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?

(b) The density of titanium 4.51 g/cm3 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 g/cm3 . Calculate the mass of 0.1500 L of benzene at this temperature.

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

c) Density of benzene = 0.8787 g/cm3

mass = ? for a volume = 0.15 L

1.40)

Carry out the following operations, and express the answer

with the appropriate number of significant figures.

(a) 320.5 - (6104.52.3)

(b) [(285.3 * 105) - (1.200 * 103)] * 2.8954

(c) (0.0045 * 20,000.0) + (2813 * 12)

(d) 863 * [1255 - (3.45 * 108)]

a)

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

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 10-3mol of ammoniumphosphate?

(b) How many moles of chloride ions are in 0.2550 g of aluminumchloride?

(c) What is the mass, in grams, of 7.7 1020molecules 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)

c) mass of 7.7 1020 molecules of caffeine (C8H10N4O2 MW=194.2 g/mol)

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) SO42-

Using 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 involvedwhen 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 Cl-

4.49)Determine the oxidation number for the indicated element ineach of the following substances:

(a) S in SO2,(+4)

(b) C in COCl2,(+4)

(c) Mn in KMnO4, (+7)

(d) Br in HBrO, (+1)

(e) As in As4, (zero) Elemental state.

(f) O in K2O2.(-1) peroxide

4.59) (a) Is the concentration of a solution an intensive or an extensiveproperty?

(b) What is the difference between 0.50 molHCl 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 molZnCl2 in exactly 150 mL of solution.

(b) How many moles ofHNO3 are present in 35.0 mL of a 4.50 M solution of nitric acid?

(c) How many milliliters of 6.00 MNaOH solution are neededto provide 0.325 mol of NaOH?

Solution

4.69) (a) Which will have the highest concentration of potassiumion: 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 MK2CrO4 or 25.0 mL of 0.080 MK3PO4?

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 thefollowing fashion:

CH3COOH (aq) + NaOH(aq)→ H2O(l) + NaC2H3O2 (aq)

If 3.45 mL of vinegar needs 42.5 mL of 0.115 M NaOH toreach the equivalence point in a titration, how many grams ofacetic acid are in a 1.00 qt sample of this vinegar?

Solution:

At end point: 3.45 mLVinegar (Acetic acid) CH3COOHneutralizes 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

4.87) A solution of 100.0 mL of 0.200 M KOH is mixed with a solutionof 200.0 mL of 0.150 M NiSO4.

(a) Write the balancedchemical equation for the reaction that occurs. (b) What precipitateforms?

(c) What is the limiting reactant?

(d) Howmany 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

Moles of KOH present = 0.02 mol

Moles of Ni(OH)2present = 0.03 mol.

d) Each 1 mole of Ni(OH)2needs 2 moles of KOH. The quantity 0.03 mol Ni(OH)2needs 0.06 mol KOH (more than initially present). So, KOH is the limiting reagent.

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