Volume 2
2.11. THE MOLE
Molarity and the Mole • Simple Calculations Using the Mole Unit • Molar Concentration • The Avogadro Constant. A Little about Gases. How Much Oxygen Does My Car Consume? • Exercises
Molarity and the Mole • Simple Calculations Using the Mole Unit • Molar Concentration • The Avogadro Constant. A Little about Gases. How Much Oxygen Does My Car Consume? • Exercises
2.11.1. Molarity and the Mole. Suppose there is a waste solution containing an unknown quantity of silver salts as well as salts of other metals. A simple and efficient way to separate the precious silver is to treat the waste with an alkali metal halide, such as LiCl or KBr. The insoluble AgCl or AgBr will then precipitate out right away, whereas the water-soluble chloride or bromide salts of the contaminating metals will stay in solution.
Since the amount of the Ag+ in the waste liquor is unknown, to precipitate as much AgCl or AgBr as possible, we want to use as much LiCl or KBr as possible. There are two aqueous solutions, 100 g each. One is 25% LiCl and the other 50% KBr. Which one of the two should we use?
It might be tempting to choose the 50% KBr solution just because it is twice as concentrated as 25% LiCl. That would be a mistake. The mistake would be likely prompted by the deceiving mass percent concentration numbers. A solution that is more concentrated mass-wise might turn out to be less concentrated in terms of the number of molecules or ions it contains. Let us recall that the mass percent concentration refers to the gram amount of a solute in 100 g of solution. A simple calculation shows that 25 g of LiCl contains 1.4 times the number of molecules/ions as 50 g of KBr, because a molecule of KBr (MW = 119.0) is 2.8 times heavier than a molecule of LiCl (MW = 42.4).
In chemistry, it is often more important to know the number of particles such as molecules or ions in a given amount of solution than the mass of the solute. Of course, using molecular weights, we can always calculate which of two given solutions contains more molecules/ions, the way we just did for the 25% LiCl and 50% KBr. It would be much more convenient, though, to have solution concentration expressed in such a way that it specified the number of molecules of a solute rather than its mass.
Unsurprisingly, such a convenient measure of concentration has been developed and widely used for a long time. This type of concentration is molarity, also known as molar concentration. Molarity refers to the number of moles of an individual chemical substance in one liter of its solution.
Moles? What moles? Certainly not those shown in Figure 2-118. In chemistry, the mole (abbreviated mol) is the unit of quantity of substance that contains 6.02 × 1023 molecules.
Since the amount of the Ag+ in the waste liquor is unknown, to precipitate as much AgCl or AgBr as possible, we want to use as much LiCl or KBr as possible. There are two aqueous solutions, 100 g each. One is 25% LiCl and the other 50% KBr. Which one of the two should we use?
It might be tempting to choose the 50% KBr solution just because it is twice as concentrated as 25% LiCl. That would be a mistake. The mistake would be likely prompted by the deceiving mass percent concentration numbers. A solution that is more concentrated mass-wise might turn out to be less concentrated in terms of the number of molecules or ions it contains. Let us recall that the mass percent concentration refers to the gram amount of a solute in 100 g of solution. A simple calculation shows that 25 g of LiCl contains 1.4 times the number of molecules/ions as 50 g of KBr, because a molecule of KBr (MW = 119.0) is 2.8 times heavier than a molecule of LiCl (MW = 42.4).
In chemistry, it is often more important to know the number of particles such as molecules or ions in a given amount of solution than the mass of the solute. Of course, using molecular weights, we can always calculate which of two given solutions contains more molecules/ions, the way we just did for the 25% LiCl and 50% KBr. It would be much more convenient, though, to have solution concentration expressed in such a way that it specified the number of molecules of a solute rather than its mass.
Unsurprisingly, such a convenient measure of concentration has been developed and widely used for a long time. This type of concentration is molarity, also known as molar concentration. Molarity refers to the number of moles of an individual chemical substance in one liter of its solution.
Moles? What moles? Certainly not those shown in Figure 2-118. In chemistry, the mole (abbreviated mol) is the unit of quantity of substance that contains 6.02 × 1023 molecules.
Figure 2-118. The mole in chemistry has nothing to do with beauty marks on skin (source) or the little animal living underground (source) or spies (source).
Why 6.02 × 1023 molecules? Because the mass of 6.02 × 1023 molecules of an individual substance expressed in grams is numerically equal to the molecular weight (MW) of this substance expressed in a.m.u. For instance, the molecular weight of H2O is 18 a.m.u. Therefore, 1 mol of H2O equals 18 g. As MW of AgNO3 is 169.9 a.m.u., 1 mol of AgNO3 is 169.9 g. Likewise, 1 mol of LiCl is 42.4 g because MW of LiCl is 42.4 a.m.u.
Why 6.02 × 1023 molecules? Because the mass of 6.02 × 1023 molecules of an individual substance expressed in grams is numerically equal to the molecular weight (MW) of this substance expressed in a.m.u. For instance, the molecular weight of H2O is 18 a.m.u. Therefore, 1 mol of H2O equals 18 g. As MW of AgNO3 is 169.9 a.m.u., 1 mol of AgNO3 is 169.9 g. Likewise, 1 mol of LiCl is 42.4 g because MW of LiCl is 42.4 a.m.u.
Digression. The mole, 6.02 × 1023 molecules, is an arbitrary constant. There are arbitrary constants and universal constants. Universal constants, such as the speed of light or the gravitational constant, are quantities that do not vary. In contrast, arbitrary constants are chosen by us. Why have we chosen 6.02 × 1023 molecules for the mole unit, not one or ten or one million molecules? The answer is simple: for convenience.
In 1805, John Dalton (1766–1844) published the first table of atomic weights with that of hydrogen defined as 1, the first atomic mass unit. That assignment certainly made sense, considering that H is the lightest element. In 1961, the atomic mass unit was adjusted, and since then has been defined as 1/12 the mass of one atom of the carbon isotope 12C, which corresponds to 1.66 × 10−24 grams. The reciprocal of this value is the number of a.m.u.'s (not molecules!) in 1 g of a substance: 1/1.66 × 10−24 = 6.02 × 1023. That is why 1 mol of a substance having the molecular weight of n a.m.u.'s is n grams.
Why are we talking grams? Why is the mole unit tied to the number of a.m.u.'s in 1 gram, not in 1 kilogram or 1 pound or 1 ton of a substance? Again, this is all about convenience. What is convenient and not convenient is determined by measurement capabilities and limitations of the instruments we have and use. Most laboratory balances are capable of accurately weighing masses ranging from a fraction of a gram to hundreds of grams. Ultimately, it is this mass range that dictates the choice of the mole as a unit tied to a mass on the order of a gram. For most substances, the mass of 1 mole varies from grams to hundreds of grams.
Chemistry is a practical science that is based on experimental studies. In chemical experiments, reagents need to be accurately measured for mixing in exact proportions. This is no different from other activities where some objects are used in a strict ratio, such as one nut per bolt in certain types of construction. We could certainly count out a dozen or so bolts and nuts one by one. What do we do however, if an accurate count is needed for hundreds, thousands, or even millions of identical objects? How did they count out the 1.2 million rivets for the construction of the two towers of San Francisco's Golden Gate Bridge? A convenient way for counting identical objects in large numbers is to determine the mass of one object and measure out the required number of the objects by weight. For example, we need 1,000 bolts and 1,000 matching nuts for a project. There is a big container full of bolts and another one full of nuts in the warehouse. As counting out 1,000 pieces of each part by hand is not practical, we first determine the masses of one nut and one bolt. Say one nut weighs 1 g and one bolt 2 g. We then weigh out 1,000 x 1 g = 1,000 g = 1 kg of the nuts and 1,000 x 2 g = 2,000 g = 2 kg of the bolts, which is easy to do. It would be even more convenient, though, if the nuts and bolts were pre-packed in, say, 100-piece boxes. We would then just get 10 boxes of each.
Molecules and atoms are incomparably smaller than the tiniest bolt or nut ever made. Due to the miniscule size and mass of molecules, there is no practical way to count them out even by the millions or billions. Like in the bolt and nut example, we use the known mass of a single molecule (from the previously determined atomic masses) to weigh out proportionally the needed number of molecules we deal with. Also similar to the bolts and nuts example, it is often more convenient in chemistry to come up with a suitable unit for counting molecules. This unit is the mole, our reference standard "pack" of 6.02 × 1023 molecules. As we will see from below, it is often most convenient to work with quantities of substances expressed in moles, and especially so with "pre-packed" molecules in the form of solutions of known molarity.
In 1805, John Dalton (1766–1844) published the first table of atomic weights with that of hydrogen defined as 1, the first atomic mass unit. That assignment certainly made sense, considering that H is the lightest element. In 1961, the atomic mass unit was adjusted, and since then has been defined as 1/12 the mass of one atom of the carbon isotope 12C, which corresponds to 1.66 × 10−24 grams. The reciprocal of this value is the number of a.m.u.'s (not molecules!) in 1 g of a substance: 1/1.66 × 10−24 = 6.02 × 1023. That is why 1 mol of a substance having the molecular weight of n a.m.u.'s is n grams.
Why are we talking grams? Why is the mole unit tied to the number of a.m.u.'s in 1 gram, not in 1 kilogram or 1 pound or 1 ton of a substance? Again, this is all about convenience. What is convenient and not convenient is determined by measurement capabilities and limitations of the instruments we have and use. Most laboratory balances are capable of accurately weighing masses ranging from a fraction of a gram to hundreds of grams. Ultimately, it is this mass range that dictates the choice of the mole as a unit tied to a mass on the order of a gram. For most substances, the mass of 1 mole varies from grams to hundreds of grams.
Chemistry is a practical science that is based on experimental studies. In chemical experiments, reagents need to be accurately measured for mixing in exact proportions. This is no different from other activities where some objects are used in a strict ratio, such as one nut per bolt in certain types of construction. We could certainly count out a dozen or so bolts and nuts one by one. What do we do however, if an accurate count is needed for hundreds, thousands, or even millions of identical objects? How did they count out the 1.2 million rivets for the construction of the two towers of San Francisco's Golden Gate Bridge? A convenient way for counting identical objects in large numbers is to determine the mass of one object and measure out the required number of the objects by weight. For example, we need 1,000 bolts and 1,000 matching nuts for a project. There is a big container full of bolts and another one full of nuts in the warehouse. As counting out 1,000 pieces of each part by hand is not practical, we first determine the masses of one nut and one bolt. Say one nut weighs 1 g and one bolt 2 g. We then weigh out 1,000 x 1 g = 1,000 g = 1 kg of the nuts and 1,000 x 2 g = 2,000 g = 2 kg of the bolts, which is easy to do. It would be even more convenient, though, if the nuts and bolts were pre-packed in, say, 100-piece boxes. We would then just get 10 boxes of each.
Molecules and atoms are incomparably smaller than the tiniest bolt or nut ever made. Due to the miniscule size and mass of molecules, there is no practical way to count them out even by the millions or billions. Like in the bolt and nut example, we use the known mass of a single molecule (from the previously determined atomic masses) to weigh out proportionally the needed number of molecules we deal with. Also similar to the bolts and nuts example, it is often more convenient in chemistry to come up with a suitable unit for counting molecules. This unit is the mole, our reference standard "pack" of 6.02 × 1023 molecules. As we will see from below, it is often most convenient to work with quantities of substances expressed in moles, and especially so with "pre-packed" molecules in the form of solutions of known molarity.
2.11.2. Simple Calculations Using the Mole Unit. To perform a chemical reaction properly, we have to use the reagents in certain proportions. Moreover, it is often critical that the reagents for a chemical transformation be used in a strict stoichiometric ratio, so that all of them are fully consumed in the reaction. Here is one example.
Picture a water tank in an area where water is scarce. The water in the tank is contaminated with 200 g of toxic BaCl2. One way to make the water potable again is to treat it with Ag2SO4 to prompt the following reaction.
BaCl2 + Ag2SO4 = BaSO4↓+ 2 AgCl↓
Both salts produced in this reaction are insoluble and will precipitate out. After filtration, the filtrate will be pure water. It is critical, however, that Ag2SO4 be used in the right quantity for the treatment. If not enough Ag2SO4 is added, there will be some toxic BaCl2 left in the water after the reaction. If too much Ag2SO4 is used, all of the BaCl2 will precipitate out, but the water will be contaminated with the silver salt used in excess. Knowing the amount of BaCl2 in the water (200 g), we need to calculate the right quantity of Ag2SO4 for the treatment.
There are two different ways to solve this problem.
Solution 1 (old method). In Volume 1, we learned how to calculate quantities of reagents for a chemical reaction using the molecular mass method. By applying this method to the current problem (Figure 2-119), we find that 299.5 g of Ag2SO4 is needed to purify the water from the 200 g of the BaCl2 contaminant.
Picture a water tank in an area where water is scarce. The water in the tank is contaminated with 200 g of toxic BaCl2. One way to make the water potable again is to treat it with Ag2SO4 to prompt the following reaction.
BaCl2 + Ag2SO4 = BaSO4↓+ 2 AgCl↓
Both salts produced in this reaction are insoluble and will precipitate out. After filtration, the filtrate will be pure water. It is critical, however, that Ag2SO4 be used in the right quantity for the treatment. If not enough Ag2SO4 is added, there will be some toxic BaCl2 left in the water after the reaction. If too much Ag2SO4 is used, all of the BaCl2 will precipitate out, but the water will be contaminated with the silver salt used in excess. Knowing the amount of BaCl2 in the water (200 g), we need to calculate the right quantity of Ag2SO4 for the treatment.
There are two different ways to solve this problem.
Solution 1 (old method). In Volume 1, we learned how to calculate quantities of reagents for a chemical reaction using the molecular mass method. By applying this method to the current problem (Figure 2-119), we find that 299.5 g of Ag2SO4 is needed to purify the water from the 200 g of the BaCl2 contaminant.
Figure 2-119. Calculating the amount of Ag2SO4 needed for the reaction with 200 g of BaCl2 by the molecular mass method.
Knowing the quantity of BaCl2 for the reaction (200 g) and the molecular weights of BaCl2 (208.2 a.m.u.) and Ag2SO4 (311.8 a.m.u.), we solve the problem in the conventional way (Figure 2-119). For the reaction of 208.2 g of BaCl2, 311.8 g of Ag2SO4 would be needed. For the reaction of the given 200 g of BaCl2, x g of Ag2SO4 is needed. Solving the proportion for x yields 299.5 g, the quantity of Ag2SO4 that should be used to decontaminate the water.
Solution 2 (new method). Now that we are familiar with the concept of the mole, we can solve the problem by an alternative method (Figure 2-120).
Knowing the quantity of BaCl2 for the reaction (200 g) and the molecular weights of BaCl2 (208.2 a.m.u.) and Ag2SO4 (311.8 a.m.u.), we solve the problem in the conventional way (Figure 2-119). For the reaction of 208.2 g of BaCl2, 311.8 g of Ag2SO4 would be needed. For the reaction of the given 200 g of BaCl2, x g of Ag2SO4 is needed. Solving the proportion for x yields 299.5 g, the quantity of Ag2SO4 that should be used to decontaminate the water.
Solution 2 (new method). Now that we are familiar with the concept of the mole, we can solve the problem by an alternative method (Figure 2-120).
Figure 2-120. Calculating the amount of Ag2SO4 needed for the reaction with 200 g of BaCl2 by the mole method.
As follows from the equation, one molecule of BaCl2 reacts with one molecule of Ag2SO4. Consequently, 1 mol (6.02 × 1023 molecules) of BaCl2 reacts with 1 mol (6.02 × 1023 molecules) of Ag2SO4. Consequently, any x number of moles of BaCl2 reacts with the same x number of moles of Ag2SO4. So, what is the x number of moles in 200 g of BaCl2?
If 1 mol of BaCl2 is 208.2 g, then 200 g of BaCl2 is 200/208.2 = 0.9606 mol (Figure 2-120). For our reaction, we need exactly the same number of moles of Ag2SO4. By multiplying 311.8 g (MW of Ag2SO4) by 0.9606 we get the answer, 299.5 g.
Note that both methods give the same result: 299.5 g of Ag2SO4 is needed for the treatment to make the water in the tank potable.
The convenience of the mole method is that once we know the amounts of reagents in moles, we can easily calculate not only their amounts in grams, but also the gram quantities of the products for a given reaction. Let us do that for the BaSO4 and AgCl formed in our reaction. Clearly, mixing 0.9606 moles of BaCl2 with 0.9606 moles of Ag2SO4 gives rise to 0.9606 moles of BaSO4 (MW = 233.4) and twice the number of moles (2 x 0.9606) of AgCl (MW = 143.3). To calculate the gram quantities of the BaSO4 and AgCl precipitated in our reaction, we just multiply their molecular weights by the corresponding number of moles (Figure 2-120).
0.9606 x 233.4 = 224.2 g for BaSO4
and
0.9606 x 2 x 143.3 = 275.3 g for AgCl
As we are already aware of, it is important to know the limiting reagent in a reaction, the one that is totally consumed after the reaction has gone to completion. It is the limiting reagent that determines the quantity of reaction products formed. If the quantities of the reagents for a chemical reaction are given in a mass unit, such as gram, a calculation is needed to identify the one that is limiting. In contrast, it takes just a quick glance at the same quantities expressed in moles to recognize the limiting reagent.
For example, what is the limiting reagent in the reaction of 135 g of NaOH with 110 g of HCl? From the mass quantities and molecular weights (Figure 2-121, top), it is not evident, without a calculation, which of the two is in excess and which is limiting. With the amounts expressed in moles, however, it is immediately clear that the limiting reagent is HCl (3.0 mol) and that NaOH (3.4 mol) is used in excess (Figure 2-121, bottom).
As follows from the equation, one molecule of BaCl2 reacts with one molecule of Ag2SO4. Consequently, 1 mol (6.02 × 1023 molecules) of BaCl2 reacts with 1 mol (6.02 × 1023 molecules) of Ag2SO4. Consequently, any x number of moles of BaCl2 reacts with the same x number of moles of Ag2SO4. So, what is the x number of moles in 200 g of BaCl2?
If 1 mol of BaCl2 is 208.2 g, then 200 g of BaCl2 is 200/208.2 = 0.9606 mol (Figure 2-120). For our reaction, we need exactly the same number of moles of Ag2SO4. By multiplying 311.8 g (MW of Ag2SO4) by 0.9606 we get the answer, 299.5 g.
Note that both methods give the same result: 299.5 g of Ag2SO4 is needed for the treatment to make the water in the tank potable.
The convenience of the mole method is that once we know the amounts of reagents in moles, we can easily calculate not only their amounts in grams, but also the gram quantities of the products for a given reaction. Let us do that for the BaSO4 and AgCl formed in our reaction. Clearly, mixing 0.9606 moles of BaCl2 with 0.9606 moles of Ag2SO4 gives rise to 0.9606 moles of BaSO4 (MW = 233.4) and twice the number of moles (2 x 0.9606) of AgCl (MW = 143.3). To calculate the gram quantities of the BaSO4 and AgCl precipitated in our reaction, we just multiply their molecular weights by the corresponding number of moles (Figure 2-120).
0.9606 x 233.4 = 224.2 g for BaSO4
and
0.9606 x 2 x 143.3 = 275.3 g for AgCl
As we are already aware of, it is important to know the limiting reagent in a reaction, the one that is totally consumed after the reaction has gone to completion. It is the limiting reagent that determines the quantity of reaction products formed. If the quantities of the reagents for a chemical reaction are given in a mass unit, such as gram, a calculation is needed to identify the one that is limiting. In contrast, it takes just a quick glance at the same quantities expressed in moles to recognize the limiting reagent.
For example, what is the limiting reagent in the reaction of 135 g of NaOH with 110 g of HCl? From the mass quantities and molecular weights (Figure 2-121, top), it is not evident, without a calculation, which of the two is in excess and which is limiting. With the amounts expressed in moles, however, it is immediately clear that the limiting reagent is HCl (3.0 mol) and that NaOH (3.4 mol) is used in excess (Figure 2-121, bottom).
Figure 2-121. What is the limiting reagent in the reaction of 135 g of NaOH with 110 g of HCl?
There are certain advantages in doing chemical calculations in moles rather than in grams or another mass unit. Ultimately, however, it does not matter if one prefers to use the molecular mass method or the mole method for calculations, so long as the result is correct.
2.11.3. Molar Concentration. While calculating quantities of reagents and reaction products in either grams or moles is largely a matter of taste, it is undeniable that molar concentration is often vastly preferred over mass percent concentration.
Many chemical reactions occur in solution. Such reactions are conveniently performed by mixing reagents in the form of pre-made solutions of known concentrations. First, let us refresh our memory on mass percent concentration, which refers to the amount of a solute in grams per 100 grams of solution.
There are certain advantages in doing chemical calculations in moles rather than in grams or another mass unit. Ultimately, however, it does not matter if one prefers to use the molecular mass method or the mole method for calculations, so long as the result is correct.
2.11.3. Molar Concentration. While calculating quantities of reagents and reaction products in either grams or moles is largely a matter of taste, it is undeniable that molar concentration is often vastly preferred over mass percent concentration.
Many chemical reactions occur in solution. Such reactions are conveniently performed by mixing reagents in the form of pre-made solutions of known concentrations. First, let us refresh our memory on mass percent concentration, which refers to the amount of a solute in grams per 100 grams of solution.
In this formula, we have grams (marked in red) in both the numerator and denominator. By cancelling out the gram units we can see that mass percent concentration is expressed in percentage (%).
As mentioned above, molarity (depicted M) refers to the number of moles of a solute in one liter (L) of solution and is measured in mol/L, as is clear from the formula below.
As mentioned above, molarity (depicted M) refers to the number of moles of a solute in one liter (L) of solution and is measured in mol/L, as is clear from the formula below.
The superior convenience of using solutions of a known molar concentration is illustrated by the following simple example. Suppose we need to make batches of Cu(OH)2 in various quantities on a regular basis, several times a day. For the preparation, a solution of CuCl2 is treated with a solution of NaOH and the blue precipitate of insoluble Cu(OH)2 formed is separated by filtration, washed with water, and dried.
CuCl2 + 2 NaOH = Cu(OH)2↓ + 2 NaCl
One way to deal with the assignment is each time we need to make a batch of Cu(OH)2, we calculate the amounts of CuCl2 and NaOH needed for making the required quantity of the product, weigh out the reagents, and do the reaction.
There is, however, another, much more efficient and easy way of dealing with the task. We prepare two stock solutions, one of NaOH and the other of CuCl2. As follows from the equation, per each molecule (mole) of CuCl2 we need two molecules (moles) of NaOH for the reaction. It would be handy to have a 1 molar (1M) solution of CuCl2 and a 2 molar (2M) solution of NaOH. The prepared solutions can be stored on the shelf and used at any moment to make Cu(OH)2 by simply mixing equal volumes of each. Moreover, it is very easy to calculate what volume of each solution is needed to make a particular quantity of Cu(OH)2. For example, if 10 g of Cu(OH)2 is needed to be made, we convert the mass amount to the molar amount by dividing 10 g by 97.6 g, the mass of 1 mole of Cu(OH)2. The result, 10/97.6 = 0.102, is the number of moles in 10 g of Cu(OH)2. Consequently, 0.102 mol of CuCl2 and 0.204 mol of NaOH are needed for the reaction.
Now that we know the number of moles of each reagent, we can calculate the volumes of the stock solutions to be mixed together to make the required amount of the product. Again, we think proportions. The molarity of our CuCl2 stock solution is 1M, meaning that 1 L of the solution contains 1 mole of CuCl2. We need the fraction volume x of this solution containing 0.102 mol of CuCl2.
1 mol – contained in 1 L
0.102 mol – contained in x L
x = (1 x 0.102)/1 = 0.102 L = 102 mL. This is the volume of the CuCl2 stock solution to be used for the reaction.
Similarly, we come up with a proportion to find out the volume y of the 2M NaOH stock solution.
2 mol – contained in 1 L
0.204 mol – contained in y L
y = (1 x 0.204)/2 = 0.102 L = 102 mL. It is not surprising that the same volume of the NaOH stock solution is needed because its molarity (2M) is twice that of the CuCl2 solution (1M).
If a different gram quantity of Cu(OH)2 should be made, we just convert it to mol and calculate the volumes of the stock solutions to be used.
It is worth to reemphasize that molarity expresses the number of moles of a solute per volume of solution. Why per volume, not mass, like for mass percent concentration? Because liquids are more convenient to measure by volume than by mass. The ease of accurately measuring volumes of solutions is the foundation for the broadly used important analytical method, called volumetric analysis.
CuCl2 + 2 NaOH = Cu(OH)2↓ + 2 NaCl
One way to deal with the assignment is each time we need to make a batch of Cu(OH)2, we calculate the amounts of CuCl2 and NaOH needed for making the required quantity of the product, weigh out the reagents, and do the reaction.
There is, however, another, much more efficient and easy way of dealing with the task. We prepare two stock solutions, one of NaOH and the other of CuCl2. As follows from the equation, per each molecule (mole) of CuCl2 we need two molecules (moles) of NaOH for the reaction. It would be handy to have a 1 molar (1M) solution of CuCl2 and a 2 molar (2M) solution of NaOH. The prepared solutions can be stored on the shelf and used at any moment to make Cu(OH)2 by simply mixing equal volumes of each. Moreover, it is very easy to calculate what volume of each solution is needed to make a particular quantity of Cu(OH)2. For example, if 10 g of Cu(OH)2 is needed to be made, we convert the mass amount to the molar amount by dividing 10 g by 97.6 g, the mass of 1 mole of Cu(OH)2. The result, 10/97.6 = 0.102, is the number of moles in 10 g of Cu(OH)2. Consequently, 0.102 mol of CuCl2 and 0.204 mol of NaOH are needed for the reaction.
Now that we know the number of moles of each reagent, we can calculate the volumes of the stock solutions to be mixed together to make the required amount of the product. Again, we think proportions. The molarity of our CuCl2 stock solution is 1M, meaning that 1 L of the solution contains 1 mole of CuCl2. We need the fraction volume x of this solution containing 0.102 mol of CuCl2.
1 mol – contained in 1 L
0.102 mol – contained in x L
x = (1 x 0.102)/1 = 0.102 L = 102 mL. This is the volume of the CuCl2 stock solution to be used for the reaction.
Similarly, we come up with a proportion to find out the volume y of the 2M NaOH stock solution.
2 mol – contained in 1 L
0.204 mol – contained in y L
y = (1 x 0.204)/2 = 0.102 L = 102 mL. It is not surprising that the same volume of the NaOH stock solution is needed because its molarity (2M) is twice that of the CuCl2 solution (1M).
If a different gram quantity of Cu(OH)2 should be made, we just convert it to mol and calculate the volumes of the stock solutions to be used.
It is worth to reemphasize that molarity expresses the number of moles of a solute per volume of solution. Why per volume, not mass, like for mass percent concentration? Because liquids are more convenient to measure by volume than by mass. The ease of accurately measuring volumes of solutions is the foundation for the broadly used important analytical method, called volumetric analysis.
Digression. There is yet another concentration measure that indicates the number of moles of a solute in 1 kg of solution. This type of concentration is called molal concentration or molality, and is not nearly as broadly used as molarity.
Now let us practice a bit. There is 250 mL of an aqueous solution containing 20 g of NaOH. What is the molar concentration of this solution? First, we need to convert the mass amount of NaOH to moles. As the molecular weight of NaOH is 40 a.m.u., 1 mole of NaOH is 40 g. By denoting the number of moles in 20 g of NaOH as x, we come up with the following proportion.
40 g of NaOH – 1 mol
20 g of NaOH – x mol
x = (20 x 1)/40 = 0.5 mol
This is the number of moles in the given amount of NaOH (20 g). To calculate the molarity, we need to divide this number of moles by the volume of our solution. However, we need to express the volume in liters because molarity is the number of moles of a solute per liter of solution. The given volume is 250 mL (milliliters). As "milli" denotes a factor of one thousandth, we convert the milliliters to liters by dividing 250 by 1,000. The volume in liters is therefore 250/1,000 = 0.25 L. Finally, we divide the number of moles of NaOH in the given solution by the volume of the solution in liters: 0.5 mol/0.25 L = 2 mol/L = 2M. The molarity of the given NaOH solution is 2M.
It is useful to learn how solutions of a certain molarity are prepared. For example, we need to make 1 L of a 1M aqueous solution of CuCl2. The desired molarity suggests that one liter of the solution should contain 1 mole of CuCl2, which is 134.5 g. Should we just weigh out 134.5 g of CuCl2 and dissolve it in 1 L of water? Not a very good idea. Why? Because molarity is the number of moles of a solute in 1 L of solution, not in 1 L of solvent (water in this case). What is the difference, you might ask? Without going into details, the volume of water (or another solvent) sometimes contracts and sometimes expands on dissolution of a substance in it. In many cases, the change in volume is insignificant and can be neglected. But to be on the safe side and make sure that the molarity of our solution to be prepared is accurate, we need to use a special vessel called a volumetric flask. On the neck of every volumetric flask there is a mark the flask should be filled up to in order to accurately measure out the specified volume of a liquid. Since we need to prepare 1 L of a solution, we need a 1-liter volumetric flask.
After placing 134.5 g of CuCl2 in a 1-L volumetric flask, most of the water is added in portions with occasional gentle swirling to have the salt dissolve. Finally, water is slowly and carefully added to the mark. Watch this video to see how this is done.
In general, the standard procedure for making up a solution of a certain molarity is pretty straightforward. However, there is a trick involved when using CuCl2 as the solute. This salt exists in two forms, anhydrous (CuCl2) and dihydrate (CuCl2•2H2O). The anhydrous salt is brown in color, whereas the dihydrate is blue (Figure 2-122). On contact with water, brown anhydrous CuCl2 is instantaneously converted to blue CuCl2•2H2O.
40 g of NaOH – 1 mol
20 g of NaOH – x mol
x = (20 x 1)/40 = 0.5 mol
This is the number of moles in the given amount of NaOH (20 g). To calculate the molarity, we need to divide this number of moles by the volume of our solution. However, we need to express the volume in liters because molarity is the number of moles of a solute per liter of solution. The given volume is 250 mL (milliliters). As "milli" denotes a factor of one thousandth, we convert the milliliters to liters by dividing 250 by 1,000. The volume in liters is therefore 250/1,000 = 0.25 L. Finally, we divide the number of moles of NaOH in the given solution by the volume of the solution in liters: 0.5 mol/0.25 L = 2 mol/L = 2M. The molarity of the given NaOH solution is 2M.
It is useful to learn how solutions of a certain molarity are prepared. For example, we need to make 1 L of a 1M aqueous solution of CuCl2. The desired molarity suggests that one liter of the solution should contain 1 mole of CuCl2, which is 134.5 g. Should we just weigh out 134.5 g of CuCl2 and dissolve it in 1 L of water? Not a very good idea. Why? Because molarity is the number of moles of a solute in 1 L of solution, not in 1 L of solvent (water in this case). What is the difference, you might ask? Without going into details, the volume of water (or another solvent) sometimes contracts and sometimes expands on dissolution of a substance in it. In many cases, the change in volume is insignificant and can be neglected. But to be on the safe side and make sure that the molarity of our solution to be prepared is accurate, we need to use a special vessel called a volumetric flask. On the neck of every volumetric flask there is a mark the flask should be filled up to in order to accurately measure out the specified volume of a liquid. Since we need to prepare 1 L of a solution, we need a 1-liter volumetric flask.
After placing 134.5 g of CuCl2 in a 1-L volumetric flask, most of the water is added in portions with occasional gentle swirling to have the salt dissolve. Finally, water is slowly and carefully added to the mark. Watch this video to see how this is done.
In general, the standard procedure for making up a solution of a certain molarity is pretty straightforward. However, there is a trick involved when using CuCl2 as the solute. This salt exists in two forms, anhydrous (CuCl2) and dihydrate (CuCl2•2H2O). The anhydrous salt is brown in color, whereas the dihydrate is blue (Figure 2-122). On contact with water, brown anhydrous CuCl2 is instantaneously converted to blue CuCl2•2H2O.
Figure 2-122. Samples of brown anhydrous CuCl2 (source) and blue CuCl2•2H2O (source).
How can we prepare a 1M solution of CuCl2 if we do not have anhydrous CuCl2 but only CuCl2•2H2O? The molecular weight of CuCl2•2H2O is 170.5, so we just use 170.5 g of the dihydrate to make 1 L of a 1M solution of CuCl2 in a 1-L volumetric flask. Would this solution be different from the one prepared from 134.5 g of CuCl2 as described above? Not at all. Of the 170.5 g of CuCl2•2H2O (1 mol), 134.5 g is CuCl2 (1 mol) and the rest, 36 g, is H2O (2 mol). The water the CuCl2•2H2O contains simply contributes to the overall volume amount of water needed to make the solution.
2.11.4. The Avogadro Constant. A Little about Gases. How Much Oxygen Does My Car Consume? The number 6.02 × 1023 molecules per mole of any substance is called the Avogadro constant, denoted as NA. The Avogadro constant is named after the Italian scientist Amedeo Carlo Avogadro (1776-1856) who first proposed that at the same temperature and pressure, equal volumes of gases contain equal number of molecules, regardless of the nature of the gas.
The same number of molecules in a given volume of any gas? That sounds quite counterintuitive since molecules of some gases are times larger in size than molecules of other gases. How can it be that a certain number of small monoatomic molecules of helium (He) occupy the same volume as the same number of much larger molecules such as ammonia (NH3), carbon dioxide (CO2), and methane (CH4)?
Imagine a developer building houses on one square mile of land with the goal of having as many houses built as possible. It is clear that to achieve the goal, houses with a smaller rather than larger footprint should be built. However, if there is a regulation stipulating that houses in that area must be separated by a distance of say at least 0.6 miles, the land will fit the same number of houses regardless of their footprint size (Figure 2-123).
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How can we prepare a 1M solution of CuCl2 if we do not have anhydrous CuCl2 but only CuCl2•2H2O? The molecular weight of CuCl2•2H2O is 170.5, so we just use 170.5 g of the dihydrate to make 1 L of a 1M solution of CuCl2 in a 1-L volumetric flask. Would this solution be different from the one prepared from 134.5 g of CuCl2 as described above? Not at all. Of the 170.5 g of CuCl2•2H2O (1 mol), 134.5 g is CuCl2 (1 mol) and the rest, 36 g, is H2O (2 mol). The water the CuCl2•2H2O contains simply contributes to the overall volume amount of water needed to make the solution.
2.11.4. The Avogadro Constant. A Little about Gases. How Much Oxygen Does My Car Consume? The number 6.02 × 1023 molecules per mole of any substance is called the Avogadro constant, denoted as NA. The Avogadro constant is named after the Italian scientist Amedeo Carlo Avogadro (1776-1856) who first proposed that at the same temperature and pressure, equal volumes of gases contain equal number of molecules, regardless of the nature of the gas.
The same number of molecules in a given volume of any gas? That sounds quite counterintuitive since molecules of some gases are times larger in size than molecules of other gases. How can it be that a certain number of small monoatomic molecules of helium (He) occupy the same volume as the same number of much larger molecules such as ammonia (NH3), carbon dioxide (CO2), and methane (CH4)?
Imagine a developer building houses on one square mile of land with the goal of having as many houses built as possible. It is clear that to achieve the goal, houses with a smaller rather than larger footprint should be built. However, if there is a regulation stipulating that houses in that area must be separated by a distance of say at least 0.6 miles, the land will fit the same number of houses regardless of their footprint size (Figure 2-123).
Figure 2-123. If the regulated house separation distance is much longer than any of the footprint dimensions of a house, the number of houses that can be built on a given piece of land will be the same regardless of the house size.
The situation with molecules in gases is similar. The "regulation" set up by Mother Nature requires that the distance separating molecules in gases be much larger than the size of the molecules (Figure 2-124). It is this long distance between the molecules that controls their number per gas volume, not their size. Although, unlike houses on land, molecules in a gas are constantly and quickly moving in 3-dimensional space, the long distances between them are maintained.
The situation with molecules in gases is similar. The "regulation" set up by Mother Nature requires that the distance separating molecules in gases be much larger than the size of the molecules (Figure 2-124). It is this long distance between the molecules that controls their number per gas volume, not their size. Although, unlike houses on land, molecules in a gas are constantly and quickly moving in 3-dimensional space, the long distances between them are maintained.
Figure 2-124. In a gas, the number of molecules per volume is controlled by the long distance separating the molecules, not their size.
If the same number of molecules of any gas occupies the same volume at the same temperature and under the same pressure, then NA = 6.02 × 1023 molecules = 1 mol of any gas should occupy a particular volume under certain conditions. This volume is 22.4 L, the volume of one mole of any gas.
Those familiar with the gas laws would immediately say that specifying a gas volume makes sense only if the pressure and temperature at which the volume is measured are specified. Indeed, the volume of a gas varies directly with the temperature (Charles' Law) and inversely with the pressure (Boyle's Law). The higher the temperature, the larger the volume of a gas. The higher the pressure, the smaller the volume of gas. The volume of 22.4 L is occupied by 1 mole of a gas at the temperature of 0 oC and pressure of 1 atmosphere.
Until 1982, 0 oC and 1 atmosphere were Standard Temperature and Pressure (STP). In 1982, the International Union of Pure and Applied Chemistry (IUPAC) changed the standard pressure from 1 atmosphere to 1 bar (= 1.01325 atm), while leaving the standard temperature the same (0 oC). This change in STP naturally prompted the change of the volume of one mole of a gas from 22.4 L to 22.7 L. Moreover, the volumes of 22.4 L or 22.7 L are for 1 mole of an ideal gas, a purely theoretical substance; the volume of 1 mole of a real gas slightly deviates from these values. In our course, we will be using the volume of 22.4 L for 1 mole of any gas at 0 oC and 1 atmosphere.
Knowing the volume of one mole of a gas under certain conditions has important benefits, including simpler chemical calculations and analytical procedures. Here is an example. Quicklime (CaO) is used in large quantities to make steel and concrete. The production of quicklime is based on the high-temperature decomposition of limestone, naturally occurring calcium carbonate.
CaCO3 = CaO + CO2↑
As quicklime is produced on a very large scale, it is important to know the purity of limestone in the deposit before mining it. One way of determining the percentage of CaCO3 in limestone is to accurately weigh out a sample of the rock, thermally decompose it, and measure the volume of the CO2 gas produced. Suppose the volume of CO2 produced as a result of the decomposition of a 50 g sample of limestone was measured at 10.2 L (at 0 oC and 1 atmosphere). What is the percentage of CaCO3 in the limestone sample? Figure 2-125 shows how this problem can be solved.
If the same number of molecules of any gas occupies the same volume at the same temperature and under the same pressure, then NA = 6.02 × 1023 molecules = 1 mol of any gas should occupy a particular volume under certain conditions. This volume is 22.4 L, the volume of one mole of any gas.
Those familiar with the gas laws would immediately say that specifying a gas volume makes sense only if the pressure and temperature at which the volume is measured are specified. Indeed, the volume of a gas varies directly with the temperature (Charles' Law) and inversely with the pressure (Boyle's Law). The higher the temperature, the larger the volume of a gas. The higher the pressure, the smaller the volume of gas. The volume of 22.4 L is occupied by 1 mole of a gas at the temperature of 0 oC and pressure of 1 atmosphere.
Until 1982, 0 oC and 1 atmosphere were Standard Temperature and Pressure (STP). In 1982, the International Union of Pure and Applied Chemistry (IUPAC) changed the standard pressure from 1 atmosphere to 1 bar (= 1.01325 atm), while leaving the standard temperature the same (0 oC). This change in STP naturally prompted the change of the volume of one mole of a gas from 22.4 L to 22.7 L. Moreover, the volumes of 22.4 L or 22.7 L are for 1 mole of an ideal gas, a purely theoretical substance; the volume of 1 mole of a real gas slightly deviates from these values. In our course, we will be using the volume of 22.4 L for 1 mole of any gas at 0 oC and 1 atmosphere.
Knowing the volume of one mole of a gas under certain conditions has important benefits, including simpler chemical calculations and analytical procedures. Here is an example. Quicklime (CaO) is used in large quantities to make steel and concrete. The production of quicklime is based on the high-temperature decomposition of limestone, naturally occurring calcium carbonate.
CaCO3 = CaO + CO2↑
As quicklime is produced on a very large scale, it is important to know the purity of limestone in the deposit before mining it. One way of determining the percentage of CaCO3 in limestone is to accurately weigh out a sample of the rock, thermally decompose it, and measure the volume of the CO2 gas produced. Suppose the volume of CO2 produced as a result of the decomposition of a 50 g sample of limestone was measured at 10.2 L (at 0 oC and 1 atmosphere). What is the percentage of CaCO3 in the limestone sample? Figure 2-125 shows how this problem can be solved.
Figure 2-125. Calculation of the percentage of CaCO3 in a 50-g limestone sample that produced 10.2 L of CO2 upon thermal decomposition.
1. We write a balanced chemical equation for the decomposition of CaCO3.
2. We calculate the number of moles in 50 g of pure CaCO3 by dividing 50 g by its molecular weight: 50/100 = 0.5 mol.
3. Next, we note that 1 mol of CaCO3 produces 1 mol of CO2 = 22.4 L. Therefore, if the sample was 100% pure CaCO3, its decomposition would produce 0.5 x 22.4 L = 11.2 L of CO2.
4. Given that 10.2 L of CO2 was produced, the percentage of CaCO3 in the sample is (10.2/11.2) x 100% = 91%. This purity should be sufficient for quicklime production.
At the end of this subsection, we will calculate both the volume and mass amount of oxygen a modern gasoline car consumes. Chances are that you will find the result of the calculation quite astonishing.
We drive the distance of 365 km (227 miles) from New York City to Washington, DC. Our car's fuel economy is 7.84 L of gasoline per 100 km (~30 miles per gallon). The total amount of gas needed for this trip is: (7.84 L x 365 km)/100 km = 28.6 L. Knowing the density of gasoline (0.7 kg/L), we can calculate its mass: 28.6 L x 0.7 kg/L = 20 kg.
To calculate how much oxygen is needed to burn 20 kg of gasoline, we need a balanced equation for the combustion reaction. For that, the formula of gasoline is required. Although gasoline is a mixture of hydrocarbons (Volume 4), it is appropriate to use the formula of octane, C8H18, for our calculation. The balanced chemical equation (Figure 2-126) and the mass of gasoline (20 kg) are all we need to know in order to perform the calculation.
1. We write a balanced chemical equation for the decomposition of CaCO3.
2. We calculate the number of moles in 50 g of pure CaCO3 by dividing 50 g by its molecular weight: 50/100 = 0.5 mol.
3. Next, we note that 1 mol of CaCO3 produces 1 mol of CO2 = 22.4 L. Therefore, if the sample was 100% pure CaCO3, its decomposition would produce 0.5 x 22.4 L = 11.2 L of CO2.
4. Given that 10.2 L of CO2 was produced, the percentage of CaCO3 in the sample is (10.2/11.2) x 100% = 91%. This purity should be sufficient for quicklime production.
At the end of this subsection, we will calculate both the volume and mass amount of oxygen a modern gasoline car consumes. Chances are that you will find the result of the calculation quite astonishing.
We drive the distance of 365 km (227 miles) from New York City to Washington, DC. Our car's fuel economy is 7.84 L of gasoline per 100 km (~30 miles per gallon). The total amount of gas needed for this trip is: (7.84 L x 365 km)/100 km = 28.6 L. Knowing the density of gasoline (0.7 kg/L), we can calculate its mass: 28.6 L x 0.7 kg/L = 20 kg.
To calculate how much oxygen is needed to burn 20 kg of gasoline, we need a balanced equation for the combustion reaction. For that, the formula of gasoline is required. Although gasoline is a mixture of hydrocarbons (Volume 4), it is appropriate to use the formula of octane, C8H18, for our calculation. The balanced chemical equation (Figure 2-126) and the mass of gasoline (20 kg) are all we need to know in order to perform the calculation.
Figure 1-126. Calculating the amount of O2 needed to burn 20 kg of gasoline.
The number of moles of gasoline consumed in the combustion is 175.7, as calculated by dividing its mass (20 kg = 20,000 g) by the molecular weight of C8H8 (114). According to the reaction stoichiometry, 25 moles of O2 is needed to burn 2 moles of octane. Therefore, to burn 175.7 moles of octane, (175.7 x 25)/2 = 2,196.25 moles of oxygen is required. Since 1 mol of any gas is 22.4 L, the calculated number of moles of O2 is 2,196.25 x 22.4 L ≈ 49,200 L = 49.2 cubic meters. Given the content of O2 in air (~21% by volume), the total amount of air needed for the reaction is ~235 m3.
For comparison, the passenger volume of a mid-size car is roughly 1/100 that of the air needed to get enough oxygen for the drive. Even more surprising might be the mass of O2 needed to burn 20 kg of gasoline, 70.28 kg, > 3.5 times the mass of the gasoline itself. The number 70.28 kg is calculated by multiplying the number of moles of O2 (2,196.25; see above) by its molecular weight (32).
Without these calculations, would we have ever guessed that as we drive, we consume that much atmospheric oxygen?
2.11.5. Exercises.
1. Calculate the number of moles in (a) 16 g of O2; (b) 5.6 L of N2 (at 0 oC and 1 atm.); (c) 10 g of NaCl; (d) 54 g of H2O; (e) 200 g of sucrose (sugar), C12H22O11; (f) 1 kg of quartz, SiO2; (g) 80 g of He; (h) 197 g of gold. Answer
2. Calculate mass amounts in grams for (a) 1.5 mol of CaCO3; (b) 0.2 mol of nitric acid; (c) 5 mol of NaOH; (d) 10 mol of magnesium; (e) 100 mol of quicklime. Answer
3. Calculate the number of moles at 0 oC and 1 atm in (a) 25 L of CO2; (b) 224 mL of Ne; (c) 1,120 L of N2. Answer
4. Calculate the density of xenon in g/L and g/cm3 at 0 oC and 1 atmosphere. Answer
5. Determine the molarity of (a) 500 mL of solution containing 112.2 g of KOH; (b) 250 mL of solution containing 98.1 g of H2SO4; (c) 1.5 L of solution containing 159 g of Na2CO3; (d) 320 mL of solution containing 55 g of BaSO4. Answer
6. To 50 g of a 20% solution of AgNO3, was added 100 mL of 0.5M solution of NaCl. Calculate how much AgCl precipitated in the reaction. Answer
7. How would you prepare 0.5 L of a 2M solution of H2SO4 using pure H2SO4 and water? Answer
8. Do you think molar concentration could be converted to mass percent concentration and vice versa? Answer
9. How much CuSO4•5H2O is needed to prepare 1 L of 1M solution of CuSO4? (a) 159.6 g; (b) 249.7 g; (c) 1,000 g. Answer
10. A chunk of lithium metal was thrown into water (excess). The volume of H2 that bubbled off as a result of the reaction 2 Li + 2 H2O = 2 LiOH + H2↑ was 22.4 L, as measured at 0 oC and 1 atm. What was the mass of the Li metal chunk? The atomic weight of Li is 6.94 a.m.u. Solve the problem by mental calculation. (a) 6.94 g; (b) 3.47 g; (c) 13.88 g. Answer
The number of moles of gasoline consumed in the combustion is 175.7, as calculated by dividing its mass (20 kg = 20,000 g) by the molecular weight of C8H8 (114). According to the reaction stoichiometry, 25 moles of O2 is needed to burn 2 moles of octane. Therefore, to burn 175.7 moles of octane, (175.7 x 25)/2 = 2,196.25 moles of oxygen is required. Since 1 mol of any gas is 22.4 L, the calculated number of moles of O2 is 2,196.25 x 22.4 L ≈ 49,200 L = 49.2 cubic meters. Given the content of O2 in air (~21% by volume), the total amount of air needed for the reaction is ~235 m3.
For comparison, the passenger volume of a mid-size car is roughly 1/100 that of the air needed to get enough oxygen for the drive. Even more surprising might be the mass of O2 needed to burn 20 kg of gasoline, 70.28 kg, > 3.5 times the mass of the gasoline itself. The number 70.28 kg is calculated by multiplying the number of moles of O2 (2,196.25; see above) by its molecular weight (32).
Without these calculations, would we have ever guessed that as we drive, we consume that much atmospheric oxygen?
2.11.5. Exercises.
1. Calculate the number of moles in (a) 16 g of O2; (b) 5.6 L of N2 (at 0 oC and 1 atm.); (c) 10 g of NaCl; (d) 54 g of H2O; (e) 200 g of sucrose (sugar), C12H22O11; (f) 1 kg of quartz, SiO2; (g) 80 g of He; (h) 197 g of gold. Answer
2. Calculate mass amounts in grams for (a) 1.5 mol of CaCO3; (b) 0.2 mol of nitric acid; (c) 5 mol of NaOH; (d) 10 mol of magnesium; (e) 100 mol of quicklime. Answer
3. Calculate the number of moles at 0 oC and 1 atm in (a) 25 L of CO2; (b) 224 mL of Ne; (c) 1,120 L of N2. Answer
4. Calculate the density of xenon in g/L and g/cm3 at 0 oC and 1 atmosphere. Answer
5. Determine the molarity of (a) 500 mL of solution containing 112.2 g of KOH; (b) 250 mL of solution containing 98.1 g of H2SO4; (c) 1.5 L of solution containing 159 g of Na2CO3; (d) 320 mL of solution containing 55 g of BaSO4. Answer
6. To 50 g of a 20% solution of AgNO3, was added 100 mL of 0.5M solution of NaCl. Calculate how much AgCl precipitated in the reaction. Answer
7. How would you prepare 0.5 L of a 2M solution of H2SO4 using pure H2SO4 and water? Answer
8. Do you think molar concentration could be converted to mass percent concentration and vice versa? Answer
9. How much CuSO4•5H2O is needed to prepare 1 L of 1M solution of CuSO4? (a) 159.6 g; (b) 249.7 g; (c) 1,000 g. Answer
10. A chunk of lithium metal was thrown into water (excess). The volume of H2 that bubbled off as a result of the reaction 2 Li + 2 H2O = 2 LiOH + H2↑ was 22.4 L, as measured at 0 oC and 1 atm. What was the mass of the Li metal chunk? The atomic weight of Li is 6.94 a.m.u. Solve the problem by mental calculation. (a) 6.94 g; (b) 3.47 g; (c) 13.88 g. Answer