.78 x 2.50 atm)
Sg = 1.3 x 10-3 mol/L
16.
mole NaOH
mole H2O
mass %
mole fraction
molarity
PPM
mol = m/MM
123 g NaOH/40.0 g = 3.08 mol NaOH
mol = m/MM
289 g H2O/18.02 g = 16.04 mol H2O
% = (msolute/mtotal) x 100
[123 g H2O/(289 + 123 g)]100 = 29.9 %
X = molsolute/moltotal
X = 3.08 mol/(3.08 mol + 16.04 mol) =.16
0.161
M
= molsolute/Vsolution(L)
M = 3.08 mol NaOH/0.300 L = 10.3 mol/L
123g/(123g + 289g) x 106= 299,000 ppm
decreasing pressure
decrease
no effect
17.
Solution (in 100g H2O)
Sat, Unsat, Supersat
40 g of KClO3 at 50oC
110 g NaNO3 at 45oC
70 g KNO3 at 60oC
80 g Pb(NO3)2 at 40oC
super saturated
super saturated
unsaturated
super saturated
22. 𝐹𝑖𝑟𝑠𝑡 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 5.10 𝑔 𝑡𝑜 𝑚𝑜𝑙𝑒𝑠
𝑀=
23. 𝐹𝑖𝑟𝑠𝑡 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 9.5 𝑔 𝑡𝑜 𝑚𝑜𝑙𝑒𝑠
24. 𝑚𝑎𝑠𝑠% =
25. 𝑚𝑎𝑠𝑠% =
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
26. 𝑀1 𝑉1 = 𝑀2 𝑉2
29. %𝑣𝑜𝑙𝑢𝑚𝑒 =
𝐿
=
4.5% =
× 100%
=
.028 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
0.1005𝐿
.13 𝑚𝑜𝑙 𝑏𝑙𝑒𝑎𝑐ℎ
1.0 𝐿
?𝑔𝑟𝑎𝑚𝑠
175 𝑔
1.0 𝑔 𝑒𝑡ℎ𝑎𝑛𝑜𝑙
(100𝑔+1.0 𝑔)
0.0005𝐿
(10𝐿+0.0005𝐿)
𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= .13𝑀
× 100% 𝑟𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑒 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑔𝑟𝑎𝑚𝑠 = 7.9𝑔 𝑙𝑎𝑐𝑡𝑜𝑠𝑒
× 100% = .99 𝑜𝑟 1%
= 50𝑝𝑝𝑚
× 100% =
×
74.1 𝑔
1𝑚𝑜𝑙
×
.25 𝐿
1
𝑀=
35𝑚𝐿
(115𝑚𝐿+35𝑚𝐿)
𝑚𝑜𝑙
𝐿
=
+/- how many g to
make saturated?
-22
-10
approx. +40
-5
= .28𝑀
(5𝑀)(𝑉) = (0.25𝑀)(0.001𝐿) 𝑟𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑒, 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑉 = 5𝑚𝐿 𝑜𝑟 .005𝐿
0.127𝑚𝑜𝑙𝐶𝑎(𝑂𝐻)2
1𝐿
𝑚𝑜𝑙
=
× 100%
30. 𝐹𝑖𝑟𝑠𝑡 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 1.55 𝑔 𝑡𝑜 𝑚𝑜𝑙𝑒𝑠
31. 𝑎)
𝐿
(16𝑀)(𝑉) = (0.1𝑀)(1.5𝐿) 𝑟𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑒, 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑉 = .0094𝐿 𝑜𝑟 9.4𝑚𝐿
27. 𝑃𝑃𝑀 = 1 × 106 ×
28. 𝑀1 𝑉1 = 𝑀2 𝑉2
𝑀=
𝑚𝑜𝑙
+/- how many °C to make
saturated?
+34
+12
-18
+5
× 100% = 23%
.013 𝑚𝑜𝑙 𝐾𝐵𝑟
1.6𝐿
= 8.1 × 10−3 𝑀
= 2.35 𝑔
b) 𝑀1 𝑉1 = 𝑀2 𝑉2 = (1𝑀)𝑉1 = (. 127𝑀)(. 25𝐿) = 31.8𝑚𝐿