Some selected Problems (from chapter one to chapter four)
1.28)
(a) A cube of osmium metal 1.500 cm on a side has a mass of 76.31 g at 25 °C.What is its density in (g/cm3) at
this temperature?
3
(b) The density of titanium 4.51 g/cm metal is at 25 °C. What mass of titanium displaces 125.0 mL of water
at 25 °C?
(c) The density of benzene at 15 °C is 0.8787
temperature.
g/cm3
. Calculate the mass of 0.1500 L of benzene at this
Solution:
a) Cube side length = 1.5 cm.
Mass of the cube = 76.31 gat 25 C.
Density = ? g/cm3
b) density of (Ti) = 4.51 g/cm3
mass = ? for a quantity that displaces 125 mL
water.
Volume of Ti = volume of water displaced = 125 cm3
mass  Volume  density 125 cm 3  4.51 g / cm 3  563.75 g
c) Density of benzene = 0.8787 g/cm3
mass = ? for a volume = 0.15 L
mass  Volume  density  0.8787 150 mL 131.805 g
1.40)
Carry out the following operations, and express the answer
with the appropriate number of significant figures.
(a) 320.5 - (6104.5>2.3)
(b) [(285.3 * 105) - (1.200 * 103)] * 2.8954
(c) (0.0045 * 20,000.0) + (2813 * 12)
(d) 863 * [1255 - (3.45 * 108)]
 6104.5 
3
3
3
a) 320.5  
  320.5  2654.130435   2.7 10  -2.3795 10  -2.4 10
 2.3 
b) [(285.3  105 ) – (1.200  103) ]  2.8954 = [285.3  102 – 1.200]  103  2.8954
28528.8  103 2.8954 = 82602287.52  82602  103
c) (0.0045  20000.0) + (2813  12) = 90 + 3.4  104 = 3.4  104
d) 863  [1255-(3.45 108)] = 863  [1255-373] = 863  882 = 761166 7.61  105
-1-
2.31)
Two isotopes 63Cu and 65Cu
Average atomic mass = 0.6917  62.9296 + 0.3083  64.9278 = 63.55 amu
2.53)
Ga and F : GaF3 : Gallium fluoride.
Li and H : LiH : Lithium hydride.
Al and I : AlI3 : Aluminium iodide.
K and S : K2S: Potassium sulfide.
2.66)
KCN: Potassium cyanide.
NaBrO2 : Sodium bromita.
Sr(OH)2 : Strontium hydroxide.
CoS : Cobalt sílfide.
Fe2(CO3) : Iron (III) carbonate or, Ferric carbonate.
Cr(NO3)3 : Chromium (III) nitrate.
(NH4)2SO3 : Ammonium sulfite.
NaH2PO4 : Sodium dihydrogen phosphate.
KMnO4 : Potassium permanganate.
Ag2Cr2O7: Silver dichromate.
3.35)
(a) What is the mass, in grams, of 2.5 10-3 mol of ammonium phosphate?
(b) How many moles of chloride ions are in 0.2550 g of aluminum chloride?
20
(c) What is the mass, in grams, of 7.7 10 molecules of caffeine, C8H10N4O2?
(d) What is the molar mass of cholesterol if 0.00105 mol has a mass of 0.406 g?
a) 2.5 10-3 mole of (NH4)3PO4 (MW = 149 g/mol)
Mass = ? g
Mass = 2.5 10-3 mole  149 g/mol = 0.3725 g
b) mol of Cl- ions = ? in 0.255 g AlCl3 (MW= 133.5 g/mol)
1 mol AlCl3 3 moles Cl 
_
No. of moles Cl  0.255 g AlCl3 

 5.73 10 3 moles Cl _
133.5 g AlCl3 1 mol AlCl3
c) mass of 7.7 1020 molecules of caffeine (C8H10N4O2 MW=194.2 g/mol)
1 mol C8 H10 N 4 O 2
194.2 g
Mass of caffeine  7.7  10 20 molecule 

 0.248 g
23
6.022  10 molecule 1 mol C8 H10 N 4 O 2
-2-
4.7) Which of the following ions will always be a spectator ion in a precipitation
reaction? (a) Cl- , (b) NO3- , (c) NH4+, (d)S2-, (e) SO42Using table : (b) NO3- , (c) NH4+ are spectator ions. The other ions form ppts.
4.23) Name the spectator ions in any reactions that may be involved when each of the
following pairs of solutions are mixed.
(a) Na2CO3(aq) and MgSO4(aq)  Spectator ions are : Na+ and SO42(b) Pb(NO3)2(aq) and Na2S(aq)  Spectator ions are : Na+ and NO3(c) (NH4)3PO4(aq) and CaCl2(aq)  Spectator ions are : NH4+ and Cl4.49) Determine the oxidation number for the indicated element in each of the
following substances:
(a) S in SO2,
(b) C in COCl2,
(c) Mn in KMnO4,
(d) Br in HBrO,
(e) As in As4,
(f) O in K2O2.
(+4)
(+4)
(+7)
(+1)
(zero) Elemental state.
(-1) peroxide
4.59) (a) Is the concentration of a solution an intensive or an extensive property?
(b) What is the difference between 0.50 mol HCl and 0.50 M HCl?
a) Concentration is an intensive property.
b) The quantity of 0.5 mole HCl is equal to 18.25 g of HCl. Whereas, 0.5 M HCl
means that 18.25 g of HCl is dissolved in a quantity of water to obtain a solution of
volume = 1.0 L.
4.61) (a) Calculate the molarity of a solution that contains 0.175 mol ZnCl2 in exactly
150 mL of solution.
(b) How many moles of HNO3 are present in 35.0 mL of a 4.50 M solution of nitric
acid?
(c) How many milliliters of 6.00 M NaOH solution are needed to provide 0.325 mol
of NaOH?
Solution
0.175 mol
a) M 
1.167 M
0.150 L
b) moles HNO 3  0.035 L  4.5 M  0.16 M
c ) moles of NaOH before dilution mles of NaOH after dilution M V ( L)
No. of moles  M V ( L)  0.325 moles  6 M V ( L) V  0.0541 L
-3-
4.69) (a) Which will have the highest concentration of potassium ion: 0.20 M KCl,
0.15 M K2CrO4, or 0.080 M K3PO4?
(b) Which will contain the greater number of moles of potassium
ion: 30.0 mL of 0.15 M K2CrO4 or 25.0 mL of 0.080 M K3PO4?
Solution:
a) The solution 0.15 M K2CrO4 has the highest [K+], it contains 0.3 mol of K+ in 1
L solution.
b) The volume of 30.0 mL of 0.15 M K2CrO4 has the greater number of moles of
K+.
4.84) The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts
with sodium hydroxide in the following fashion:
CH3COOH (aq) + NaOH (aq)→ H2O (l) + NaC2H3O2 (aq)
If 3.45 mL of vinegar needs 42.5 mL of 0.115 M NaOH to reach the equivalence point
in a titration, how many grams of acetic acid are in a 1.00 qt sample of this vinegar?
Solution:
At end point: 3.45 mLVinegar (Acetic acid) CH3COOH neutralizes with 42.5 mL of
0.115 M NaOH
(MV)Acetic acid= (MV)NaOH
MAA 0.00345 L = 0.0425  0.115  MAA= 1.4167 M.
So, 1.4167 moles (or 85.002 g) of acetic acid in 1 L.
(MW AA= 60 g/mol)
1 L = 1.057 qt
85.002 g AA 1.0 L solution
No. grams in one qt 

 80.42 g
1 L solution
1.057 qt
4.87) A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL
of 0.150 M NiSO4.
(a) Write the balanced chemical equation for the reaction that occurs. (b) What
precipitate forms?
(c) What is the limiting reactant?
(d) How many grams of this precipitate form?
(e) What is the concentration
of each ion that remains in solution?
Solution
a) 2 KOH (aq) + NiSO4 (aq) → K2SO4 (aq) + Ni(OH)2 (s)
b) precipitate : Ni(OH)2
c) Limiting reagent
-4-
Moles of KOH present = 0.02 mol
Moles of Ni(OH)2 present = 0.03 mol.
d) Each 1 mole of Ni(OH)2 needs 2 moles of KOH. The quantity 0.03 mol Ni(OH)2
needs 0.06 mol KOH (more than initially present). So, KOH is the limiting reagent.
g Ni(OH) 2 produced  0.02 mol KOH 
1 mol Ni(OH) 2 92.7 g Ni(OH) 2

 0.93 g Ni(OH) 2
2 mol KOH
1 mol Ni(OH) 2
e ) Excess reagent is NiSO4
Excess Ni 2 ions  Amount initially present  reacted amount
Excess Ni 2 ions  0.03  0.01  0.02 mol Ni 2
[ Ni 2 ]excess 
and Total volume  0.3 L
0.02
 0.067 M
0.3
For SO 42 :
Moles SO 42 initially moles SO 42 finally (excess )
[ SO 42 ]excess 
0.15 M  0.2 L
 0.1 M
0 .3 L
-5-