NAME _KEY_
CHEM 162w14 Ch. 13 Book Related Problems
Lab ______
score:
____/21
1. Which of the following likely to be more soluble in hexane, C6H14?
a) CCl4 Explain: NP-NP
b) benzene Explain: NP-NP
c) octanoic acid, Explain: length of molecule
3
2. Which of the following likely to be more soluble in water? NP(H-bonding)
a) C6H12O6 Explain: H-bonding; other np only dispersion
b) CH3CH2COONa Explain: Ionic; incr entropy when ionic solid dissolves
c) HCl
Explain: HCl is SA, complete dissociation, incr entropy
3
3. Rank in order of decreasing boiling point: 0.120 m glucose, 0.050 m LiBr, 0.050 m Zn(NO3)2?
Zn – Glu – Li (3 ions – higher [ ] – 2 ions @ 0.5)
1
4. Rank in order of highest to lowest freezing point: 0.040 m C3H8O3, 0.020 m KBr, 0.030 m C6H5OH?
C6H5OH – GLY & KBr
0.03 - 0.04 & 2(0.02) glycol & KBr have same concentration of 0.040 m
1
5. Check all answers that apply. 0.5 point deduction for each wrong pick or missing correct pick.
a) Water is known as a polar molecule because:
a) the oxygen atom has a greater attraction for electrons than the hydrogen atom does
e) the electrons of the covalent bond are not shared equally between the hydrogen and oxygen atoms
3
b) which interactions and processes contribute to the dissolution of ionic compounds in water?
a) Affinity of hydrogen towards anions
c) Ion-dipole interactions
d) Hydration
e) Affinity of oxygen towards cations
6. Rank the organic compounds from most soluble to least soluble, overlap if same.
CH3CH2CH2COOH - CH3CH2CH2CH2OH - CH3CH2OCH2CH3 - CH3CH2CH2CH2CH3
1
6.5
7. Correctly classify each of the following compounds as “highly soluble” or “slightly to insoluble” in water.
Highly
Slightly or Insoluble
NaCl KOH KNO3 CH3OH
C6H6 C6H14 CCl4 CH2Cl2 CH3(CH2)5OH CHCl3
CH3CH2OH CH3COOH Na2SO4
The properties of water make it a versatile solvent. A wide range of organic and inorganic compounds are soluble in water. For
this reason water is sometimes referred to as the universal solvent.
Many organic compounds which contain polar functional groups are soluble in water. However their solubility is inversely
proportional to the length of carbon chain. Inorganic compounds which readily dissociate to form ions are water soluble.
Solubility of organic compounds is mainly due to hydrogen bonding and dipole-dipole interactions, while inorganic compounds
which dissociate into ions dissolve due to ion-dipole interactions
1.5
8. Classify the steps involved in the formation of a solution as being endothermic or exothermic.
separation of solute molecules
separation of solvent molecules
formation of solute-solvent interactions
ENDO
ENDO
EXO
9. Arrange the solns in order of decreasing molal concentration based on the fp of 3 glucose solns.
A-1.45oC B-5.69oC C-3.04oC
B-C-A
1
NAME _KEY_
CHEM 162w14 Ch. 13 Book Related Problems
Lab ______
score:
____/15
10. What is the freezing point of an aqueous soln that boils at 104.1oC?
Kb = 0.51 Kf = 1.86 Tb = 4.10oC Tb = Kb*m => m = (T)/Kb = 4.1/0.51 = 8.01 m
Tf = Kf*m = (1.86)*(8.01) = 14.9oC f.p. = 0.0 – 14.9 = -14.9 = -15oC
2
11. At 25oC, the molar solubility of calcium phosphate in water is 1.10*10-7 M.
a) solubility grams per liter. Ca3(PO4)2 = 310 g/mol (1.10*10-7 mol/L)*(310 g/mol) = 3.41*10-5 g/L
2
b) What is the solubility expressed in ppm?
dilute solution, ppm = mg/L so, g to mg gives 3.41*10-2 mg/L = 3.41*10-2 ppm or 0.0341 ppm
12. table shows solubility of CO2 in water
C
(mol/L )
3
c)
P
k
(atm )
(mol/L⋅atm )
3.80×10−2
1.00
9.40×10−2
b) 2.47
3.4*10-2
1.00
a) What Henry’s constant CO2 at 20 C?
o
a) 3.80×10−2
T
C
(∘
20.0
20.0
3.40×10−2
25.0
C=kP so k=C/P = (.0380 M)/1-L = 3.80×10−2 M/atm
-2
b) What pressure to achieve [CO2] of 9.40*10 M?
C=kP so P=C/k = (0.094/0.038) = 2.47 atm
c) 1 atm, moles of CO2 released increasing 1-L water to 25oC?
@25: C=kP = (0.034)*(1 atm) = 0.034 M
@20: we calculated 0.038 M so, difference 0.038 – 0.034 = 0.004 mol
13. A solution is prepared by dissolving 41.0 g KCl in 225 g of water.
a) mass percent KCl in soln?
m% = (mass KCl)/(mass soln)*100 = (41.0 g)/(41.0+225) = 15.4%
b) mole fraction KCl? mol KCl=41/74.6 =.550 mol
mol water=225/18.0=12.5 mol
X=(mol KCl)/(mol soln)=(.550)/(.550+12.5) = .55/13.05= 0.042
2
14.
a) 0.600 mol of a nonvolatile nonelectrolyte is dissolved in 3.50 mol if water, what is the VP of water when the
VP of pure water is 23.8 torr @ 25oC? X=(mol water)/(mol water+non)=(3.5)/(3.5+0.6)=0.854
PA = XAPA=(0.854)*(23.8 torr) =20.3 torr (0.027 atm)
6
b) solution has 1.40 mol cyclohexane (P= 0.128 atm) & 2.40 mol acetone (P=0.302 atm). What is the tot VP?
above soln.
Find cyclo VP: Xcyc=(1.4)/(1.4+2.4)=0.368 Pcyc=XP=(0.368)*(0.128 atm)=0.0472 atm
Find ace VP: Xace=(2.4)/(1.4+2.4)=0.632 Pace=XP=(0.632)*(0.302 atm)=0.191 atm
Ptotal = Pace + Pcyc = 0.0472 + 0.191 = 0.238 atm
c) mole fraction of vapor above soln is cyclohexane? Pcyc=XP=(0.368)*(0.128 atm)=0.0472 atm
Xcyc = (Pcyc)/(Ptot) = (0.0472 atm)/(0.238 atm) = 0.199