Solutions are composed of two parts. The substance that has been dissolved is the solute. The substance in which the solute dissolves is the solvent. It is not necessary to have a liquid as a solvent, even though the term solution probably brings to mind a solid solute, like sugar, dissolved in a liquid solvent, like water. For instance, air is a solution where the solvent is N2 (nearly 80 percent of the air) in which the O2 (nearly 20 percent) is dissolved (mixed). Another example is gold used in jewelry; it is an alloy (a mixture of metals) of gold (solvent) and one or more metals (solutes), including copper and nickel.
When physical units are used, the concentrations of solutions are generally expressed in one of the following ways:
1. By number of mass units of solute per volume units of solution (as, 20 g KCl per liter solution).
2. By the percentage composition, or the number of mass units of solute per 100 mass units of solution.
EXAMPLE 1 A 10% aqueous NaCl solution contains 10 g NaCl for each 100 g solution. Ten grams NaCl are mixed with 90 g H2O to form 100 g solution.
The interesting part of this method of expressing concentration is that there is no consideration of the solvent or the solute identities. Note that 10 grams of anything that dissolves in 90 g of any solvent results in a 10% solution. The point is that the chemistry of a particular 10% solution is very likely to be different from another 10% solution because the number of molecules of solute is not the same in solutions produced and measured in this manner.
3. By the mass of solute per mass of solvent (as, 5.2 g NaCl in 100 g H2O).
4. In some areas of study, such as air and water pollution, there is an interest in very small concentrations of solute. These are frequently specified as parts per million (ppm), parts per billion (ppb), and lesser concentrations based on mass relationships.
Molar concentration (M) is defined as the number of moles solute in one liter of solution.
Molarity is a measure of concentration in which the same number of molecules of solute are found in any solution numerically the same. For instance, 1 L of a 3 molar NaCl solution contains the same number of molecules as does 1 L of a 3 molar solution of H2SO4 or 1 L of a 3 MCH3OH solution.
EXAMPLE 2 A 0.500 M H2SO4 solution contains 49.04 g H2SO4 per liter solution, since 49.04 is half the molar mass of H2SO4 (98.08 g/mol). A 1.00 M solution contains 98.08 g H2SO4 per liter solution.
Cautions: (a) M is the symbol for the quantity, the molar concentration, and M is the symbol for a unit, mol/L. The term often used for molar concentrations is molarity. There is another solution concentration called molality (alternate name for molal concentration, m). Be careful to avoid confusion associated with mispronunciation or use of the wrong symbol (M instead of m or the reverse).
The normality of a solution (N) is the number of equivalents of the solution contained in one liter of solution. The equivalent mass is that fraction of the molar mass which corresponds to the defined unit of chemical reaction, and an equivalent (eq) is this same fraction of a mole. Equivalent masses are determined as follows:
1. The defined unit of reaction for acids and bases is the neutralization reaction
The equivalent mass of an acid is that fraction of the molar mass which contains or can supply one mole of H+. A simple way of looking at the equivalent mass is that it is the mass of the acid divided by the number of H’s per molecule, assuming that complete ionization occurs.
EXAMPLE 3 The equivalent masses of HCl and HC2H3O2 are the same as their molar masses because each mole of either acid will provide one mole of H+. On the other hand, one equivalent mass of H2SO4 is one-half the molar mass, making an eq 
mol H2SO4. This works pretty well until we come across an acid that does not ionize (disassociate) well, like H3PO4. In the case of phosphoric acid, one eq can be the mass of 1 H3PO4, molar mass, or 
molar mass, depending on the extent to which it dissociates (1, 2, or 3 H+ per mole acid). One equivalent of H3BO3 is always 1 mol because only one hydrogen is replaceable in neutralization reactions. The equivalent mass of SO3 is one-half the molar mass, since SO3 can react with water to give 2H+.
There aren’t any simple rules for predicting how many hydrogens of an acid will be replaced in a given neutralization.
2. The equivalent mass of a base is that fraction of the molar mass which contains or can supply one mole of OH–, or can react with one mole of H+.
EXAMPLE 4 The equivalent masses of NaOH, NH3 (reacts with H2O to give NH+ + OH–), Mg(OH)2, and Al(OH)3 are equal to 
, , , and of their molar masses, respectively, if ionization is complete.
3. The equivalent mass of an oxidizing or reducing agent for a particular reaction is equal to its molar mass divided by the total number of moles of electrons gained or lost when the reaction of one mole occurs. A given oxidizing or reducing agent may have more than one equivalent mass, depending on the reaction in which it is used. You must determine the electrons moved during each reaction.
Equivalent masses are so defined because equal numbers of equivalents of two substances react exactly with each other. This is true for neutralization because one H+ neutralizes one OH–, and for oxidation-reduction reaction because the number of electrons lost by the reducing agent equals the number gained by the oxidation agent (electrons cannot be eliminated—by the law of conservation of matter).
EXAMPLE 5 One mole HCl, mol H2SO4, and 
mol K2Cr2O7 (as an oxidizing agent), each in 1 L of solution, give 1 N solutions of these substances. A 1 N solution of HCl is also a molar (1 M) solution. A 1 N solution of H2SO4 is also a one-half molar (0.5) solution.
Note that N is the symbol for a quantity, the normality, and N is the symbol for a unit, eq/L.
The molality of a solution is the number of moles of solute per kilogram of solvent in the solution. The molality (m) cannot be computed from the molar concentration (M) unless the density of the solution is known (see Problem 12.88).
EXAMPLE 6 A solution made up of 98.08 g H2SO4 and 1000 g H2O would be a 1.000 molal solution. (As with N and N, this book uses m for the quantity, whereas “m” is the unit, mol solute/kg solvent.)
The mole fraction (X or x, depending on the author) of any component in a solution is defined as the number of moles (n) of that component, divided by the total number of moles of all components in the solution. The sum of the mole fractions of all components of a solution in mole fractions is 1. In a two-component solution, the mole fraction of a component is calculated by
The mole percent is the same calculation as X, then multiplied by 100 to convert to percent.
Molar concentration and normality scales are useful for volumetric experiments, in which the amount of solute in a given portion of solution is related to the measured volume of solution. We will see this in later chapters, where the normality scale is very convenient for comparing the relative volumes required for two solutions to react chemically with each other (avoiding a limiting reactant situation). A limitation of the normality scale is that a given solution may have more than one normality, depending on the reaction for which it is used. As an example, different ions in a compound that is not in a 1:1 ratio, such as Ag2SO4, will have a different normality when focusing on silver than when sulfate is the important ion for consideration. On the other hand, the molar concentration of a solution is a fixed number because the molar mass of a substance does not depend on the reaction for which the substance is used.
The molality scale is useful for experiments in which physical measurements (freezing point, boiling point, vapor pressure, osmotic pressure, etc.) are made over a wide range of temperatures. The molality of a given solution, which is determined solely by the masses of solution components, is independent of temperature. In contrast, the molar concentration (or the normality) of a solution is defined in terms of volume; it may vary appreciably as the temperature is changed, because of the temperature-dependence of the volume. As a point of interest, in dilute aqueous solutions (less than 0.1 M), the molality is very close numerically to the molarity.
The mole fraction scale is useful when one is concerned with physical properties of solutions (Chapter 14), which are expressed most clearly in terms of relative numbers of solvent and solute molecules. Sometimes, the physical properties are affected by the number of particles in solution. Then, the molality of ions becomes important, not just that of the molecules—NaCl (Na+ and Cl– in solution) and Na2SO4 (2Na+ and 
in solution) affect some physical measurements differently.
Molar concentration and normality are expressed in a specific amount of solute per a fixed volume of solvent. Both of these can be expressed in terms of the amount of solute by algebraic manipulation.
If a solution is diluted by the addition of solvent, the volume will increase and the concentration decrease. In the process of dilution, the amount of solute does not change. If we have two solutions of different concentration, but containing the same amount of solute (different volumes), they will be related by the following expression:
where the subscripts refer to the conditions of 1 (before dilution) and 2 (after dilution). If any three of the variables are known, then we can calculate the fourth.
12.1. Explain how you would prepare 60 mL of an aqueous solution of AgNO3 that is 0.030 g AgNO3 per mL.
Since each mL of solution is to contain 0.030 g AgNO3, the calculation is
The solution can be prepared by dissolving 1.8 g AgNO3 in much less than 60 mL H2O (
of the final volume works well). Stir until dissolved, then add water to bring up to 60 mL final volume while stirring. The stirring is to make certain that the solution is homogeneous (consistent throughout).
If you had used 60 mL of water, there is no guarantee that the final volume will be 60 mL! The only way to make certain of the final volume of 60 mL is to dilute to 60 mL, not add 60 mL H2O.
12.2. What mass of 5.0% by weight NaCl solution is necessary to yield 3.2 g NaCl?
A 5.0% NaCl solution contains 5.0 g NaCl in 100 g solution. Then,
and
Another setup for the problem takes advantage of the ratio and proportion (w is the mass desired):
12.3. How much NaNO3 do you need make 50 mL of an aqueous solution containing 70 mg Na+/mL?
The mass of Na+ in 50 mL of solution = (50 mL)(70 mg/mL) = 3500 mg = 3.5 g Na+. The molar mass of NaNO3 is 85, of which sodium is 23. The line of logic involved is
and
The use of quantitative factors can also be used to produce a setup. We use w for the mass.
12.4. A 500-mL sample of the water treated by a water softener required 6 drops of standard soap solution to produce a permanent lather. The soap solution had been calibrated against an artificial hard water solution containing 0.136 g CaCl2 per liter. On the average, it required 28 drops of standard soap solution to lather 500 mL of artificial solution. Calculate the “hardness” of the sample expressed as ppm of CaCO3. Note: CaCO3 is very insoluble and does not actually exist in hard water. The measure of hardness is really the amount of CaCO3 that would be formed if all the Ca2+ were precipitated as CaCO3.
Noticing the effluent’s 6-drop requirement against the standard of 28 drops to lather the same volume of hard water, the hardness of the effluent sample is 6/28 of that of the CaCl2 solution. The conversion to standard units per liter, assuming that each mole of CaCl2 is equivalent to 1 mol CaCO3, is
One liter of this nearly pure water weighs 1000 g; the conversion to ppm is
The hardiness of the effluent is calculated as (123)(6/28) = 26 ppm. This value is lower than most natural waters, but pretty poor for treated water; the cartridge needs to be replaced or recharged.
12.5. Describe how to prepare 50 g of a 12.0% solution, starting with BaCl2 · 2H2O and distilled or deionized water.
A 12.0% BaCl2 solution contains 12.0 g BaCl2 per 100 g solution, which is 6.00 g BaCl2 in 50.0 g solution. However, you are to start with the hydrate and must account for the water in the molecule when you weigh the substance. The molar mass of BaCl2 is 108, but that of BaCl2 · 2H2O is 224. Therefore,
and
then,
The solution is prepared by dissolving 7.0 g BaCl2 · 2H2O in 43 g (43 mL) H2O while stirring.
Note: A small part of the solvent water came with the hydrated salt.
12.6. Calculate the mass of anhydrous HCl in 5.00 mL of concentrated hydrochloric acid (density 1.19 g/mL) containing 37.23% HCl by weight.
The mass of 5.00 mL of solution is (5.00 mL)(1.19 g/mL) = 5.95 g. Since the solution contains 37.23% HCl by weight, the calculation of the mass of HCl required is (0.3723)(5.95 g) = 2.22 g HCl.
12.7. Calculate the volume of concentrated sulfuric acid (density 1.84 g/mL) containing 98% H2SO4 by weight that would be produced from 40.0 g pure H2SO4.
One mL solution has a mass of 1.84 g and contains (0.98)(1.84 g) = 1.80 g pure H2SO4. Then, 40 g H2SO4 are contained in
A setup can be produced by the use of conversion factors.
12.8. Exactly 4.00 g of a sulfuric acid solution were diluted with water, then an excess of BaCl2 was added. The dried BaSO4 weighed 4.08 g. Find the percent H2SO4 in the original acid solution.
First, we determine the mass of H2SO4 required to precipitate 4.08 g BaSO4 referring to the reaction below and placing the equation information and problem information.
The next step is to set up the ratio and proportion, then solve for the mass of H2SO4.
and
12.9. (a) How many grams solute are required to prepare 1 L of 1 M Pb(NO3)2? (b) What is the molar concentration of the solution with respect to each of the ions?
(a) A one molar solution contains 1 mol of solute in 1 liter of solution. The molar mass of Pb(NO3)2 is 331.2, meaning that 331.2 g Pb(NO3)2 are required for l L of 1 M Pb(NO3)2 solution.
(b) A 1 M solution of Pb(NO3)2 is 1 M Pb2+ and 2M 
.
12.10. What is the molar concentration of a solution containing 16.0 g CH3OH in 200 mL solution?
The molar mass of CH3OH is 32.0, and the calculation to determine molarity is
12.11. Determine the molar concentration of these two solutions: (a) 18.0 g AgNO3 per liter solution and (b) 12.00 g AlCl3 · 6H2O per liter of solution.
(a) 
(b) 
12.12. How much (NH4)2SO4 is required to prepare 400 mL of M/4 (M/4 = 
M) solution?
The molar mass of ammonium sulfate is 132.1. One liter of M/4 solution contains
Then, the 400-mL volume of M/4 solution requires
12.13. What is the molality of a solution containing 20.0 g of cane sugar, C12H22O11, in 125 g H2O?
12.14. The molality of a solution of ethyl alcohol, C2H5OH, in water is 1.54 mol/kg. How many grams of alcohol are dissolved in 2.5 kg H2O?
The molar mass of ethyl alcohol is 46.1. Since the molality is 1.54, 1 kg of water dissolves 1.54 mol C2H5OH. The amount of C2H5OH that 2.5 kg water dissolves is
(2.5)(1.54) = 3.85 mol C2H5OH, which weighs (3.85)(46.1 g/mol) = 177 g C2H5OH
12.15. Calculate the (a) molar concentration and (b) molality of a sulfuric acid solution of density 1.198 g/mL containing 27.0% H2SO4 by weight.
It is recommended, when quantities of substances are not stated, to select an arbitrary quantity as the basis of calculation. In this case, let the basis be 1.000 L solution.
(a) One liter has a mass of 1198 g and contains (0.270)(1198) = 323 g H2SO4 of molar mass 98.1,
(b) From (a), there is 323 g (3.29 mol) solute per liter solution. The amount of water in 1 L of solution is 1198 g solution – 323 g solute = 875 g H2O. The molality is
12.16. Determine the mole fractions of both substances in a solution containing 36.0 g H2O and 46 g C3H5(OH)3, glycerin (molar masses: 18.0 for water and 92 for glycerin).
The mole fraction requires the expression of the components in moles.
Check: The sum of the mole fractions should be 1, as it is in this problem (0.80 + 0.20 = 1).
12.17. How many equivalents of solute are contained in (a) 1 L of 2 N solution; (b) 1 L of 0.5 N solution; (c) 0.5 L of 0.2 N solution? (Hint: Using “of” usually indicates multiplication.)
(a) 1 L × 2 eq/L = 2 eq; (b) 1L × 0.5 eq/L = 0.5 eq; (c) 0.5 L × 0.2 eq/L = 0.1 eq
12.18. How many (a) equivalents and (b) milliequivalents (meq) of solute are in 60 mL of a 4.0 N solution? (Refer to the hint in Problem 12.17.)
(a) 
(b) 
Alternate setup for (b): meq = (number mL)(normality) = (60 mL)(4.0 meq/mL) = 240 meq
12.19. What mass solute is required to prepare 1 L of 1 N solution of (a) LiOH; (b) Br2(as an oxidizing agent); (c) H3PO4 (all three H’s are replaceable)?
(a) One liter of 1 N LiOH requires (23.95/1) g = 23.95 g LiOH.
(b) Note from the partial equation Br2 + 2e– → 2Br– that two electrons react per Br2. The equivalent mass of Br2 is half its molar mass. One liter of 1 N Br2 requires (159.8/2) g = 79.9 g Br2.
(c) One liter of 1 N H3PO4 requires (98.00/3) g = 32.67 g H3PO4, assuming total ionization.
12.20. Calculate the normality of each of the following solutions: (a) 7.88 g of HNO3 per liter solution and (b) 26.5 g of Na2CO3 per liter solution when neutralized to form CO2.
(a) Equivalent mass of HNO3 for the H+, not an oxidizing agent (for the N), is the molar mass, 63.02.
(b) The reaction is: 
+ 2H+ → CO2 + H2O, and the equivalent mass of Na2CO3 is ()(molar mass = ()(106.0) = 53.0 Na2CO3.
12.21. How many milliliters of 2.00 M Pb(NO3)2 contain 600 mg Pb2+?
One liter of 1 M Pb(NO3)2 contains 1 mol Pb2+, or 207g. Then, a 2M solution contains 2 mol Pb2+ or 414g Pb2+ per liter. As 1 mL is 1/1000th of a liter, there would be 1/1000th of the mass of Pb2+ in that liter solution, which is 414 mg/mL. Then, 600 mg Pb+ are contained in
An alternate method recognizes that a 2 M solution contains 2 mmol/mL and,
12.22. Given the unbalanced equation
(a) How many grams of KMnO4 are needed to make 500 mL of a 0.250 N solution?
(b) How many grams of KI are needed to make 25 mL of a 0.36 N solution?
(a) The Mn in KMnO4 changes in oxidation number from +7 to +2, requiring 5 electrons.
The 0.500-L solution of 0.250 N KMnO4 requires
(0.500 L)(0.250eq/L)(31.6g/eq) = 3.95g KMnO4
As a different scenario, if the KMnO4 made above were to be used in the reaction:
the label 0.250 N would no longer be appropriate because the oxidation number change in this case is 3, not 5 (Mn+7 → Mn+4). The appropriate normality would be
(b) The oxidation state changes from –1 in I– to 0 in I2.
Then, 0.025 L (25 mL) of 0.36 N solution requires
(0.025 L)(0.36eq/L)(166g/eq) = 1.49 g KI
12.23. Calculate the molar concentration of a 2.28 m NaBr solution (density of 1.167 g/cm3).
The molar mass of NaBr is 102.9. For each kilogram of water there are (2.28)(102.9) = 235 g NaBr, or a total mass of 1000 + 235 = 1235 g. This mass of solution occupies a volume of 1235 g/(1.167 g/cm3), which is 1058 cm3. Since 1 cm3 is 1 mL, the volume is 1.058 L. The molar concentration is
12.24. A solution of an organic halide in benzene, C6H6, has a mole fraction of halide of 0.0821. Express its concentration in terms of molality.
Benzene’s molar mass is 78.1. A total of 0.0821 mol halide is mixed with 1 – 0.0821 = 0.9179 mol benzene, which has a mass of (0.9179 mol)(78.1 g/mol) = 71.7 g = 0.0717 kg. The molality halide is
12.25. To what extent must a solution containing 40 mg AgNO3 per mL be diluted to yield one that contains 16 mg AgNO3 per mL?
Let V be the volume to which 1 mL of the original solution must be diluted to yield a solution that is 16 mg AgNO3 per mL. (Note: The solute amount does not change with dilution; the volume changes.)
Note that 2.5 mL is not the volume of water to be added; it is the final volume after water has been added to 1 mL of the original solution. Also, notice that there is no provision made for other than the final volume being the result of adding volumes. This works well with dilute solutions (less than 0.1 M), but may not work so well with concentrated solutions. If more concentrated solutions are used, the final volume will probably have to be determined by experimentation.
12.26. How can a 0.50 M BaCl2 solution be diluted to yield one that contains 20.0 mg Ba2+/mL?
One liter of the original solution contains 0.50 mol BaCl2 (or of Ba2+). The mass of Ba2+ in 0.50 mol is
The problem now is to find the extent to which a solution of 68.7 mg Ba2+/mL must be diluted to yield one containing 20.0 mg Ba2+/mL. At this point, the problem is the same as the previous one.
This means that each milliliter of 0.50 M BaCl2 must be diluted with water to 3.43 mL.
12.27. A procedure calls for 100 mL of 20% H2SO4, density 1.14 g/mL. How much of the concentrated acid, density 1.84 g/mL, and containing 98% H2SO4 by weight, must be diluted with water to prepare 100 mL of the required acid?
The concentrations must first be changed from a mass basis to a volumetric basis, so that the dilution equation will apply. Concentrations include mass, but are not a direct statement of the mass.
We can now let V2 be the volume of 98% acid required for 100 mL of 20% H2SO4.
12.28. What volumes of N/2 and of N/10 HCl must be mixed to give 2L of N/5 HCl?
Let v = volume N/2 required and let 2L – v be the volume of N/10 required.
and, substituting 0.5 for v, the requirements are 0.5L of N/2 and 1.5 L of N/10 solutions.
12.29. How many mL of concentrated H2SO4, density 1.84 g/mL, containing 98% H2SO4 by weight do we need to make (a) 1 L of 1 N solution; (b) 1 L of 3.00 N solution; (c) 200 mL of 0.500 N solution?
Equivalent mass H2SO4 = (molarmass) = (98.1) = 49.0g/eq H2SO4
The H2SO4 content of 1 L concentrated acid is (0.98)(1000 mL)(1.84g/mL) = 1800 g H2SO4. The normality of the concentrated acid is calculated by
allowing us to use the dilution formula: Vconc × Nconc = Vdilute × Ndilute. Solving for Vconc in each case,
(a) 
(b) 
(c) 
12.30. How much NH4Cl is required to prepare 100 mL of a solution containing 70 mg NH4Cl per mL?
Ans. 7.0 g
12.31. What mass of crystalline Na2SO4 do you need in order to mix 1.5 L of a solution containing 0.375 moles of salt per liter of solution?
Ans. 79.899 g Na2SO4 required (3-place accuracy matches most chemistry lab balances)
12.32. Among the uses for cartridge brass (30% zinc and 70% copper by weight) are ammunition and radiators for cars. What is the number of moles of each component in an 18-g sample?
Ans. 0.083 mol Zn and 0.198 mol Cu
12.33. Concrete drills can be produced from a steel that includes 0.50 wt% carbon, 1.35 wt% chromium, 0.28 wt% nickel, and 0.22 wt% vanadium. A 1-kg sample of the steel received from a potential supplier is being analyzed by the Quality Control Department. Express each of the components of this sample in terms of moles.
Ans. 0.416 mol C, 0.147 mol Cr, 0.048 mol Ni, and 0.043 mol V
12.34. What mass (g) of concentrated solution containing 37.9% HCl by weight will give 5.0 g HCl?
Ans. 13.2 g
12.35. An experiment requires 100 g of a 19.7% solution of NaOH. How many grams each of NaOH and H2O do we need to mix the solution?
Ans. 19.7 g NaOH and 80.3 g H2O
12.36. How much CrCl3 · 6H2O is needed to prepare 1 L solution containing 20 mg Cr3+/mL?
Ans. 102 g
12.37. Determine the volume of dilute HNO3, density 1.11 g/cm3 and 19% HNO3, by weight, that contains 10 g HNO3.
Ans. 47 cm3 or 47 mL
12.38. What is the percent composition Na2SO4 (density 1.03) if it contains 0.001 g salt per mL of solution?
Ans. 0.097% Na2SO4
12.39. How many cubic centimeters of a solution containing 40 g CaCl2/L are needed to react with 0.642 g of pure Na2CO3? CaCO3 is formed in the metathesis reaction.
Ans. 16.8 cm3 or 16.8 mL
12.40. What volume of a NaCl solution that is 0.5 g/mL is required to just react with 25 g of a AgNO3 (AgCl is a precipitate produced by this reaction).
Ans. 17.1 mL
12.41. Ammonia gas is passed into water, yielding a solution of density 0.93 g/mL and containing 18.6% NH3 by weight. What is the mass of NH3 (mg) per milliliter of solution?
Ans. 173 mg/mL
12.42. Given 100 mL of pure water at 4°C, what volume of a hydrochloric acid solution, density 1.175 g/mL and containing 34.4% HCl by weight, could be prepared?
Ans. 130 mL
12.43. A liter of milk weighs 1.032 kg. The butterfat, which is 4% by volume, has a density of 0.865 g/mL. What is the density of the fat-free milk?
Ans. 1.039 g/mL
12.44. A benzene-soluble cement is produced by melting 49 g of rosin in an iron pan and adding 28 g each of shellac and beeswax. How much of each component should be taken to make 75 kg of cement?
Ans. 35 kg rosin, 20 kg shellac, 20 kg beeswax
12.45. How much CaCl2 · 6H2O and water must be weighed out to make 100 g of a solution that is 5.0% CaCl2?
Ans. 9.9 g CaCl2 · 6H2O and 90.1 g water
12.46. What mass of BaCl2 would be needed to make 250 mL of a solution having the same concentration of Cl– as one containing 3.78 g of NaCl per 100 mL?
Ans. 16.8 g BaCl2
12.47. Although not usually calculated, the molality and molarity of an alloy (a metal in a metal solution) can be calculated. Nickel steel contains nickel in small amounts mixed with iron. (a) Express the molality of the 2.5 g Ni (atomic mass = 58.69) dissolved in 1000 mL Fe (atomic mass = 55.85, density = 7.66 g/cm3 under the lab conditions). (b) Express the molarity of the metal solution (no volume change).
Ans. (a) 0.0056 m; (b) 0.0426 M
12.48. Using data from the preceding problem, calculate the mole fraction Ni and the mole fraction Fe.
Ans. 0.000311 for Ni and 0.9997 for Fe
12.49. The sulfate content of 6.00 liters of drinking water is determined by first evaporating off some water, which forms a more concentrated solution of less volume. This volume was then treated with BaCl2 solution, which resulted in the precipitation of 0.0965 g BaSO4. Express the sulfate ion concentration in ppm.
Ans. 6.62 ppm
12.50. Silver nitrate, AgNO3, can be added to drinking water as a test reagent for the concentration of chlorine that has been used to kill bacteria and other disease-causing organisms. Insoluble AgCl is produced as a heavy, white precipitate. A 10-mL water sample required 1.35 mL of AgNO3 containing 0.00366 g Ag+ per mL to use up all the chlorine. What is the concentration of Cl– in the sample stated in ppm (sample density = 1.000 g/mL)?
Ans. 162 ppm
12.51. An air sample from a chemical plant was analyzed by a mass spectrometer. The analysis revealed the presence of 1.2 × 10–8 moles benzene per mole of air. Express the concentration of benzene, C6H6, in ppb by weight. Assume the average molar mass of air is 29.
Ans. 32 ppb
12.52. (a) Chromium plating of surfaces for protection from corrosion can be done using a solution of Cr(NO3)3. What is the mass of Cr3+ in 25 L of a 1.75 M solution? (b) Gold plating can be performed using a 3.50 M Au(NO3)3 solution. What is the mass of Au3+ in 12.75 L of this solution?
Ans. (a) 2275 g Cr3+;(b) 8790 g Au3+
12.53. (a) What mass of silver is weighed to produce 10 L of 6 M AgNO3?(b) What mass of gold is required for 10 L of 6M Au(NO3)3?
Ans. (a) 6.5kg Ag; (b) 11.8 kg Au
12.54. What is the molarity of a solution containing 37.5 g Ba(MnO4)2 per liter, and what is the molar concentration with respect to each type of ion?
Ans. 0.100 M Ba(MnO4)2; 0.100 M Ba2+; 0.200 M 
12.55. A solution is labeled “0.100 M Ba(MnO4)2.” With what normality should it be labeled if used as (a) an oxidizing agent in strong acid (Mn2+ produced), (b) an oxidizing agent in slightly acid solution (MnO2 produced), or (c) a precipitant for BaSO4?
Ans. (a) 0.500 N; (b) 0.300 N; (c) 0.100 N
12.56. How many grams of solute are required to prepare 1 L of 1 M CaCl2 · 6H2O?
Ans. 219.1 g
12.57. The presence of waters of hydration changes the amounts of materials needed to mix up solutions. (a) Copper(II) ion in solution is occasionally used to kill bacteria in water. What is the mass of CuSO4 that is required to mix 1 L of 5 M CuSO4? (b) What is the mass of CuSO4 · 5H2O needed to produce the same solution? Assume no change in volume on mixing.
Ans. (a) 800 g CuSO4; (b) 1250 g CuSO4 · 5H2O.
12.58. We can estimate an amount of substance needed to produce a solution as a check on our calculations. (a) Using mass numbers 7 for lithium, 12 for carbon and 16 for oxygen, estimate the amount of lithium carbonate needed to produce 1 L of a 3 M Li2CO3 solution. (b) Now, perform the calculation using the atomic masses as a comparison.
Ans. (a) 222 g; (b) 221.674 g (0.674 g difference—less than 0.4% error)
12.59. What mass (g) of anhydrous CaCl2 is required to prepare (a) 1 L of 1 M CaCl2; (b) 2.50 L of 0.200 M CaCl2 · 2H2O; (c) 650 mL of 0.600 M CaCl2?
Ans. (a) 111 g; (b) 55.5 g; (c) 43.3 g
12.60. A single drop of dimethyl mercury, CH3—Hg—CH3, can be absorbed through the skin resulting in death. What is the minimum molarity of dimethyl mercury that will deliver 5.0 × 10–5 gHg2+ in one drop, assuming that there are 20 drops/mL?
Ans. 5 × 10–3 M
12.61. A 6.00-g sample of a polymer was dissolved in 280 mL of a solvent. Osmotic pressure measurements show that its concentration was 2.12 × 10–4 M. Calculate the molar mass of the polymer.
Ans. M = 1.01 × 105 g/mol
12.62. Exactly 100 g NaCl are dissolved in sufficient water to give 1500 mL of solution. What is the molar concentration of the solution prepared in this manner?
Ans. 1.14 M
12.63. Suppose you were responsible for producing 2.25 L of a solution that was to be 0.082 M copper(II) sulfate. What is the mass of CuSO4 you must weigh? (Density = 1.00 g/mL and no volume change.)
Ans. 29.5 g CuSO4
12.64. Calculate the molality of (a) 0.65 mol glucose, C6H12O6, in 250 g H2O; (b) 45 g glucose in 1 kg water; (c)18g glucose in 200 g water.
Ans. (a) 2.6 m; (b) 0.25 m; (c) 0.50 m
12.65. How many grams of CaCl2 should be added to 300 mL of water to make up 2.46 m CaCl2?
Ans. 82 g
12.66. Three compounds can be produced by starting with two C4H8O4 molecules and dehydrating (removing a H2O) to produceC8H14O7, then dehydrating again between the8-carbon and4-carbon molecules, ending up withC12H20O10. Three solutions were produced: 50 g of C4H8O4 in 2 L water, 50 g of C8H14O7 in 2 L water, and 50 g ofC12H20O10 in 2 L solution. What are the molalities of these solutions?
Ans. 0.208 m; 0.113 m; 0.077 m
12.67. A solution contains 57.5mL ethanol, C2H5OH, and 600 mL benzene, C6H6. How many grams of alcohol are there per 1000 g benzene? What is the molality of the solution? Densities are 0.80 g/cm3 for ethanol and 0.90 g/cm3 for benzene.
Ans. 85 g, 1.85 mol/kg or 1.85molal
12.68. Benzoic acid, C6H5COOH, is soluble in benzene, C6H6. What is the molality of a solution in which 3.55 g benzoic acid are dissolved in 75 mL of benzene? (Density benzene is 0.866 g/mL at the temperature of the experiment.)
Ans. 0.45 m
12.69. A solution contains 10.0 g HC2H3O2 (alternate formula format—CH3COOH), acetic acid, in 125 g H2O. Express as (a) mole fractions HC2H3O2 and H2O, (b) molality of the acid.
Ans. (a) X(acid) = 0.024 and X(water) = 0.976; (b) 1.33 molal
12.70. A solution contains 116 g acetone (CH3COCH3), 138 g ethanol (C2H5OH), and 126 g H2O. Determine the mole fraction of each.
Ans. X(acetone) = 0.167; X (alcohol) = 0.250; X (water) = 0.583
12.71. What is the mole fraction of the solute in a 1.00 molal aqueous solution?
Ans. 0.0177
12.72. An aqueous solution labeled as 35.0% HClO4 had a density of 1.251 g/mL. Calculate molarity and molality of the solution.
Ans. 4.36 M, 5.36m
12.73. A sucrose solution was prepared by dissolving 13.5 g C12H22O11 in sufficient water to make exactly 100 mL, which was then found to have a density of 1.050 g/mL. Compute the molar concentration and the molality of the solution.
Ans. 0.395 M, 0.431 m
12.74. For a solute of molar mass, M, show that the molar concentration, M, and molality, m, of the solution are related by
where d is the solution density in g/cm3 (g/mL). (Hint : show that each cubic centimeter of solution contains MM/1000g of solution and M/m g of solvent.) Use this relationship to check the answers to Problems 12.70 and 12.71.
12.75. What volume of 0.232 N solution contains (a) 3.17 meq solute, (b) 6.5 eq solute?
Ans. (a)13.7mL; (b)28.0L
12.76. Determine the molarity of each of the following solutions: (a)166 g KI/L solution; (b)33.0(NH4)2SO4 in 200 mL solution; (c) 12.5 g CuSO4 · 5H2O in 100 cm3 solution; (d) 10.0 mg Al3+ per cubic centimeter solution.
Ans. (a) 1.00 M; (b) 1.25 M; (c) 0.500 M; (d) 0.370 M
12.77. What volume of 0.200 M Ni(NO3)2 · 6H2O contains 500 mg Ni2+?
Ans. 42.6 mL
12.78. What is the volume of concentrated H2SO4 (density 1.835 g/cm3, 93.2% H2SO4 by weight) required to make up 500 mL of 3.00 N acid solution.
Ans. 43.0 mL
12.79. Calculate the volume of concentrated HCl (density 1.19 g/cm3, 38% HCl by weight) to obtain when making up 18 L of N/50 acid.
Ans. 29 cm3 or 29 mL
12.80. Determine the mass of KMnO4 required to make 80 mL of N/8 KMnO4 when the final solution is used to act as an oxidizing agent in an acid solution and Mn2+ is a product of the reaction.
Ans. 0.316 g
12.81. We are given this unbalanced equation: 
(a) What is the normality of a K2Cr2O7 solution 35.0 mL of which contains 3.87 g K2Cr2O7?
(b) What is the normality of an FeSO4 solution 750 mL of which contains 96.3 g FeSO4?
Ans. (a) 2.25 N (b) 0.845 N
12.82. What mass of Na2S2O3 · 5H2O is needed to make up 500 mL of 0.200 N solution, assuming
Ans. 24.8 g
12.83. A 4.51-g sample of a compound was dissolved in 98.0 g of solvent. By close observation of its freezing point, the solution was found to be 0.388 m. What is the molar mass of the unknown compound?
Ans. 119g/mol or 119 M
12.84. An analyst planned to use BaCl2 to make up 60.0 mL of 0.500 M Ba2+ solution, but his only source of BaCl2 in stock was 2.66 g of BaCl2 · 2H2O. Knowing that nitrate ion would not interfere with his procedure, he decided to make up the difference using Ba(NO3)2. How much Ba(NO3)2 is needed?
Ans. 4.99g Ba(NO3)2
12.85. A solution contains 75 mg NaCl per milliliter. To what extent must it be diluted to give a solution that is 15 mg NaCl per mL?
Ans. Each milliliter of original solution is diluted with water to a volume of 5 mL.
12.86. How many cubic centimeters of solution that is 100 mg Co2+ per cubic centimeter are needed to prepare 1.5 L of solution that is 20 mg Co2+ per cubic centimeter?
Ans. 300 cm3
12.87. Calculate the approximate volume of water that must be added to 250 mL of 1.25 N solution to make it 0.500 N (neglecting volume changes).
Ans. 375 mL
12.88. What is the volume of 6 M HNO3 that you must measure to produce 175 mL of 4.5 M HNO3?
Ans. 131 mL
12.89. What is the resultant molar concentration of HCl when 15 mL of 6 M HCl and 15 mL of 3 M NaOH are mixed?
Ans. 1.5 M HCl
12.90. Determine the volume of dilute nitric acid (density 1.11 g/mL, 19.0% HNO3 by weight) that can be prepared by diluting with water 50 mL of concentrated acid (density 1.42 g/mL, 69.8% HNO3 by weight). Calculate the molar concentrations and molalities of the concentrated and dilute acids.
Ans. 234 mL; molar concentrations 15.7 and 3.35; molal concentrations 36.7 and 3.72
12.91. What volume of 95% alcohol by weight (density 0.809 g/cm3) must be used to prepare 150 cm3 of 30.0% alcohol by weight (density 0.957 g/cm3)?
Ans. 56.0 cm3
12.92. What volumes of 12 N and 3 N HCl must be mixed to produce 1 L of 6 N HCl?
Ans. liter of 12 N, 
liter of 3 N
12.93. What is the molar concentration of a solution made by mixing 300 mL of 0.0200 M H2SO4 with 200 mL of 0.0300 M H2SO4?
Ans. 0.024 M
12.94. A chemist boiled down 500 mL of 0.0865 M NaCl until only 127 mL remained. Calculate the molar concentration of the remaining solution.
Ans. 0.341 M NaCl
