# 19 Molarity, Solutions, and Dilutions (M4Q6)

## Introduction

Learning Objectives for Solution Stoichiometry

- Calculate molarity of a solution and solve stoichiometry problems using solution molarities.

| Solutions & Molarity | Dilution of Solutions | Solution Stoichiometry |

| Key Concepts and Summary | Key Equations | Glossary | End of Section Exercises |

In preceding sections, we focused on the composition of *substances*: samples of matter that contain only one type of element or compound. However, *mixtures*—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (Figure 1). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.

## Solutions & Molarity

We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology.

The relative amount of a given solution component is known as its **concentration**. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the **solvent** and may be viewed as the medium in which the other components are dispersed, or **dissolved**. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an **aqueous solution **(indicated by (aq) in a chemical reaction).

A **solute** is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as **dilute** (of relatively low concentration) and **concentrated** (of relatively high concentration).

Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. **Molarity ( M)**, or molar concentration, is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:

M = [latex]\frac{\text{mol solute}}{\text{L solution}}[/latex]

Students often get confused with the use of the terms *molarity* and *molar*. The terms molarity and molar are synonymous, but have different uses. For example, if you have a solution that is labeled “3.5 M”. We could say that the *molarity *(or molar concentration) of the solution is 3.5, or we could say that the solution is 3.5 *molar*. We would *not* typically read “3.5 M” as “3.5 molarity”.

Also notice that due to the definition of molarity, if you are provided with a molarity, you are inherently provided with a conversion factor between moles of solute and liters of solution (this is similar to how we can think about density as a conversion factor, or ratio, of mass to volume of a substance).

### Example 1

**Calculating Molar Concentrations**

A 355 mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?

**Solution**

Since the moles of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:

**Check Your Learning**

A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL?

### Answer:

0.05 *M*

### Example 2

**Deriving Moles and Volumes from Molar Concentrations**

How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example 1?

**Solution**

In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 1, 0.375 *M*:

M = [latex]\frac{\text{mol solute}}{\text{L solution}}[/latex]

mol solute = M × L solution

mol solute = 0.375 [latex]\frac{\text{mol sugar}}{\text{L}}[/latex] × [latex](10 \;\text{mL} \times \frac{1\ \text{L}}{1000 \;\text{mL}})[/latex] = 0.004 mol sugar

Here we are using molarity as a conversion factor—since we know molarity and volume, we can find moles of substance.

**Check Your Learning**

What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?

### Answer:

80 mL

### Example 3

**Calculating Molar Concentrations from the Mass of Solute**

Distilled white vinegar (Figure 2) is a solution of acetic acid, CH_{3}CO_{2}H, in water. A 0.500 L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?

**Figure 2.**Distilled white vinegar is a solution of acetic acid in water.

**Solution**

As in the previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of the amount of moles, so we must use the solute’s molar mass to obtain the amount of solute in moles:

M = [latex]\frac{\text{mol solute}}{\text{L solution}} = \frac{25.2 \;\text{g CH}_3\text{CO}_2\text{H}\;\times\;\frac{1 \;\text{mol CH}_2\text{CO}_2\text{H}}{60.052 \;\text{g CH}_2\text{CO}_2\text{H}}}{0.500 \;\text{L solution}}[/latex] = 0.839 [latex]\frac{\text{mol solute}}{\text{L solution}}[/latex]

= 0.839 M

**Check Your Learning**

Calculate the molarity of 6.52 g of CoCl_{2} (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.

### Answer:

0.674 M

### Example 4

**Determining the Mass of Solute in a Given Volume of Solution**

How many grams of NaCl are contained in 0.250 L of a 5.30 *M* solution?

**Solution**

The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 2:

M = [latex]\frac{\text{mol solute}}{\text{L solution}}[/latex]

mol solute = M × L solution

mol solute = 5.30 [latex]\frac{\text{mol NaCl}}{\text{L}}[/latex] × 0.250 L = 1.325 mol NaCl

Finally, this molar amount is used to derive the mass of NaCl:

1.325 mol NaCl × [latex]\frac{58.44\;\text{g NaCl}}{\text{mol NaCl}}[/latex] = 77.4 g NaCl

**Check Your Learning**

How many grams of CaCl_{2} (110.98 g/mol) are contained in 250.0 mL of a 0.200 *M* solution of calcium chloride?

### Answer:

5.55 g CaCl_{2}

When performing calculations stepwise, as in Example 4, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 4, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.

In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (see Example 5). This eliminates intermediate steps so that only the final result is rounded.

### Example 5

**Determining the Volume of Solution Containing a Given Mass of Solute**

In Example 3, we found the typical concentration of vinegar to be 0.839 *M*. What volume of vinegar contains 75.6 g of acetic acid?

**Solution**

First, use the molar mass to calculate moles of acetic acid from the given mass:

g solute × [latex]\frac{\text{mol solute}}{\text{g solute}}[/latex] = mol solute

Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:

mol solute × [latex]\frac{\text{L solution}}{\text{mol solute}}[/latex] = L solution

Combining these two steps into one yields:

g solute × [latex]\frac{\text{mol solute}}{\text{g solute}}[/latex] × [latex]\frac{\text{L solution}}{\text{mol solute}}[/latex] = L solution

75.6 g CH_{3}CO_{2}H × [latex]\frac{1\;\text{mol}\;\text{CH}_{3}\text{CO}_{2}\text{H}}{60.05\;\text{g}\;\text{CH}_{3}\text{CO}_{2}\text{H}}[/latex] × [latex]\frac{\text{L solution}}{0.839\;\text{mol}\;\text{CH}_{3}\text{CO}_{2}\text{H}}[/latex] = 1.50 L solution

**Check Your Learning**

What volume of a 1.50 M KBr solution contains 66.0 g KBr?

### Answer:

0.370 L

## Dilution of Solutions

**Dilution** is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, we might say that a glass of iced tea becomes diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 3).

**Figure 3.**Both solutions contain the same mass of copper nitrate. The solution on the right is more dilute because the copper nitrate is dissolved in more solvent. (credit: Mark Ott)

Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated *stock solution*, we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents (Figure 4).

**Figure 4.**A solution of KMnO

_{4}is prepared by mixing water with 4.74 g of KMnO

_{4}in a flask. (credit: modification of work by Mark Ott)

A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the amount of moles of solute (*n*) in a solution is equal to the product of the solution’s molarity (M) and its volume in liters (L):

*n*= M × L

Expressions like these may be written for a solution before and after it is diluted:

*n*

_{1}= M

_{1 }× L

_{1}

*n*

_{2}= M

_{2}× L

_{2}

where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process *does not change the amount of solute in the solution, **n*_{1} = *n*_{2}. Thus, these two equations may be set equal to one another:

M_{1 }× L_{1} = M_{2} × L_{2}

This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the dimensional analysis method. Reflecting this versatility, the dilution equation is often written in the more general form:

C_{1} × V_{1} = C_{2} × V_{2}

where C and V are concentration and volume, respectively.

Use the simulation to explore the relationship between solute amount, solution volume, and concentration and to confirm the dilution equation.

### Example 6

**Determining the Concentration of a Diluted Solution**

If 0.850 L of a 5.00 M solution of copper nitrate, Cu(NO_{3})_{2}, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?

**Solution**

We are given the volume and concentration of a stock solution, *V*_{1} and *C*_{1}, and the volume of the resultant diluted solution, *V*_{2}. We need to find the concentration of the diluted solution, *C*_{2}. We thus rearrange the dilution equation in order to isolate *C*_{2}:

C_{1} × V_{1} = C_{2} × V_{2}

C_{2} = [latex]\frac{\text{C}_{1}\text{V}_{1}}{\text{V}_{2}}[/latex]

Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution’s concentration to be less than one-half 5 *M*. We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:

_{2 }= [latex]\frac{0.850\;\text{L}\;\times\; 5.00\;\frac{\text{mol}}{\text{L}}}{1.80\;\text{L}}[/latex] = 2.36 M

This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M).

**Check Your Learning**

What is the concentration of the solution that results from diluting 25.0 mL of a 2.04 M solution of CH_{3}OH to 500.0 mL?

### Answer:

0.102 M CH_{3}OH

### Example 7

**Volume of a Diluted Solution**

What volume of 0.12 M HBr can be prepared from 11 mL (0.011 L) of 0.45 M HBr?

**Solution**

We are given the volume and concentration of a stock solution, *V*_{1} and *C*_{1}, and the concentration of the resultant diluted solution, *C*_{2}. We need to find the volume of the diluted solution, *V*_{2}. We thus rearrange the dilution equation in order to isolate *V*_{2}:

C_{1} × V_{1} = C_{2} × V_{2}

V_{2} = [latex]\frac{\text{C}_{1}\text{V}_{1}}{\text{C}_{2}}[/latex]

Since the diluted concentration (0.12 *M*) is slightly more than one-fourth the original concentration (0.45 *M*), we would expect the volume of the diluted solution to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields:

V_{2} = [latex]\frac{(0.45\;\text{M})(0.011\;\text{L})}{0.12\;\text{M}}[/latex]

V_{2} = 0.041 L

The volume of the 0.12 M solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate.

**Check Your Learning**

A laboratory experiment calls for 0.125 M HNO_{3}. What volume of 0.125 M HNO_{3} can be prepared from 0.250 L of 1.88 M HNO_{3}?

### Answer:

3.76 L

### Example 8

**Volume of a Concentrated Solution Needed for Dilution**

What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH?

**Solution**

We are given the concentration of a stock solution, *C*_{1}, and the volume and concentration of the resultant diluted solution, *V*_{2} and *C*_{2}. We need to find the volume of the stock solution, *V*_{1}. We thus rearrange the dilution equation in order to isolate *V*_{1}:

C_{1} × V_{1} = C_{2} × V_{2}

V_{1} = [latex]\frac{\text{C}_{2}\text{V}_{2}}{\text{C}_{1}}[/latex]

Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:

V_{1 }= [latex]\frac{(0.100\;\text{M})(5.00\;\text{L})}{1.59\;\text{M}}[/latex]

V_{1} = 0.314 L

Thus, we would need 0.314 L of the 1.59 M solution to prepare the desired solution. This result is consistent with our rough estimate.

**Check Your Learning**

What volume of a 0.575 M solution of glucose, C_{6}H_{12}O_{6}, can be prepared from 50.00 mL of a 3.00 M glucose solution?

### Answer:

0.261 L

## Solution Stoichiometry

In an earlier section, we learned that the coefficients of a balanced chemical equation provide the relative numbers of the chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. The process of using the coefficients in a balanced chemical equation to relate the amount of reactants and products can be summarized as in Figure 5.

**Figure 5.**A flowchart for stoichiometric calculations involving pure substances. The molar masses of the reactants and the products are used as conversion factors so that you can calculate the mass of product from the mass of reactant and vice versa.

Since solutions offer a convenient medium for carrying out chemical reactions, it is often necessary to know how much of one solution will react with a given quantity of another. We can perform calculations similar to those outlined in Figure 5 by first using the volume of the solution and the concentration of the solution to determine how many moles of solute are present. This process is outlined in Figure 6.

**Figure 6.**Flowchart for stoichiometric calculations of reactions in solution. The volume and concentration of a solution can be used to determine the moles of a substance, which can then be used to calculate the amount of another substance required or produced in the reaction.

The initial goal is the same for both procedures—convert to moles. You can accomplish this by either using mass and molar mass of a substance or by using volume and concentration of a solution. Once you have determined how many moles of a reactant or product are present, you can use the stoichiometric mole ratio to determine the quantity of the other reactants or products in the reaction.

### Example 9

**Solution Stoichiometry**

Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN)_{2}]^{−} ion. Gold is then recovered by reduction with metallic zinc according to the following equation:

Zn(s) + 2 [Au(CN)_{2}]^{–}(aq) → [Zn(CN)_{4}]^{2-}(aq) + 2 Au(s)

What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10^{−4} M solution of [Au(CN)_{2}]^{−}?

**Solution**

First, we check the chemical equation to make sure it is balanced as written. Then we calculate the number of moles of [Au(CN)_{2}]^{−} present by multiplying the volume of the solution by its concentration. From the balanced chemical equation, we then use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, we multiply the number of moles of gold by its molar mass.

We can adapt Figure 6 to show this process as follows:

**Check Your Learning**

What mass of solid lanthanum(III) oxalate nonahydrate [La_{2}(C_{2}O_{4})_{3}·9H_{2}O] (molar mass = 703.87 g/mol) can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl_{3} by adding a stoichiometric amount of sodium oxalate (Na_{2}C_{2}O_{4})? The *unbalanced* reaction is as follows:

LaCl_{3}(aq) + Na_{2}C_{2}O_{4}(s) + H_{2}O(l) → La_{2}(C_{2}O_{4})_{3}·9H_{2}O(s) + NaCl(aq)

### Answer:

3.89 g

### Example 10

**Limiting Reactants in Solution**

When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows:

2 AgNO_{3}(aq) + K_{2}Cr_{2}O_{7}(aq) → Ag_{2}Cr_{2}O_{7}(s) + 2 KNO_{3}(aq)

What mass of Ag_{2}Cr_{2}O_{7} is formed when 500 mL of 0.17 M K_{2}Cr_{2}O_{7} are mixed with 250 mL of 0.57 M AgNO_{3}?

**Solution**

First, calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity.

mol K_{2}Cr_{2}O_{7} = 500 mL × [latex]\frac{1\;\text{L}}{1000\;\text{mL}}[/latex] × [latex]\frac{0.17\;\text{mol}\;\text{K}_{2}\text{Cr}_{2}\text{O}_{7}}{1\;\text{L}}[/latex] = 0.085 mol K_{2}Cr_{2}O_{7}

mol AgNO_{3} = 250 mL × [latex]\frac{1\;\text{L}}{1000\;\text{mL}}[/latex] × [latex]\frac{0.57\;\text{mol}\;\text{AgNO}_{3}}{1\;\text{L}}[/latex] = 0.14 mol AgNO_{3}

Then, determine the limiting reactant by calculating the moles of Ag_{2}Cr_{2}O_{7} produced if each reactant were completely reacted.

0.085 mol K_{2}Cr_{2}O_{7} × [latex]\frac{1\;\text{mol}\;\text{Ag}_{2}\text{Cr}_{2}\text{O}_{7}}{1\;\text{mol}\;\text{K}_{2}\text{Cr}_{2}\text{O}_{7}}[/latex] = 0.085 mol Ag_{2}Cr_{2}O_{7}

0.014 mol AgNO_{3} [latex]\frac{1\;\text{mol}\;\text{Ag}_{2}\text{Cr}_{2}\text{O}_{7}}{2\;\text{mol}\;\text{AgNO}_{3}}[/latex] = 0.071 mol Ag_{2}Cr_{2}O_{7}

Since complete reaction of AgNO_{3} produces less product (Ag_{2}Cr_{2}O_{7}), AgNO_{3} is the limiting reactant and K_{2}Cr_{2}O_{7} is the excess reactant.

To find what mass of Ag_{2}Cr_{2}O_{7 }is formed, we use the molar mass (431.72 g/mol):

0.071 mol Ag_{2}Cr_{2}O_{7} × [latex]\frac{431.72\;\text{g}\;\text{Ag}_{2}\text{Cr}_{2}\text{O}_{7}}{1\;\text{mol}\;\text{Ag}_{2}\text{Cr}_{2}\text{O}_{7}}[/latex] = 30.8 g Ag_{2}Cr_{2}O_{7}

**Check Your Learning**

Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation:

2 NaHCO_{3}(aq) + H_{2}SO_{4}(aq) → 2 CO_{2}(g) + Na_{2}SO_{4}(aq) + 2 H_{2}O(l)

If 13.0 mL of 3.0 M H_{2}SO_{4} are added to 732 mL of 0.112 M NaHCO_{3}, what mass of CO_{2} is produced?

### Answer:

3.4 g

## Key Concepts and Summary

Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.

We can perform stoichiometric calculations for reactions that occur in solution by using the volume and concentration of the solution to determine the moles of substance present. Stoichiometric factors then work the same as before to convert between moles of reactants and products.

## Key Equations

- M = [latex]\frac{\text{mol solute}}{\text{L solution}}[/latex]
- C
_{1}× V_{1}= C_{2}× V_{2} - Percent by mass = [latex]\frac{\text{mass of solute}}{\text{mass of solution}}[/latex] × 100

## Glossary

- aqueous solution
- solution for which water is the solvent

- concentrated
- qualitative term for a solution containing solute at a relatively high concentration

- concentration
- quantitative measure of the relative amounts of solute and solvent present in a solution

- dilute
- qualitative term for a solution containing solute at a relatively low concentration

- dilution
- process of adding solvent to a solution in order to lower the concentration of solutes

- dissolved
- describes the process by which solute components are dispersed in a solvent

- mass percentage
- ratio of solute-to-solution mass expressed as a percentage

- mass-volume percent
- ratio of solute mass to solution volume, expressed as a percentage

- molarity (
*M*) - unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution

- parts per billion (ppb)
- ratio of solute-to-solution mass multiplied by 10
^{9}

- parts per million (ppm)
- ratio of solute-to-solution mass multiplied by 10
^{6}

- solute
- solution component present in a concentration less than that of the solvent

- solvent
- solution component present in a concentration that is higher relative to other components

- volume percentage
- ratio of solute-to-solution volume expressed as a percentage

### Chemistry End of Chapter Exercises

### Molarity

- Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.
- What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different?
- Determine the molarity for each of the following solutions:
- 0.444 mol of CoCl
_{2}in 0.654 L of solution - 98.0 g of phosphoric acid, H
_{3}PO_{4}, in 1.00 L of solution - 0.2074 g of calcium hydroxide, Ca(OH)
_{2}, in 40.00 mL of solution - 10.5 kg of Na
_{2}SO_{4}·10H_{2}O in 18.60 L of solution - 7.0 × 10
^{−3}mol of I_{2}in 100.0 mL of solution - 1.8 × 10
^{4}mg of HCl in 0.075 L of solution

- 0.444 mol of CoCl
- Consider this question: What is the mass of the solute in 0.500 L of 0.30
*M*glucose, C_{6}H_{12}O_{6}, used for intravenous injection?- Outline the steps necessary to answer the question.
- Answer the question.

- Calculate the number of moles and the mass of the solute in each of the following solutions:
- 2.00 L of 18.5
*M*H_{2}SO_{4}, concentrated sulfuric acid - 100.0 mL of 3.8 × 10
^{−5}*M*NaCN, the minimum lethal concentration of sodium cyanide in blood serum - 5.50 L of 13.3
*M*H_{2}CO, the formaldehyde used to “fix” tissue samples - 325 mL of 1.8 × 10
^{−6}*M*FeSO_{4}, the minimum concentration of iron sulfate detectable by taste in drinking water

- 2.00 L of 18.5
- Consider this question: What is the molarity of KMnO
_{4}in a solution of 0.0908 g of KMnO_{4}in 0.500 L of solution?- Outline the steps necessary to answer the question.
- Answer the question.

- Calculate the molarity of each of the following solutions:
- 0.195 g of cholesterol, C
_{27}H_{46}O, in 0.100 L of serum, the average concentration of cholesterol in human serum - 4.25 g of NH
_{3}in 0.500 L of solution, the concentration of NH_{3}in household ammonia - 1.49 kg of isopropyl alcohol, C
_{3}H_{7}OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol - 0.029 g of I
_{2}in 0.100 L of solution, the solubility of I_{2}in water at 20 °C

- 0.195 g of cholesterol, C
- There is about 1.0 g of calcium, as Ca
^{2+}, in 1.0 L of milk. What is the molarity of Ca^{2+}in milk? - If 0.1718 L of a 0.3556-
*M*C_{3}H_{7}OH solution is diluted to a concentration of 0.1222*M*, what is the volume of the resulting solution? - What volume of a 0.33-
*M*C_{12}H_{22}O_{11}solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025*M*? - What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?
- 1.00 L of a 0.250-
*M*solution of Fe(NO_{3})_{3}is diluted to a final volume of 2.00 L - 0.5000 L of a 0.1222-
*M*solution of C_{3}H_{7}OH is diluted to a final volume of 1.250 L - 2.35 L of a 0.350-
*M*solution of H_{3}PO_{4}is diluted to a final volume of 4.00 L - 22.50 mL of a 0.025-
*M*solution of C_{12}H_{22}O_{11}is diluted to 100.0 mL

- 1.00 L of a 0.250-
- What is the final concentration of the solution produced when 225.5 mL of a 0.09988-
*M*solution of Na_{2}CO_{3}is allowed to evaporate until the solution volume is reduced to 45.00 mL? - A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?
- An experiment in a general chemistry laboratory calls for a 2.00-
*M*solution of HCl. How many mL of 11.9*M*HCl would be required to make 250 mL of 2.00*M*HCl? - The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg/L. If an industry is discharging hexavalent chromium as potassium dichromate (K
_{2}Cr_{2}O_{7}), what is the maximum permissible molarity of that substance?

### Solution Stoichiometry

- In a beaker, 150. mL of a 0.500 M silver(I) nitrate solution is mixed with 250. mL of a 0.250 M potassium chromate solution and a solid is collected.
- Write the complete balanced equation for this reaction.
- What is the name and formula for the solid collected?
- How much solid is collected (in grams)?

- How much 0.496 M sodium hydroxide is required to completely neutralize 20.0 mL of 0.809 M hydrochloric acid?
- How many grams of lead(II) iodide (molar mass = 461.00 g/mol) is collected if 60.0 mL of 0.300 M potassium iodide is mixed with 25.0 mL of 0.540 M lead(II) nitrate? Assume 100% yield for this reaction.

**Answers to Chemistry End of Chapter Exercises**

- Molarity of NaCl changes; Moles of NaCl remains the same
- These two samples have the same mole:volume ratio; These two samples have different total volume and different individual amount of moles per volume
- (a) 0.679
*M*; (b) 1.00*M*; (c) 0.06998*M*; (d) 1.75*M*; (e) 0.070*M*; (f) 6.6*M* - (a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g
- (a) 37.0 mol H
_{2}SO_{4}; 3.63 × 10^{3}g H_{2}SO_{4}; (b) 3.8 × 10^{−6}mol NaCN; 1.9 × 10^{−4}g NaCN; (c) 73.2 mol H_{2}CO; 2.20 kg H_{2}CO; (d) 5.9 × 10^{−7}mol FeSO_{4}; 8.9 × 10^{−5}g FeSO_{4} - (a) Determine the molar mass of KMnO
_{4}; determine the number of moles of KMnO_{4}in the solution; from the number of moles and the volume of solution, determine the molarity; (b) 1.15 × 10^{−3}*M* - (a) 5.04 × 10
^{−3}*M*; (b) 0.499*M*; (c) 9.92*M*; (d) 1.1 × 10^{−3}*M* - 0.025
*M* - 0.4999 L
- 1.9 mL
- (a) 0.125
*M*; (b) 0.04888*M*; (c) 0.206*M*; (e) 0.0056*M* - 0.501 M
- 11.9
*M* - 42.0 mL
- Maximum permissible molarity: 4.81×10
^{-6}M - (a) 2 AgNO
_{3}(aq) + K_{2}CrO_{4}(aq) → Ag_{2}CrO_{4}(s) + 2 KNO_{3}(aq)

(b) silver(I) chromate, Ag_{2}CrO_{4}

(c) 12.4 g Ag_{2}CrO_{4}(AgNO_{3}is the limiting reactant) - 32.6 mL
- 4.15 g