Dissociation of Weak Acid and Base Problems

Some problems are solved using both approximation and the quadratic equation. Values in the parenthesis are obtained using the quadratic equation.

Approximation is made by assuming that the amount of dissociated weak acid or base is negligible such that the given amount of undissociated acid or base remains practically the same.

When approximation introduces large % error, the quadratic equation alone is used as indicated in the Answer part of the problem.

For convenience, problems are grouped into aqueous solutions containing:

a weak acid or a weak base

a strong acid or a strong base

a salt of a weak acid or a salt of a weak base

a weak acid and its salt (weak acid/conjugate base composition) or a weak base and its salt (weak base/conjugate acid composition)

For a more detailed discussion of this topic and other chemistry topics, go to www.chemistrypartner.blogspot.com.

Aqueous Solution of a Weak Acid or a Weak Base

The pH of an aqueous solution containing a weak acid or a weak base is calculated from their dissociation constant equations shown below:

1.	If a solution of HCOOH has a concentration of 0.49 M, what is its pH? Ka for formic acid is 1.8 x 10-4.
	Answer: pH = 2.03 ( 2.03 )
2.	What is the pH of an aqueous solution of CH3(CH2)2CH2NH2 having a concentration of 0.04 M? The dissociation constant of 1-butylamine is 4 x 10-4.
	Answer: pH = 11.58 (Quadratic equation is used.)
3.	What is the pH of a 0.85 M HNO2 solution. ( Ka = 4.5 x 10-4 )
	Answer: pH = 1.71; ( 1.71)
4.	What is the pH of an aqueous solution of (CH3)2NH having a concentration of 0.08 M? The dissociation constant of dimethylamine is 5.9 x 10-4.
	Answer: pH = 11.84 ( 11.82)
5.	What is the pH of a 0.13 M HC2H2ClO2 solution. ( Ka = 1.36 x 10-3 )
	Answer: pH = 1.9 (Quadratic equation is used.)

Aqueous Solution of a Strong Acid or a Strong Base

The [H₃O⁺] or pH of an aqueous solution containing a strong acid or base is calculated using the ion-product constant of water, K_w.

K_w = [H₃O⁺] [OH^-]

1 x 10^-14 = [H₃O⁺] [OH^-]

6.	Find the [H3O+] and [OH-] of a 0.48 M of HCN solution. Ka = 6.2 x 10-10.
	Answer: [H3O+] = 1.73 x 10-5 M; [OH-] = 5.8 x 10-10 M ( 1.73 x 10-5; 5.8 x 10-10 )
7.	A one liter aqueous solution contains 3.37 g KOH. What is the pH of this solution? The dissociation constant for water, Kw, is 1 x 10-14. ( KOH = 56.1056 g )
	Answer: pH = 12.78
8.	A one liter aqueous solution contains 6 g NaOH. What is the pH of this solution? The dissociation constant for water, Kw, is 1 x 10-14. ( NaOH = 39.99707 g )
	Answer: pH = 13.18

Aqueous Solution of a Salt of a Weak Acid or a Salt of a Weak Base

The general relationship of the conjugate acid/base pairs is used to solve problems 9 to 14.

This relationship is expressed as follows:

K_aK_b = K_w

K_aK_b = 1 x 10^-14

9.	Calculate the pH of a 0.71 M NaC6H5COO solution. (Ka for benzoic acid is 6.3 x 10-5; Kw = 1 x 10-14)
	Answer: pH = 9.03 ( 9.03 )
10.	What is the pH of a solution of 0.36 M NaF? Calculate the percent hydrolysis. (Ka = 6.7 x 10-4; Kw = 1 x 10-14)
	Answer: pH = 8.37; 6.44 x 10-4 % ( 8.37; 6.44 x 10-4 % )
11.	Calculate the dissociation constant of the conjugate base of HC2H2ClO2. Ka = 1.36 x 10-3.
	Answer: 7.35 x 10-12
12.	What is the pH of an aqueous solution of 0.1 M CH3CH2NH3Cl? ( Given: Kb = 4.28 x 10-4; Kw = 1 x 10-14)
	Answer: pH = 5.82 ( 5.82 )
13.	What is the dissociation constant of the conjugate acid of NH3. Kb = 1.8 x 10-5.
	Answer: 5.56 x 10-10
14.	Find the pH of a 0.49 M NaCH3COO solution. (Ka for acetic acid is 1.8 x 10-5; Kw = 1 x 10-14)
	Answer: pH = 9.22 ( 9.22 )

Aqueous Solution of a Weak Acid And Its Salt or a Weak Base And Its Salt

Calculation of [H₃O⁺] or pH of a buffer solution is accomplished using the dissociation constant equation of a weak acid or a weak base.

For an aqueous solution containing a weak acid and its salt:

Inverting the last term of the equation gives us: The last equation is the general form of the Henderson-Hasselbach equation.

15.	270 mL of 0.03 M NaOH solution is added to 270 mL of 0.04 M HF solution. Find the pH of the resulting solution. Given: K for hydrofluoric acid is 6.7 x 10-4.
	Answer: pH = 3.65 ( 3.68)
16.	If 3 L of 0.02 M CH3NH2 solution has a pH of 10.41, find the weight of CH3NH3Cl dissolved in the solution. ( Kb for methylamine is 4.8 x 10-4; mol. wt.methylamine = 67.5181 g/mol )
	Answer: 7.57 g
17.	What is the pH of an aqueous solution having a concentration of 0.03 M HF and 0.02 M NaF? ( Ka = 6.7 x 10-4)
	Answer: pH = 3.00 ( 3.03 )
18.	An aqueous solution of 0.11 M HNO2 and 0.11 M NaNO2 has a volume of 530 mL. If 1.2 g of NaOH is added to this solution, find the change in pH. Assume no change in the volume of solution. ( Ka = 4.5 x 10-4; NaOH = 39.99707 g )
	Answer: pH = 3.84 ( 3.84)
19.	An aqueous solution has a strength of 0.07 M HF and 0.04 M NaF. Calculate the pH of this solution. ( Ka = 6.7 x 10-4)
	Answer: pH = 2.93 ( 2.95 )
20.	In a buffer solution of 0.07 M C6H5COOH and 0.07 M NaC6H5COOH, HCl is added to it such that its calculated concentration in the solution is 0.06 M. Assuming a constant volume of solution, find the change in pH of the solution. ( Ka = 6.3 x 10-5)
	Answer: pH = 3.09 ( 3.12 )

Colligative Properties of Dilute Solutions of Non-volatile Solute

Together with the freezing point depression, boiling point elevation and osmotic pressure are properties of dilute solutions that depend on the amount of solute present. The relationship of boiling point elevation with the amount of solute present is expressed by the equation: For osmotic pressure, the equation is as follows:

Boiling Point Elevation Problems

1.	If a solution of 163.65 g of an unknown compound in 1330 g of benzene has a boiling point of 82.75° C, what is the molecular weight of the solute? The boiling point of benzene is 80.2° C. (Kb = 2.53°C-kg/mol)
	Answer: 122.08 g/mol
2.	Determine the boiling point of a benzoic acid solution containing 151.86 g of benzoic acid and 1.09 kg of benzene. Assume standard atmospheric pressure. (mol. wt. of benzoic acid = 122.1232 ; Kb = 2.53°C-kg/mol; benzeneTb = 80.2° C)
	Answer: 83.09° C
3.	An aqueous solution of succinic acid has a concentration of 3.09 molal. Calculate the approximate boiling point of the solution. Water boils at 100°C at standard atmospheric pressure. Kb for water is 0.512°C-kg/mol.
	Answer: 101.58° C
4.	Determine the boiling point of a beta-naphthol solution containing 282.83 g of beta-naphthol and 1.34 kg of ethanol. Assume standard atmospheric pressure. (mol. wt. of beta-naphthol = 144.1726 ; Kb = 1.22°C-kg/mol; ethanolTb = 78.5° C)
	Answer: 80.29° C

5.	Find the boiling point of a solution of 63.76 g alpha-naphthol in 1930 g benzene. The boiling point of benzene is 80.2° C. (mol. wt. of alpha-naphthol = 144.1726 ; Kb = 2.53°C-kg/mol)
	Answer: 80.78° C
6.	Determine the boiling point of an alpha-naphthol solution containing 97.62 g of alpha-naphthol and 0.29 kg of ethanol. Assume standard atmospheric pressure. (mol. wt. of alpha-naphthol = 144.1726 ; Kb = 1.22°C-kg/mol; ethanolTb = 78.5° C)
	Answer: 81.35° C
7.	Determine the boiling point of an aniline solution containing 172.76 g of aniline and 5.59 kg of benzene. Assume standard atmospheric pressure. (mol. wt. of aniline = 93.128 ; Kb = 2.53°C-kg/mol; benzeneTb = 80.2° C)
	Answer: 81.04° C
8.	Find the boiling point of a solution of 63.29 g glucose in 140 g ethanol. The boiling point of ethanol is 78.5° C. (mol. wt. of glucose = 180.1572 ; Kb = 1.22°C-kg/mol)
	Answer: 81.56° C

9.	If a solution of 151.33 g of an unknown compound in 770 g of ethanol has a boiling point of 79.83° C, what is the molecular weight of the solute? The boiling point of ethanol is 78.5° C. (Kb = 1.22°C-kg/mol)
	Answer: 180.28 g/mol
10.	Determine the boiling point of a glucose solution containing 396.10 g of glucose and 2.71 kg of ethanol. Assume standard atmospheric pressure. (mol. wt. of glucose = 180.1572 ; Kb = 1.22°C-kg/mol; ethanolTb = 78.5° C)
	Answer: 79.49° C
11.	If a solution of 223.85 g of an unknown compound in 3040 g of ethanol has a boiling point of 79.33° C, what is the molecular weight of the solute? The boiling point of ethanol is 78.5° C. (Kb = 1.22°C-kg/mol)
	Answer: 108.23 g/mol
12.	If a solution of 325.83 g of an unknown compound in 1850 g of ethanol has a boiling point of 79.99° C, what is the molecular weight of the solute? The boiling point of ethanol is 78.5° C. (Kb = 1.22°C-kg/mol)
	Answer: 144.21 g/mol

13.	Determine the boiling point of a succinic acid solution containing 123.46 g of succinic acid and 0.34 kg of water. Assume standard atmospheric pressure. (mol. wt. of succinic acid = 118.089 ; Kb = 0.512°C-kg/mol; waterTb = 100° C)
	Answer: 101.57° C
14.	An aqueous solution of 1,4-dioxane has a concentration of 2.27 molal. Calculate the approximate boiling point of the solution. Water boils at 100°C at standard atmospheric pressure. Kb for water is 0.512°C-kg/mol.
	Answer: 101.16° C
15.	If a solution of 205.38 g of an unknown compound in 770 g of water has a boiling point of 100.76° C, what is the molecular weight of the solute? The boiling point of water is 100° C. (Kb = 0.512°C-kg/mol)
	Answer: 179.69 g/mol
16.	Determine the boiling point of a 1,4-dioxane solution containing 71.62 g of 1,4-dioxane and 0.44 kg of water. Assume standard atmospheric pressure. (mol. wt. of 1,4-dioxane = 88.106 ; Kb = 0.512°C-kg/mol; waterTb = 100° C)
	Answer: 100.95° C
17.	Find the boiling point of a solution of 253.02 g diphenyl ether in 1600 g ethanol. The boiling point of ethanol is 78.5° C. (mol. wt. of diphenyl ether = 170.2104 ; Kb = 1.22°C-kg/mol)
	Answer: 79.63° C

18.	If a solution of 116.26 g of an unknown compound in 210 g of water has a boiling point of 102.14° C, what is the molecular weight of the solute? The boiling point of water is 100° C. (Kb = 0.512°C-kg/mol)
	Answer: 132.45 g/mol
19.	An aqueous solution of glycerol has a concentration of 3.12 molal. Calculate the approximate boiling point of the solution. Water boils at 100°C at standard atmospheric pressure. Kb for water is 0.512°C-kg/mol.
	Answer: 101.6° C
20.	Find the boiling point of a solution of 255.1 g pyrene in 1600 g benzene. The boiling point of benzene is 80.2° C. (mol. wt. of pyrene = 202.255 ; Kb = 2.53°C-kg/mol)
	Answer: 82.19° C
21.	Determine the boiling point of a solution containing 178.17 g of benzyl alcohol and 1.14 kg of ethanol. Assume standard atmospheric pressure. (mol. wt. of benzyl alcohol = 108.1396 ; Kb = 1.22°C-kg/mol; ethanolTb = 78.5° C)
	Answer: 80.26° C

22.	Find the boiling point of a solution of 420.41 g phenanthrene in 2170 g benzene. The boiling point of benzene is 80.2° C. (mol. wt. of phenanthrene = 178.233 ; Kb = 2.53°C-kg/mol)
	Answer: 82.95° C
23.	If a solution of 431.32 g of an unknown compound in 2490 g of benzene has a boiling point of 82.66° C, what is the molecular weight of the solute? The boiling point of benzene is 80.2° C. (Kb = 2.53°C-kg/mol)
	Answer: 178.15 g/mol
24.	Find the boiling point of a solution of 91.33 g anthracene in 580 g benzene. The boiling point of benzene is 80.2° C. (mol. wt. of anthracene = 178.233 ; Kb = 2.53°C-kg/mol)
	Answer: 82.44° C
25.	Determine the boiling point of a phenol solution containing 184.34 g of phenol and 2.42 kg of benzene. Assume standard atmospheric pressure. (mol. wt. of phenol = 94.1128 ; Kb = 2.53°C-kg/mol; benzeneTb = 80.2° C)
	Answer: 82.25° C

Osmotic Pressure Problems

1.	Calculate the osmotic pressure of a 0.871 L solution containing 16.62 g of diethylene glycol at 4° C. ( mol. wt. of diethylene glycol = 106.1212; R = 0.082057 L-atm/K-mol )
	Answer: 4.09 atm
2.	A solution containing 6.48 g of a non-electrolyte in 0.539 kg of water has an osmotic pressure of 2.306 atm at 3° C. What is the molecular weight of the solute? ( R = 0.082057 L-atm/K-mol )
	Answer: 118.07 g/mol
3.	A solution containing 27.17 g of an organic solute in 738 g of water has an osmotic pressure of 9.349 atm at 12° C. What is the molecular weight of the solute? ( R = 0.082057 L-atm/K-mol )
	Answer: 92.09 g/mol

4.	0.851 L of an aqueous solution has 14.31 g 1,4-dioxane. What is the osmotic pressure of this solution at 1° C? ( mol. wt. of 1,4-dioxane = 88.106; R = 0.082057 L-atm/K-mol )
	Answer: 4.29 atm
5.	705 mL of an aqueous solution has 8.98 g glucose. What is the osmotic pressure of this solution at 9° C? ( mol. wt. of glucose = 180.1572; R = 0.082057 L-atm/K-mol )
	Answer: 1.64 atm
6.	Calculate the osmotic pressure of a 680 mL solution containing 45.39 g of 1,4-dioxane at 1° C. ( mol. wt. of 1,4-dioxane = 88.106; R = 0.082057 L-atm/K-mol )
	Answer: 17.03 atm

7.	Calculate the osmotic pressure of a 865 mL solution containing 6.53 g of sucrose at 5° C. ( mol. wt. of sucrose = 342.2992; R = 0.082057 L-atm/K-mol )
	Answer: 0.5 atm
8.	A solution containing 36.14 g of an organic compound in 0.288 kg of water has an osmotic pressure of 22.524 atm at 16° C. What is the molecular weight of the solute? ( R = 0.082057 L-atm/K-mol )
	Answer: 132.12 g/mol

9.	Calculate the osmotic pressure of a 953 cm3 solution containing 6.78 g of glycol at 2° C. ( mol. wt. of glycol = 62.0682; R = 0.082057 L-atm/K-mol )
	Answer: 2.59 atm
10.	556 mL of an aqueous solution has 29.96 g glutaric acid. What is the osmotic pressure of this solution at 16° C? ( mol. wt. of glutaric acid = 132.1158; R = 0.082057 L-atm/K-mol )
	Answer: 9.67 atm

Freezing Point Depression Problems

The freezing point depression is a colligative property of dilute solutions. It depends on the number of solute particles present in the solution. This relationship is expressed by the following equation: The structural formulas of the organic solvents and organic solutes used in this set of freezing point depression problems are given below along with their common names, derived names or IUPAC names.

1.	Calculate the weight of ethyl ether required to mix with 0.8 L of water to make an aqueous solution which has a freezing point of -1.5° C. Freezing point of water is 0° C. Kf = 1.86°C-kg/mol.
	Answer: 47.82 g
2.	Determine the freezing point depression of an aqueous solution which contains 168.64 g glycerin and 0.2 L of water. Freezing point of water is 0° C. Kf = 1.86°C-kg/mol.
	Answer: -17.03° C
3.	What is the freezing point of an aqueous solution containing 0.46 mol diethylene glycol and 170 g of water? (Given: waterTf = 0° C; Kf = 1.86°C-kg/mol.)
	Answer: -5.03° C

4.	Calculate the freezing point of a phenylethyl alcohol-benzene mixture which has 12 % phenylethyl alcohol by weight. (Given:  benzene has a freezing point of 5.45° C; Kf = 4.9°C-kg/mol.)
	Answer: -0.02° C
5.	If a solution of 38.23 g of an unknown compound in 0.12 kg benzene has a freezing point of -4.81° C, what is the molecular weight of the solute? (Given: benzeneTf = 5.45° C; Kf = 4.9°C-kg/mol.)
	Answer: 152.15 g
6.	What is the concentration in molality (m) of an aqueous solution of glycerin if the freezing point of the solution is -1.93° C. (Given: waterTf = 0° C; Kf of water is 1.86°C-kg/mol.)
	Answer: 1.04 m
7.	What is the concentration in molality (m) of a solution of bromobenzene in benzene if the freezing point of the solution is -2.36° C. (Given: benzeneTf = 5.45° C; Kf = 4.9°C-kg/mol.)
	Answer: 1.59 m

8.	Calculate the freezing point of a solution of 53.69 g methyl n-propyl ether in 0.09 kg water. Freezing point of water is 0° C. Kf = 1.86°C-kg/mol.
	Answer: -14.97° C
9.	What is the freezing point of a solution of 114.27 g glycol in 0.47 kg water? The observed melting point of a sample of pure water is 0° C. Kf for water is 1.86°C-kg/mol.
	Answer: -7.29° C
10.	Calculate the freezing point of a solution of 21.82 g ethyl ether in 0.08 kg water. Freezing point of water is 0° C. Kf = 1.86°C-kg/mol.
	Answer: -6.84° C
11.	If a solution of 60.08 g of an unknown compound in 0.39 kg toluene has a freezing point of -100.45° C, what is the molecular weight of the solute? (Given: tolueneTf = -95° C; Kf = 3.33°C-kg/mol.)
	Answer: 94.13 g
12.	If a solution of 106.33 g of an unknown compound in 0.27 kg water has a freezing point of -11.8° C, what is the molecular weight of the solute? (Given: waterTf = 0° C; Kf = 1.86°C-kg/mol.)
	Answer: 62.08 g
13.	An aqueous solution, which has a composition of 52.19 g of an organic compound and 0.16 kg water, freezes at -20.21° C. Calculate the molecular weight of the solute. (Given:  water has a freezing point of 0° C; Kf = 1.86°C-kg/mol.)
	Answer: 30.02 g

14.	Pure benzene has a freezing point of 5.45° C. When a sample of 0.15 kg of benzene was mixed with 102.94 g of an unidentified compound, the sample was observed to freeze at -11.18° C. Determine the molecular weight of the compound. (Given: Kf = 4.9°C-kg/mol.)
	Answer: 202.21 g
15.	Pure naphthalene has a freezing point of 80.22° C. When a sample of 0.59 kg of naphthalene was mixed with 315.35 g of an unidentified compound, the sample was observed to freeze at 59.68° C. Determine the molecular weight of the compound. (Given: Kf = 6.85°C-kg/mol.)
	Answer: 178.25 g
16.	If a solution of 235.48 g of an unknown compound in 5.43 kg naphthalene has a freezing point of 78.75° C, what is the molecular weight of the solute? (Given: naphthaleneTf = 80.22° C; Kf = 6.85°C-kg/mol.)
	Answer: 202.08 g
17.	Pure toluene has a freezing point of -95° C. When a sample of 0.35 kg of toluene was mixed with 237.85 g of an unidentified compound, the sample was observed to freeze at -106.19° C. Determine the molecular weight of the compound. (Given: Kf = 3.33°C-kg/mol.)
	Answer: 202.23 g

18.	What is the freezing point of an aqueous solution containing 0.95 mol glutaric acid and 180 g of water? (Given: waterTf = 0° C; Kf = 1.86°C-kg/mol.)
	Answer: -9.82° C
19.	What is the freezing point of a solution of 198.14 g xylene in 1.18 kg naphthalene? The observed melting point of a sample of pure naphthalene is 80.22° C. Kf for naphthalene is 6.85°C-kg/mol.
	Answer: 69.39° C
20.	Calculate the freezing point of a glutaric acid-water mixture which has 6 % glutaric acid by weight. (Given:  water has a freezing point of 0° C; Kf = 1.86°C-kg/mol.)
	Answer: -0.9° C

These sets of problems are about the concentrations and dilution of solutions from one concentration to another concentration.

Concentration Problems

1.	What is the molarity of a 18.0 mL solution if it contains: (a) 5.34 g of Zn(NO3)2, (b) 7.58 g of CaC2O4.H2O, (c) 4.39 g of H3BO3? (mol. wt.: Zn(NO3)2=189.3898, CaC2O4.H2O=146.1148, H3BO3=61.8319)
	Answer: (a) 1.6 M, (b) 2.9 M, (c) 3.9 M
2.	Determine the weight in grams of the solute contained in a given volume and a given concentration of the following solutions: (a) 5.0 L of 5.76 M CuSO4 solution, (b) 12.0 mL of 0.83 M Mg(OH)2 solution, (c) 24.0 mL of 2.3 M CuSCN solution. (mol. wt.: CuSO4=159.6036, Mg(OH)2=58.3196, CuSCN=121.6237)
	Answer: (a) 4596.6 g, (b) 0.6 g, (c) 6.7 g
3.	Calculate the weight of the solute required in order to prepare the following solutions: (a) 12.0 mL of 3.36 M Zn(NO3)2.6H2O solution, (b) 1.0 L of 1.35 M NaIO3 solution, (c) 6.0 L of 2.78 M NaCl solution. (mol. wt.: Zn(NO3)2.6H2O=297.481, NaIO3=197.89247, NaCl=58.44277)
	Answer: (a) 12 g, (b) 267.2 g, (c) 974.8 g
4.	What is the molarity (M) of the following solutions: (a) 33% (w/w) sodium hydroxide solution with a density of 1.36 g/mL, (b) 40% (w/w) hydrobromic acid solution with a density of 1.38 g/mL. (mol. wt.: NaOH=39.99707, HBr=80.9119)
	Answer: (a) 11.22 M, (b) 6.82 M
5.	Determine the weight in grams of the solute contained in the following solutions: (a) 10.0 mL of 0.58 M K2S2O5 solution, (b) 2.0 L of 1.78 M CaF2 solution, (c) 5.0 L of 0.48 M NaOH solution. (mol. wt.: K2S2O5=222.3136, CaF2=78.0768, NaOH=39.99707)
	Answer: (a) 1.3 g, (b) 278 g, (c) 96 g
6.	Calculate the resulting molarity (M) of CH3COO-1 from the mixture of 18.00 mL of 4.79 M NaCH3COO solution and 21.00 mL of 3.11 M Ba(CH3COO)2 solution, assuming their volumes are additive.
	Answer: 5.56 M
7.	Determine the weight in grams of the solute contained in the following solutions: (a) 10.0 mL of 0.92 M KClO3 solution, (b) 2.0 L of 1.38 M KSCN solution, (c) 4.0 L of 0.32 M ZnBr2 solution. (mol. wt.: KClO3=122.5495, KSCN=97.176, ZnBr2=225.188)
	Answer: (a) 1.1 g, (b) 268.2 g, (c) 288.2 g
8.	Calculate the final molarity (M) of a mixed solution of 23.0 mL of 4.29 M CdC2O4.3H2O solution and 16.0 mL of 1.8 M CdC2O4.3H2O solution, assuming their volumes are additive.
	Answer: 3.3 M
9.	Determine the weight in grams of the solute contained in a given volume and a given concentration of the following solutions: (a) 5.0 L of 0.13 M Ca(NO3)2.4H2O solution, (b) 14.0 mL of 5.73 M Na2CO3.H2O solution, (c) 22.0 mL of 5.89 M Na4P2O7 solution. (mol. wt.: Ca(NO3)2.4H2O=236.1506, Na2CO3.H2O=124.00394, Na4P2O7=265.9024)
	Answer: (a) 153.5 g, (b) 9.9 g, (c) 34.5 g
10.	Calculate the resulting molarity (M) of C2O4-2 from the mixture of 6.00 mL of 1.9 M K2C2O4 solution and 9.00 mL of 4.78 M BaC2O4 solution, assuming their volumes are additive.
	Answer: 3.6 M

Dilution Problems

1.	What is the volume of the diluted solution of 25.0 mL of 4.77 M Mg(NO3)2.6H2O if its final concentration is 0.41 M Mg(NO3)2.6H2O?
	Answer: 290.9 mL
2.	What is the volume of the diluted solution of 17.0 mL of 3.91 M AgNO3 if its final concentration is 1.4 M AgNO3?
	Answer: 47.5 mL
3.	14 mL of an unknown concentration of Co(NO3)2.6H2O solution was diluted to 21.9 mL to make a concentration of 1.46 M Co(NO3)2.6H2O. Determine the initial concentration of the diluted solution.
	Answer: 2.28 M
4.	36.3 mL of 2.62 M K2CrO4 solution was prepared from an unknown volume of 4.32 M K2CrO4 solution. What was the initial volume of the solution?
	Answer: 22.0 mL
5.	What is the volume of the diluted solution of 18.0 mL of 1.56 M NaClO3 if its final concentration is 1.44 M NaClO3?
	Answer: 19.5 mL
6.	What is the volume of the diluted solution of 10.0 mL of 2.27 M KIO3 if its final concentration is 2.07 M KIO3?
	Answer: 11 mL
7.	If 7.0 mL of 3.5 M NaHCO3 solution is diluted to 35.0 mL, calculate the new molarity of the solution.
	Answer: 0.7 M
8.	18.9 mL of 2.34 M KI solution was prepared from an unknown volume of 4.03 M KI solution. What was the initial volume of the solution?
	Answer: 11.0 mL
9.	5 mL of an unknown concentration of K2SO4 solution was diluted to 19.4 mL to make a concentration of 0.7 M K2SO4. Determine the initial concentration of the diluted solution.
	Answer: 2.72 M
10.	327 mL of 0.1 M NaNO2 solution was prepared from an unknown volume of 5.45 M NaNO2 solution. What was the initial volume of the solution?
	Answer: 6.0 mL

This set of problems is still about the different gas laws: Boyle's Law, Charles' Law, Gay-Lussac's Law, Combined Gas , Ideal Gas Law. This time I included problems about the van der Waals equation of state. This is the van der Waals equation of state: The equation takes into account what the Ideal Gas Law does not: the intermolecular attraction of gas molecules (this is the correction factor for pressure) and the collective volume occupied by the individual gas molecules (this is the correction factor for volume).

Boyle's Law Problems

1.	Calculate the pressure required to reduce the volume of 8.0 L of gas at a pressure of 1.306 atm to 14.0 L with the temperature remaining constant.
	Answer: 0.746 atm
2.	Under constant temperature, 3.0 L of a sample gas was allowed to expand to a final volume of 18.0 L. If the initial pressure of the gas was 0.815 atm, what was the final pressure of the gas after its expansion?
	Answer: 0.136 atm
3.	A sample of a gas which occupies 4.1 L has a pressure of 3 atm. Determine the resulting volume of the gas if the pressure is changed to 11 atm and the temperature is maintained constant.
	Answer: 1.12 L
4.	At a pressure of 9.192 atm, a sample of gas has a volume of 0.304 L. If the pressure is reduced to 3.446 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 0.811 L
5.	At a pressure of 8.75 atm, a sample of gas has a volume of 0.685 L. If the pressure is reduced to 5.5 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 1.09 L
6.	Calculate the pressure required to reduce the volume of 4000 mL of gas at a pressure of 1139 torr to 3000 mL with the temperature remaining constant.
	Answer: 1519 torr
7.	Under constant temperature, 9000 mL of a sample gas was allowed to expand to a final volume of 24000 mL. If the initial pressure of the gas was 5104 torr, what was the final pressure of the gas after its expansion?
	Answer: 1914 torr
8.	A sample of a gas which occupies 1.74 L has a pressure of 6 atm. Determine the resulting volume of the gas if the pressure is changed to 7 atm and the temperature is maintained constant.
	Answer: 1.49 L
9.	Calculate the pressure required to reduce the volume of 8000 mL of gas at a pressure of 454 torr to 6000 mL with the temperature remaining constant.
	Answer: 605 torr
10.	Under constant temperature, 9.0 L of a sample gas was allowed to expand to a final volume of 25.0 L. If the initial pressure of the gas was 31.417 atm, what was the final pressure of the gas after its expansion?
	Answer: 11.31 atm

Charles' Law Problems

1.	Calculate the final volume of a 3.52 L gas at 79° C when the temperature is brought down to 26° C as the pressure remains unchanged.
	Answer: 2.99 L
2.	A sample of gas with a volume of 12.0 L at 425° C was allowed to expand to 17.0 L while maintaining a constant pressure. What was the new temperature of the gas?
	Answer: 716° C
3.	A sample of gas with a volume of 3.0 L at -255° C was allowed to expand to 23.0 L while maintaining a constant pressure. What was the new temperature of the gas?
	Answer: -135° C
4.	Maintaining a constant pressure, a sample of gas having a volume of 10.0 L at -130° C is heated to 85° C. What is the final volume of the gas?
	Answer: 25.0 L
5.	22 L of gas at -119° C was compressed to 13 L at constant pressure. Determine the resulting temperature of the gas.
	Answer: -182° C
6.	A sample of gas with a volume of 7.0 L at -224° C was allowed to expand to 19.0 L while maintaining a constant pressure. What was the new temperature of the gas?
	Answer: -140° C
7.	Calculate the final volume of a 50.98 L gas at 85° C when the temperature is brought down to 20° C as the pressure remains unchanged.
	Answer: 41.72 L
8.	20 L of gas at -150° C was compressed to 6 L at constant pressure. Determine the resulting temperature of the gas.
	Answer: -236° C
9.	Maintaining a constant pressure, a sample of gas having a volume of 7.0 L at -224° C is heated to -70° C. What is the final volume of the gas?
	Answer: 29.0 L
10.	Calculate the final volume of a 5.98 L gas at 75° C when the temperature is brought down to 15° C as the pressure remains unchanged.
	Answer: 4.95 L

Gay-Lussac's Law Problems

1.	A sample of a gas sealed in a vessel was determined to have a pressure of 1.091 atm at 25° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 72° C?
	Answer: 1.355 atm
2.	A gas cylinder containing a certain gas is found to have a pressure of 0.886 atm at 41° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 10° C, calculate the resulting pressure inside the cylinder.
	Answer: 0.858 atm
3.	A sample of a gas has a pressure of 13 atm and a temperature of 117° C. Find the corresponding change in the temperature if the pressure is (a) reduced to 6 atm, (b) tripled. Assume the volume of the gas is constant.
	Answer: (a) -93° C, (b) 897° C
4.	A sample of a gas sealed in a vessel was determined to have a pressure of 2.223 atm at 25° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 59° C?
	Answer: 2.663 atm
5.	A gas cylinder containing a certain gas is found to have a pressure of 10.461 atm at 41° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 25° C, calculate the resulting pressure inside the cylinder.
	Answer: 9.628 atm
6.	A sample of a gas has a pressure of 16 atm and a temperature of -238° C. Find the corresponding change in the temperature if the pressure is (a) reduced to 5 atm, (b) tripled. Assume the volume of the gas is constant.
	Answer: (a) -262° C, (b) -168° C
7.	A sample of a gas sealed in a vessel was determined to have a pressure of 0.911 atm at 8° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 77° C?
	Answer: 1.161 atm
8.	A gas cylinder containing a certain gas is found to have a pressure of 1.298 atm at 81° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 29° C, calculate the resulting pressure inside the cylinder.
	Answer: 1.192 atm
9.	A sample of a gas has a pressure of 8 atm and a temperature of 841° C. Find the corresponding change in the temperature if the pressure is (a) reduced to 3 atm, (b) tripled. Assume the volume of the gas is constant.
	Answer: (a) 145° C, (b) 3069° C
10.	A gas cylinder containing a certain gas is found to have a pressure of 35.133 atm at 75° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 6° C, calculate the resulting pressure inside the cylinder.
	Answer: 34.527 atm

Combined Gas Law Problems

1.	21.73 L of a sample of gas has a pressure of 3.66 atm and a temperature of 20° C. After compressing the volume down to 1.1 L and increasing the temperature to 43° C, calculate the change in the pressure of the gas.
	Answer: 77.977 atm
2.	148.39 L of a sample of gas has a pressure of 1.07 atm and a temperature of 20° C. After compressing the volume down to 11.21 L and increasing the temperature to 59° C, calculate the change in the pressure of the gas.
	Answer: 16.049 atm
3.	An expandable air balloon containing 67.9 L of a sample of a gas has a pressure of 2.507 atm at 76° C. What will be the resulting volume of the gas if the pressure is changed to 10 atm and the temperature is increased by 10° C?
	Answer: 17.5 L
4.	21.28 L of a sample of gas has a pressure of 3.64 atm and a temperature of 18° C. After compressing the volume down to 5.8 L and increasing the temperature to 69° C, calculate the change in the pressure of the gas.
	Answer: 15.696 atm
5.	What will be the final temperature of a 12.12 L gas at 29° C if its volume is changed to 24.005 L and its pressure changed from 3.775 atm to 1.855 atm?
	Answer: 21° C
6.	What will be the final temperature of a 64.592 L gas at 18° C if its volume is changed to 11.988 L and its pressure changed from 3.637 atm to 13.676 atm?
	Answer: -70° C
7.	16.06 L of a sample of gas has a pressure of 3.6 atm and a temperature of 15° C. After compressing the volume down to 3.19 L and increasing the temperature to 59° C, calculate the change in the pressure of the gas.
	Answer: 20.893 atm
8.	An expandable air balloon containing 159.1 L of a sample of a gas has a pressure of 1.07 atm at 54° C. What will be the resulting volume of the gas if the pressure is changed to 9 atm and the temperature is increased by 31° C?
	Answer: 20.7 L
9.	What will be the final temperature of a 27.361 L gas at 14° C if its volume is changed to 5.121 L and its pressure changed from 1.053 atm to 15.409 atm?
	Answer: 513° C
10.	An expandable air balloon containing 95.8 L of a sample of a gas has a pressure of 1.301 atm at 82° C. What will be the resulting volume of the gas if the pressure is changed to 6 atm and the temperature is increased by 15° C?
	Answer: 21.7 L

Ideal Gas Law Problems

1.	Given a weight of 9.89 g of each of the following gases: (a) F2, (b) Ar, (c) O2, calculate the volume of each gas at STP. (mol. wt.: F2=37.9968, Ar=39.948, O2=31.9988)
	Answer: (a) 5.83 L, (b) 5.55 L, (c) 6.92 L
2.	Given a volume of 13.79 L of the following gases at STP : (a) Ne, (b) O2, (c) N2, find the weight in grams of each gas. (mol. wt.: Ne=20.179, O2=31.9988, N2=28.0134)
	Answer: (a) 12.42 g, (b) 19.7 g, (c) 17.25 g
3.	Calculate the volume of 7.055 mol of a gas confined in a sealed vessel with a pressure of 3.381 atm and a temperature of 19° C.
	Answer: 49.998 L
4.	A 39.886 L tank containing a sample of gas has a pressure of 4.059 atm and a temperature of 24° C. Find the moles of gas contained inside the tank.
	Answer: 6.643 mol
5.	Given a weight of 9.36 g of each of the following gases: (a) He, (b) N2O, (c) N2, calculate the volume of each gas at STP. (mol. wt.: He=4.0026, N2O=44.0128, N2=28.0134)
	Answer: (a) 52.38 L, (b) 4.76 L, (c) 7.48 L
6.	Given a volume of 12.48 L of the following gases at STP : (a) CH4, (b) Ne, (c) NH3, find the weight in grams of each gas. (mol. wt.: CH4=16.0426, Ne=20.179, NH3=17.0304)
	Answer: (a) 8.94 g, (b) 11.24 g, (c) 9.49 g
7.	A 13.818 L tank containing a sample of gas has a pressure of 9.377 atm and a temperature of 13° C. Find the moles of gas contained inside the tank.
	Answer: 5.521 mol
8.	Calculate the volume of 0.693 mol of a gas confined in a sealed vessel with a pressure of 1.638 atm and a temperature of 15° C.
	Answer: 9.998 L
9.	A 36.344 L tank containing a sample of gas has a pressure of 1.181 atm and a temperature of 73° C. Find the moles of gas contained inside the tank.
	Answer: 1.512 mol
10.	Calculate the volume of 1.064 mol of a gas confined in a sealed vessel with a pressure of 13.184 atm and a temperature of 29° C.
	Answer: 2 L

Van Der Waals Equation Problems

1.	What is the pressure inside a 13.00 L capacity pressure tank containing 8.49 mol of NH3 at 21° C using the (a) Ideal Gas equation, (b) the Van Der Waals equation? (Given: a = 4.17 L2-atm/mol2, b = 0.0371 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: (a) 15.76 atm; (b) 14.37 atm
2.	Determine the volume occupied by 2.2 mol NH3 at 5.596 atm and 68° C as predicted by the (a) Ideal Gas equation, (b) Van Der Waals equation. (Given: a = 4.17 L2-atm/mol2, b = 0.0371 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: (a) 11.00 L, (b) 10.76 L
3.	Calculate the amount in moles of CO2 which has a volume of 9.00 L at 118.718 atm and 17° C using Van Der Waals equation. (Given: a = 3.59 L2-atm/mol2, b = 0.0427 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: 61.93 mol
4.	How many moles of CO will occupy a volume of 5.00 L at 34.324 atm and 75° C? Use the Van Der Waals equation. (Given: a = 1.49 L2-atm/mol2, b = 0.0399 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: 6.08 mol
5.	0.18 mol of N2 occupies a volume of 18.00 L at 28° C. Calculate the pressure of the gas using (a) Ideal Gas equation, (b) Van Der Waals equation. (Given: a = 1.39 L2-atm/mol2, b = 0.0391 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: (a) 0.25 atm; (b) 0.25 atm
6.	What volume will be occupied by 0.58 mol of HBr at 1.263 atm and 72° C? Calculate the volume using (a) Ideal Gas equation, (b) Van Der Waals equation. (Given: a = 4.45 L2-atm/mol2, b = 0.0443 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: (a) 13.00 L, (b) 12.94 L
7.	What is the pressure inside a 22.00 L capacity pressure tank containing 122.36 mol of O2 at 20° C using the (a) Ideal Gas equation, (b) the Van Der Waals equation? (Given: a = 1.36 L2-atm/mol2, b = 0.0318 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: (a) 133.72 atm; (b) 120.38 atm
8.	17.18 mol of CO occupies a volume of 21.00 L at 42° C. Calculate the pressure of the gas using (a) Ideal Gas equation, (b) Van Der Waals equation. (Given: a = 1.49 L2-atm/mol2, b = 0.0399 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: (a) 21.15 atm; (b) 20.86 atm
9.	Determine the volume occupied by 39.91 mol CO2 at 119.534 atm and 19° C as predicted by the (a) Ideal Gas equation, (b) Van Der Waals equation. (Given: a = 3.59 L2-atm/mol2, b = 0.0427 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: (a) 8.00 L, (b) 6.28 L
10.	How many moles of O2 will occupy a volume of 21.00 L at 41.694 atm and 38° C? Use the Van Der Waals equation. (Given: a = 1.36 L2-atm/mol2, b = 0.0318 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: 35.36 mol
11.	Calculate the amount in moles of Ne which has a volume of 16.00 L at 33.865 atm and 57° C using Van Der Waals equation. (Given: a = 0.211 L2-atm/mol2, b = 0.0171 L/mol, R = 0.082057 L-atm/mol-K)
	Answer: 19.77 mol

These current sets of problems are about the gas laws, namely, Boyle's Law, Charles' Law, Gay-Lussac's Law, Combined Gas Law, and the Ideal Gas Law.

Boyle's Law Problems

1.	Calculate the pressure required to reduce the volume of 25.0 L of gas at a pressure of 0.594 atm to 22.0 L with the temperature remaining constant.
	Answer: 0.675 atm
2.	Under constant temperature, 15.0 L of a sample gas was allowed to expand to a final volume of 19.0 L. If the initial pressure of the gas was 0.149 atm, what was the final pressure of the gas after its expansion?
	Answer: 0.118 atm
3.	A sample of a gas which occupies 20 L has a pressure of 5 atm. Determine the resulting volume of the gas if the pressure is changed to 7 atm and the temperature is maintained constant.
	Answer: 14.29 L
4.	A sample of a gas which occupies 2.5 L has a pressure of 4 atm. Determine the resulting volume of the gas if the pressure is changed to 7 atm and the temperature is maintained constant.
	Answer: 1.43 L
5.	At a pressure of 6.125 atm, a sample of gas has a volume of 0.457 L. If the pressure is reduced to 2 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 1.4 L
6.	Calculate the pressure required to reduce the volume of 17000 mL of gas at a pressure of 670 torr to 9000 mL with the temperature remaining constant.
	Answer: 1266 torr
7.	Under constant temperature, 4000 mL of a sample gas was allowed to expand to a final volume of 19000 mL. If the initial pressure of the gas was 2850 torr, what was the final pressure of the gas after its expansion?
	Answer: 600 torr

Charles' Law Problems

1.	Maintaining a constant pressure, a sample of gas having a volume of 2.0 L at -197° C is heated to 749° C. What is the final volume of the gas?
	Answer: 26.9 L
2.	Calculate the final volume of a 0.05 L gas at 98° C when the temperature is brought down to 15° C as the pressure remains unchanged.
	Answer: 0.04 L
3.	Calculate the final volume of a 9.28 L gas at 79° C when the temperature is brought down to 10° C as the pressure remains unchanged.
	Answer: 7.46 L
4.	18 L of gas at 410° C was compressed to 2 L at constant pressure. Determine the resulting temperature of the gas.
	Answer: -197° C
5.	Calculate the final volume of a 4.58 L gas at 74° C when the temperature is brought down to 5° C as the pressure remains unchanged.
	Answer: 3.67 L
6.	A sample of gas with a volume of 14.0 L at 325° C was allowed to expand to 18.0 L while maintaining a constant pressure. What was the new temperature of the gas?
	Answer: 496° C
7.	22 L of gas at -266° C was compressed to 4 L at constant pressure. Determine the resulting temperature of the gas.
	Answer: -272° C
8.	Maintaining a constant pressure, a sample of gas having a volume of 12.0 L at -28° C is heated to 455° C. What is the final volume of the gas?
	Answer: 35.6 L
9.	Calculate the final volume of a 17.43 L gas at 83° C when the temperature is brought down to 15° C as the pressure remains unchanged.
	Answer: 14.1 L
10.	25 L of gas at 237° C was compressed to 10 L at constant pressure. Determine the resulting temperature of the gas.
	Answer: -69° C

Gay-Lussac's Law Problems

1.	A sample of a gas sealed in a vessel was determined to have a pressure of 0.993 atm at 1° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 72° C?
	Answer: 1.254 atm
2.	A gas cylinder containing a certain gas is found to have a pressure of 11.2 atm at 63° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 15° C, calculate the resulting pressure inside the cylinder.
	Answer: 10.7 atm
3.	A sample of a gas sealed in a vessel was determined to have a pressure of 0.961 atm at 14° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 64° C?
	Answer: 1.175 atm
4.	A gas cylinder containing a certain gas is found to have a pressure of 4 atm at 47° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 12° C, calculate the resulting pressure inside the cylinder.
	Answer: 3.85 atm
5.	A sample of a gas has a pressure of 8 atm and a temperature of -29° C. Find the corresponding change in the temperature if the pressure is (a) reduced to 3 atm, (b) tripled. Assume the volume of the gas is constant.
	Answer: (a) -182° C, (b) 459° C
6.	A sample of a gas has a pressure of 8 atm and a temperature of 527° C. Find the corresponding change in the temperature if the pressure is (a) reduced to 5 atm, (b) tripled. Assume the volume of the gas is constant.
	Answer: (a) 227° C, (b) 2127° C
7.	A sample of a gas has a pressure of 11 atm and a temperature of -164° C. Find the corresponding change in the temperature if the pressure is (a) reduced to 2 atm, (b) tripled. Assume the volume of the gas is constant.
	Answer: (a) -253° C, (b) 54° C

Combined Gas Law Problems

1.	An expandable air balloon containing 57.6 L of a sample of a gas has a pressure of 1.341 atm at 93° C. What will be the resulting volume of the gas if the pressure is changed to 8 atm and the temperature is increased by 28° C?
	Answer: 10.4 L
2.	81.28 L of a sample of gas has a pressure of 0.72 atm and a temperature of 19° C. After compressing the volume down to 8.42 L and increasing the temperature to 83° C, calculate the change in the pressure of the gas.
	Answer: 8.47 atm
3.	What will be the final temperature of a 89.817 L gas at 16° C if its volume is changed to 6.422 L and its pressure changed from 1.06 atm to 4.958 atm?
	Answer: -176° C
4.	9.2 L of a sample of gas has a pressure of 3.03 atm and a temperature of 30° C. After compressing the volume down to 3.89 L and increasing the temperature to 55° C, calculate the change in the pressure of the gas.
	Answer: 7.76 atm
5.	What will be the final temperature of a 154.139 L gas at 28° C if its volume is changed to 24.467 L and its pressure changed from 0.797 atm to 15.435 atm?
	Answer: 652° C

Ideal Gas Law Problems

1.	A 40.116 L tank containing a sample of gas has a pressure of 3.393 atm and a temperature of 64° C. Find the moles of gas contained inside the tank.
	Answer: 4.922 mol
2.	Calculate the volume of 1.232 mol of a gas confined in a sealed vessel with a pressure of 3.916 atm and a temperature of 65° C.
	Answer: 8.726 L
3.	Given a volume of 20.86 L of the following gases at STP : (a) NH3, (b) H2S, (c) Ne, find the weight in grams of each gas. (mol. wt.: NH3=17.0304, H2S=34.0758, Ne=20.179)
	Answer: (a) 15.86 g, (b) 31.73 g, (c) 18.79 g
4.	Given a weight of 2.84 g of each of the following gases: (a) CH4, (b) O2, (c) F2, calculate the volume of each gas at STP. (mol. wt.: CH4=16.0426, O2=31.9988, F2=37.9968)
	Answer: (a) 3.97 L, (b) 1.99 L, (c) 1.67 L

Formula Calculation Problems

1.	Give the name of the following compounds: (a) Li2SO4, (b) CaCl2, (c) Na2HPO4.2H2O, (d) KHSO4, (e) PbF2, (f) NH4Br, (g) MnSO4.H2O
	Answer: (a) Lithium sulfate, (b) Calcium chloride, (c) Sodium hydrogen phosphate dihydrate, (d) Potassium hydrogen sulfate (e) Plumbous fluoride or Lead(II) fluoride (f) Ammonium bromide (g) Manganese sulfate monohydrate
2.	Give the formula of the following compounds: (a) Magnesium chloride, (b) Barium peroxodisulfate tetrahydrate, (c) Lithium hydroxide monohydrate, (d) Ammonium bromide, (e) Nickel chloride, (f) Lithium bromide, (g) Aluminum nitrate nonahydrate
	Answer: (a) MgCl2, (b) BaS2O8.4H2O, (c) LiOH.H2O, (d) NH4Br (e) NiCl2 (f) LiBr (g) Al(NO3)3.9H2O
3.	What is the formula weights of the following compounds: (a) CaC4H4O6.2H2O, (b) Na2S.9H2O, (c) NH4NO3, (d) Sr(NO3)2, (e) NH4SCN, (f) KCl, (g) CuS
	Answer: (a) 224.1824 g , (b) 240.17634 g, (c) 80.0432 g, (d) 211.6298 g, (e) 76.116 g, (f) 74.5513 g, (g) 95.606 g
4.	Calculate the number of moles of LiNO3 that will contain 6.02 x 1023 atoms of O.
	Answer: 0.33 mol LiNO3
5.	Calculate the number of moles of UO2(NO3)2 that will contain 6.02 x 1023 atoms of O.
	Answer: 0.125 mol UO2(NO3)2
6.	Calculate the number of moles of AlK(SO4)2.12H2O that will contain 6.02 x 1023 atoms of O.
	Answer: 0.05 mol AlK(SO4)2.12H2O
7.	How many number of Co atoms are contained in 43.3 g of CoSO4?
	Answer: 1.68 x 1023 Co atoms
8.	How many number of F atoms are contained in 7.81 g of BaF2?
	Answer: 5.36 x 1022 F atoms
9.	How many number of Cr atoms are contained in 63.75 g of Hg2CrO4?
	Answer: 7.42 x 1022 Cr atoms
10.	How many number of I atoms are contained in 2.54 g of Hg2I2?
	Answer: 4.67 x 1021 I atoms
11.	Determine the number of molecules of (NH4)2S present in 59.86 g of (NH4)2S.
	Answer: 5.29 x 1023 molecules
12.	Determine the number of molecules of Rb2SO4 present in 83.38 g of Rb2SO4.
	Answer: 1.88 x 1023 molecules
13.	Determine the number of molecules of NH4Br present in 12.3 g of NH4Br.
	Answer: 7.56 x 1022 molecules
14.	Calculate the weight in grams of 2.62 mol Cd(NO3)2.4H2O.
	Answer: 808.22 g
15.	Calculate the weight in grams of 0.92 mol NH4NO3.
	Answer: 73.64 g
16.	Calculate the weight in grams of 3.8 mol NaCH3COO.
	Answer: 311.73 g
17.	Determine the mass of 8.04 x 10-9 mol of CaCl2. Calculate the number of Ca atoms in 8.04 x 10-9 mol of CaCl2.
	Answer: 8.92 x 10-7 g; 4.84 x 1015 atoms
18.	Determine the mass of 10 x 10-3 mol of MgSO4.7H2O. Calculate the number of O atoms in 10 x 10-3 mol of MgSO4.7H2O.
	Answer: 2.46 g; 6.62 x 1022 atoms
19.	Determine the mass of 9.27 x 10-8 mol of KNO3. Calculate the number of O atoms in 9.27 x 10-8 mol of KNO3.
	Answer: 9.37 x 10-6 g; 1.67 x 1017 atoms
20.	How many grams of each of the constituent elements are contained in 1 mol of: (a) LiI, (b) Li2SO4.H2O, (c) Rb2SO4, (d) CaC4H4O6.2H2O, (e) CsCl, (f) CoSO4, (g) Ba(NO3)2.H2O
	Answer: (a) 1 mol Li = 6.941 g Li, 1 mol I = 126.9045 g I; (b) 2 mol H = 2.0158 g H, 5 mol O = 79.997 g O, 2 mol Li = 13.882 g Li, 1 mol S = 32.06 g S; (c) 2 mol Rb = 170.9356 g Rb, 1 mol S = 32.06 g S, 4 mol O = 63.9976 g O; (d) 8 mol H = 8.0632 g H, 8 mol O = 127.9952 g O, 1 mol Ca = 40.08 g Ca, 4 mol C = 48.044 g C; (e) 1 mol Cs = 132.9054 g Cs, 1 mol Cl = 35.453 g Cl; (f) 1 mol Co = 58.9332 g Co, 1 mol S = 32.06 g S, 4 mol O = 63.9976 g O; (g) 2 mol H = 2.0158 g H, 7 mol O = 111.9958 g O, 1 mol Ba = 137.33 g Ba, 2 mol N = 28.0134 g N
21.	Calculate the number of moles in 8.25 g CuCl2.
	Answer: 0.06 mol CuCl2
22.	Calculate the number of moles in 46.24 g NaH2PO4.
	Answer: 0.39 mol NaH2PO4
23.	Calculate the number of moles in 48.82 g Sr(NO3)2.
	Answer: 0.23 mol Sr(NO3)2
24.	Determine the number of moles in 3.49 g of Be.
	Answer: 0.39 mol Be
25.	Determine the number of moles in 68.94 g of O.
	Answer: 4.31 mol O
26.	Determine the number of moles in 98.05 g of Si.
	Answer: 3.49 mol Si
27.	Calculate the weight of one molecule of (a) Hg2CrO4, (b) FeSO4.7H2O, (c) ZnSO4.H2O, (d) Cu(OH)2, (e) MgCl2.6H2O, (f) Cu(NO3)2, (g) Al(OH)3
	Answer: (a) 8.59 x 10-22 g/molecule, (b) 4.62 x 10-22 g/molecule, (c) 2.98 x 10-22 g/molecule, (d) 1.62 x 10-22 g/molecule, (e) 3.38 x 10-22 g/molecule, (f) 3.12 x 10-22 g/molecule, (g) 1.3 x 10-22 g/molecule
28.	How much BaCO3 can be produced from 62.98 g of O?
	Answer: 258.94 g BaCO3
29.	How much MnS can be produced from 68.56 g of Mn?
	Answer: 108.57 g MnS
30.	How much Al(OH)3 can be produced from 62.1 g of Al?
	Answer: 179.53 g Al(OH)3