Chapter 9.3: Units for Concentration
 Page ID
 42101
Prince George's Community College 

Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry 
Learning Objective
 To describe the concentration of a solution in the way that is most appropriate for a particular problem or application.
All of us have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration. The quantity of solute that is dissolved in a particular quantity of solvent or solution. of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for reactions that occur in solution. Chemists use many different ways to define concentrations, some of which are described in this section.
Molarity
The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) is a common unit of concentration and is the number of moles of solute present in exactly \(1 L\) of solution \((mol/L)\) of a solution is the number of moles of solute present in exactly \(1 L\) of solution. Molarity is also the number of millimoles of solute present in exactly 1 mL of solution:
\( molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \tag{9.3.1}\)
The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as \(M\). An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the concentration of a solute. So
\([\rm{sucrose}] = 1.00\: M\)
is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either
\( V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \tag{9.3.2}\)
or
\( V_{mL} M_{mmol/mL} = \cancel{mL} \left( \dfrac{mmol} {\cancel{mL}} \right) = mmoles \tag{9.3.3}\)
Mole Fraction
We can also describe the concentration of a molecule as the mole fraction, We had previously defined mole fraction in describing mixtures of gases in Section 6.5.
The mole fraction (X) of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture (n_{t}):
\[x_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\tag{9.3.4}\]
The mole fraction is a dimensionless quantity between 0 and 1. If X_{A} = 1.0, then the sample is pure A, not a mixture. If X_{A} = 0, then no A is present in the mixture. The sum of the mole fractions of all the components present must equal 1.
Example 9.3.1
Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL.
 What is its molarity?
 What is its mole fraction?
Given: mass of substance and mass and density of solution
Asked for: molarity and mole fraction
Strategy:
A Calculate the number of moles of acetic acid in the sample. Then calculate the number of liters of solution from its mass and density. Use these results to determine the molarity of the solution.
B Determine the mass of the water in the sample and calculate the number of moles of water. Then determine the mole fraction of acetic acid by dividing the number of moles of acetic acid by the total number of moles of substances in the sample.
Solution:

A The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass. The volume of the solution equals its mass divided by its density. The calculations follow:
\( moles \; CH_{3}CO_{2}H=\dfrac{3.78\;\cancel{g}\;CH_{3}CO_{2}H }{60.05\;\cancel{g}/mol}=0.0629\; mol \)
\( volume=\dfrac{mass }{density}=\dfrac{100.0\; \cancel{g} \; solution}{1.00\;\cancel{g}/mL} =100.00\;mL \)
\( molarity \;of \; CH_{3}CO_{2}H=\dfrac{moles\;CH_{3}CO_{2}H }{liter\;solution}= \dfrac{0.0629\; mol \; CH_{3}CO_{2}H}{\left (100\;\cancel{mL} \right )\left ( 1L/\cancel{mL} \right )}=0.629\;M \;CH_{3}CO_{2}H \) 
This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than 1/2 mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL.

B To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have
\( moles\;H_{2}O=\dfrac{96.2 \;\cancel{g} \;H_{2}O }{18.02\;\cancel{g}/mol }=5.34\;mol\;H_{2}O \)The mole fraction X of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present:
\( X_{CH_{3}CO_{2}H}=\dfrac{moles\; \; CH_{3}CO_{2}H }{moles\; \; CH_{3}CO_{2}H+moles\;H_{2}O}=\dfrac{0.0629\;mol}{0.0629\;mol+5.34\;mol}=0.0116 \)This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01.
Exercise
A solution of HCl gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of HCl per 100.0 g of solution, and its density is 1.10 g/mL.
 What is its molarity?
 What is its mole fraction?
Answer:
 6.10 M HCl
 X_{HCl} = 0.111
Molality
While molarity is ideal for preparing solutions using volumetric flasks it has the disadvantage that volume varies with temperature and also can be affected by composition and pressure. The concentration of a solution can also be described by its molality (m)The number of moles of solute present in exactly 1 kg of solvent., the number of moles of solute per kilogram of solvent:
\( molality\left ( m \right )=\dfrac{\left (moles \; solute \right ) }{\left (kilograms \; solvent \right )} \tag{9.3.5} \)
Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). Molality is independent of temperature, pressure and composition.
For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of H_{2}O under these conditions is very close to 1.0 L, and a 0.50 M solution of KBr in water, for example, has approximately the same concentration as a 0.50 m solution.
Mass Percent
Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentageThe ratio of the total mass of the solute to the total mass of the solution., parts per million (ppm)Milligrams of solute per kilogram of solvent., or parts per billion (ppb)Micrograms of solute per kilogram of solvent.:
\( mass \; percentage = \dfrac{\left (mass \; of \; solute\right )}{\left (mass \; of \; solution \right )} \times 100 \tag{9.3.6} \)
\( parts \; per \; million \left (ppm \right ) = \dfrac{\left (mass \; of \; solute \right ) }{\left (mass \; of \; solution \right )} \times 10^{6} \tag{9.3.7} \)
\( parts \; per \; billion \left (ppb \right ) = \dfrac{\left (mass \; of \; solute \right ) }{\left (mass \; of \; solution \right )} \times 10^{9} \tag{9.3.8} \)
In the health sciences, the concentration of a solution is typically expressed as parts per thousand (ppt)Grams of solute per kilogram of solvent, primarily used in the health sciences., indicated as a proportion. For example, adrenalin, the hormone produced in highstress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution.
The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of H_{2}SO_{4} per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).
Mixing Ratio
To confuse things, many fields use mixing ratios, the ratio of the number of molecules or moles of one species to the total number in a fixed volume. For example, the concentration of CO_{2} in the atmosphere today is 400 ppm, or there are 400 CO_{2} molecules in every million molecules in the air. More properly this should be written as 400 ppmV, that is 400 parts per million by volume, because Avogadro's principle tells us that equal volumes of gas contain equal numbers of molecules.
Example 9.3.2
Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 13.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.
 What is the molarity of the solution?
 What is the mass of benzene in the sample?
Given: volume of sample, solute concentration, and density of solution
Asked for: molarity of solute and mass of solute in 250 mL
Strategy:
A Use the concentration of the solute in parts per million to calculate the molarity.
B Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.
Solution:

A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 13.7 ppm of benzene. Because 13.7 ppm is equivalent to 13.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 13.7 mg of benzene per liter (1000 mL). The molarity is therefore
\( molarity =\dfrac{moles}{liter}=\dfrac{\left (12.7\;\cancel{mg} \right )\left ( 1\;\cancel{g}/1000\;\cancel{mg} \right )\left ( 1\;mol/78.114\;\cancel{g} \right )}{1000\;mL} \)=1.63\times 10^{4}\;M \) 
B We are given that there are 13.7 mg of benzene per 1000 g of solution, which is equal to 13.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is
\( mass\;of\;benzene=\dfrac{\left ( 12.7\; mg\; benzene \right )\left ( 250\;\cancel{mL} \right )}{1000\;\cancel{mL}}= 3.18\;mg\;benzene \)
Exercise
The maximum allowable concentration of lead in drinking water is 9.0 ppb. What is the molarity of Pb^{2+} in a 9.0 ppb aqueous solution? Use your calculated concentration to determine how many grams of Pb^{2+} are in an 8 oz glass of water.
Answer: 4.3 × 10^{−8} M; 2 × 10^{−6} g
Choosing Units to Use
How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature.
Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as we saw in Section 6.5. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies.
Table 9.3.1 summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example 6.
Table 9.3.1 Different Units for Expressing the Concentrations of Solutions*
Unit  Definition  Application 

molarity (M)  moles of solute/liter of solution (mol/L)  Used for quantitative reactions in solution and titrations; mass and molecular mass of solute and volume of solution are known. 
mole fraction (X)  moles of solute/total moles present (mol/mol)  Used for partial pressures of gases and vapor pressures of some solutions; mass and molecular mass of each component are known. 
molality (m)  moles of solute/kg of solvent (mol/kg)  Used in determining how colligative properties vary with solute concentration; masses and molecular mass of solute are known. 
mass percentage (%)  [mass of solute (g)/mass of solution (g)] × 100  Useful when masses are known but molecular masses are unknown. 
parts per thousand (ppt)  [mass of solute/mass of solution] × 10^{3} (g solute/kg solution)  Used in the health sciences, ratio solutions are typically expressed as a proportion, such as 1:1000. 
parts per million (ppm)  [mass of solute/mass of solution] × 10^{6} (mg solute/kg solution)  Used for trace quantities; masses are known but molecular masses may be unknown. 
parts per billion (ppb)  [mass of solute/mass of solution] × 10^{9} (µg solute/kg solution)  Used for trace quantities; masses are known but molecular masses may be unknown. 
*The molarity of a solution is temperature dependent, but the other units shown in this table are independent of temperature. 
Example 9.3.3
Vodka is essentially a solution of pure ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80proof vodka.
 the mass percentage
 the mole fraction
 the molarity
 the molality
Given: volume percent and density
Asked for: mass percentage, mole fraction, molarity, and molality
Strategy:
A Use the density of the solute to calculate the mass of the solute in 100.0 mL of solution. Calculate the mass of water in 100.0 mL of solution.
B Determine the mass percentage of solute by dividing the mass of ethanol by the mass of the solution and multiplying by 100.
C Convert grams of solute and solvent to moles of solute and solvent. Calculate the mole fraction of solute by dividing the moles of solute by the total number of moles of substances present in solution.
D Calculate the molarity of the solution: moles of solute per liter of solution. Determine the molality of the solution by dividing the number of moles of solute by the kilograms of solvent.
Solution:
The key to this problem is to use the density of pure ethanol to determine the mass of ethanol (CH_{3}CH_{2}OH), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units.
A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density:
\( mass\;of\;EtOH=\left ( 40.0\;\cancel{mL} \right )\left ( \dfrac{0.789\;g}{\cancel{mL}} \right )= 31.6\;g\;EtOH \)
If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution.
B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage:
\( \% \;EtOH=\left ( \dfrac{mass\;EtOH}{mass\;of\;solution} \right )\left ( 100 \right )= \left ( \dfrac{31.6\;\cancel{g}\;EtOH}{31.6\;\cancel{g}\;EtOH+60.0\;\cancel{g}\;H_{2}O} \right )\left ( 100 \right )=34.5 \:\% \)
C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL:
\( moles\;EtOH=\left (31.6\;\cancel{g\;EtOH} \right ) \left ( \dfrac{1\;mol}{46.07\;\cancel{g\;EtOH}} \right )=0.686\;mol\;EtOH \)
Similarly, the number of moles of water is
\( moles\;H_{2}O=\left (60.0\;\cancel{g\;H_{2}O} \right ) \left ( \dfrac{1\;mol}{18.02\;\cancel{g\;H_{2}O}} \right )=3.33\;mol\;H_{2}O \)
The mole fraction of ethanol is thus
\( X_{EtOH}=\left ( \dfrac{0.686 \;\cancel{mol}}{0.686\;\cancel{mol}+3.33\;\cancel{mol}} \right ) =0.171 \)
D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is
\( M_{EtOH}=\left ( \dfrac{0.686 \;mol}{100\;\cancel{mL}} \right )\left ( \dfrac{1000 \;\cancel{mL}}{L} \right ) =6.86\; M \)
The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward:
\( m_{EtOH}=\left ( \dfrac{0.686 \;mol\;EtOH}{60.0\;\cancel{g}\;H_{2}O} \right )\left ( \dfrac{1000 \;\cancel{g}}{kg} \right ) = \dfrac{11.4\;mol\;EtOH}{kg\;H_{2}O }=11.46.86\; m \)
Exercise
A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene.
 mass percentage
 mole fraction
 molarity
 molality
Answer:
 mass percentage toluene = 24.8%
 X_{toluene} = 0.219
 2.35 M toluene
 3.59 m toluene
Summary
The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent.
Key Takeaway
 Different units are used to express the concentrations of a solution depending on the application.
Key Equations
molarity
Equation 9.3.1: \[ molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \]
molality
Equation 9.3.5: \[ molality\left ( m \right )=\dfrac{\left (moles \; solute \right ) }{\left (kilograms \; solvent \right )} \]
mass percentage
Equation 9.3.6: \[ mass \; percentage = \dfrac{\left (mass \; of \; solute\right )}{\left (mass \; of \; solution \right )} \times 100 \]
parts per million by mass
Equation 9.3.7: \[ parts \; per \; million \left (ppm \right ) = \dfrac{\left (mass \; of \; solute \right ) }{\left (mass \; of \; solution \right )} \times 10^{6} \]
relationship among volume, molarity, and moles
Equation 9.3.2: \[ V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \]
Conceptual Problems

Does the molality have the same numerical value as the molarity for a highly concentrated aqueous solution of fructose (C_{6}H_{12}O_{6}) (approximately 3.2 M)? Why or why not?

Explain why the molality and molarity of an aqueous solution are not always numerically identical. Will the difference between the two be greater for a dilute or a concentrated solution? Explain your answer.

Under what conditions are molality and molarity likely to be equal? Is the difference between the two greater when water is the solvent or when the solvent is not water? Why?

What is the key difference between using mole fraction or molality versus molarity to describe the concentration of a solution? Which unit(s) of concentration is most appropriate for experiments that must be carried out at several different temperatures?

An experiment that relies on very strict control of the reaction stoichiometry calls for adding 50.0 mL of a 0.95 M solution of A to 225 mL of a 1.01 M solution of B, followed by heating for 1 h at 60°C. To save time, a student decided to heat solution B to 60°C before measuring out 225 mL of solution B, transferring it to the flask containing solution A, and proceeding normally. This change in procedure caused the yield of product to be less than usual. How could such an apparently minor change in procedure have resulted in a decrease in the yield?
Numerical Problems

Complete the following table for aqueous solutions of the compounds given.
Compound Molarity (M) Solution Density (g/mL) Mole Fraction (X) H_{2}SO_{4} 18.0 1.84 CH_{3}COOH 1.00 7.21 × 10^{−3} KOH 3.60 1.16 
Complete the following table for each compound given.
Compound Mass (g) Volume of Solution (mL) Molarity (M) Na_{2}SO_{4} 7.80 225 KNO_{3} 125 1.27 NaO_{2}CCH_{3} 18.64 0.95 
How would you prepare 100.0 mL of an aqueous solution with 0.40 M KI? a solution with 0.65 M NaCN?

Calculate the molality of a solution with 775 mg of NaCl in 500.0 g of water. Do you expect the molarity to be the same as the molality? Why or why not?

What is the molarity of each solution?
 12.8 g of glucose (C_{6}H_{12}O_{6}) in water, total volume 150.0 mL
 9.2 g of Na_{3}PO_{4} in water, total volume 200.0 mL
 843 mg of I_{2} in EtOH, total volume 150.0 mL

A medication used to treat abnormal heart rhythms is labeled “Procainamide 0.5 g/250 cc.” Express this concentration in parts per thousand.

Meperidine is a medication used for pain relief. A bottle of meperidine is labeled as 50 mg/mL. Express this concentration in parts per thousand.

An aqueous solution that is 4.61% NaOH by mass has a density of 1.06 g/mL. Calculate the molarity of the solution, the mole fraction of NaOH, and the molality of the solution.

A solution of concentrated phosphoric acid contains 85.0% H_{3}PO_{4} by mass and has a density of 1.684 g/mL. Calculate the following.
 the molarity of the solution
 the mole fraction of H_{3}PO_{4}
 the molality of the solution

A solution of commercial concentrated nitric acid is 16 M HNO_{3} and has a density of 1.42 g/mL. What is the percentage of HNO_{3} in the solution by mass? What is the molality?

A commercial aqueous ammonia solution contains 28.0% NH_{3} by massand has a density of 0.899 g/mL. Calculate the following.
 the molarity
 the mole fraction

Concentrated, or glacial, acetic acid is pure acetic acid and has a density of 1.053 g/mL. It is widely used in organic syntheses, in the manufacture of rayon and plastics, as a preservative in foods, and occasionally to treat warts. What volume of glacial acetic acid is required to prepare 5.00 L of a 1.75 M solution of acetic acid in ethanol?

Solutions of sodium carbonate decahydrate, also known as washing soda, are used as skin cleansers. The solubility of this compound in cold water is 21.52 g/100 mL. If a saturated solution has a density of 1.20 g/mL, what is its molarity? What is the mole fraction of sodium carbonate decahydrate in this solution?

Hydrogen peroxide (H_{2}O_{2}) is usually sold over the counter as an aqueous solution that is 3% by mass. Assuming a solution density of 1.01 g/mL, what is the molarity of hydrogen peroxide? What is the molar concentration of a solution that is 30% hydrogen peroxide by mass (density = 1.112 g/mL)? How would you prepare 100.0 mL of a 3% solution from the 30% solution?

Determine the concentration of a solution with 825 mg of Na_{2}HPO_{4} dissolved in 450.0 mL of H_{2}O at 20°C in molarity, molality, mole fraction, and parts per million. Assume that the density of the solution is the same as that of water. Which unit of concentration is most convenient for calculating vapor pressure changes? Why?

How many moles of Cl^{−} are there in 25.0 mL of a 0.15 M CaCl_{2} solution?

How many moles of Na^{+} are there in 25.0 g of a 1.33 × 10^{−3} m Na_{2}HPO_{4} solution? What is the sodium concentration of this solution in ppb?

How many grams of copper are there in 30.0 mL of a 0.100 M CuSO_{4} solution?

How many grams of nitrate ion are there in 75.0 g of a 1.75 × 10^{−4} m Pb(NO_{3})_{2} solution? What is the nitrate concentration of the solution in ppb?

How many milliliters of a 0.750 M solution of K_{2}CrO_{4} are required to deliver 250 mg of chromate ion?

How many milliliters of a 1.95 × 10^{−6} M solution of Ag_{3}PO_{4} are required to deliver 14.0 mg of Ag^{+}?

Iron reacts with bromine according to the following equation:
2Fe(s) + 3Br_{2}(aq) → 2FeBr_{3}(aq)How many milliliters of a 5.0 × 10^{−2} M solution of bromine in water are required to react completely with 750.0 mg of iron metal?

Aluminum reacts with HCl according to the following equation:
2Al(s) + 6HCl(aq) → 2AlCl_{3}(aq) + 3H_{2}(g)If 25.0 mL of a solution of HCl in water is required to react completely with 1.05 g of aluminum metal, what is the molarity of the HCl solution?

The precipitation of silver chloride is a diagnostic test for the presence of chloride ion. If 25.0 mL of 0.175 M AgNO_{3} are required to completely precipitate the chloride ions from 10.0 mL of an NaCl solution, what was the original concentration of NaCl?

Barium sulfate is virtually insoluble. If a 10.0 mL solution of 0.333 M Ba(NO_{3})_{2} is stirred with 40.0 mL of a 0.100 M Na_{2}SO_{4}, how many grams of barium sulfate will precipitate? Which reactant is present in excess? What is its final concentration?
Answers

Compound Molarity (M) Solution Density (g/mL) Mole Fraction (X) H_{2}SO_{4} 18.0 1.84 0.82 CH_{3}COOH 0.393 1.00 7.21 × 10^{−3} KOH 3.60 1.16 6.33 × 10 ^{ −2 } 

100.0 ml of 0.40 M KI: dissolve 6.64 g of KI in enough water to make 100.0 mL of solution; 100.0 ml of 0.65 M NaCN: dissolve 3.18 g of NaCN in enough water to make 100.0 mL of solution.


 0.474 M glucose
 0.28 M Na_{3}PO_{4}
 0.0221 M I_{2}




 14.6 M
 X = 0.510
 57.7 m


 14.8 M
 X = 0.292


The molarity is 0.745 M, and the mole fraction is 0.0134.


The molarity is 0.0129 M, the molality is 0.0129 m, the mole fraction is 2.33 × 10^{−4}, and the solution contains 1830 ppm Na_{2}HPO_{4}. Mole fraction is most useful for calculating vapor pressure, because Raoult’s law states that the vapor pressure of a solution containing a nonvolatile solute is equal to the mole fraction of solvent times the vapor pressure of the pure solvent. The mole fraction of the solvent is just one minus the mole fraction of solute.


6.65 × 10^{−5} mol sodium; 6.14 × 10^{4} ppb


1.63 × 10^{−3} g; 2.17 × 10^{4} ppb


2.22 × 10^{4} mL or 22.2 L


4.68 M HCl


0.777 g BaSO_{4;} Na_{2}SO_{4;} 0.0134 M Na_{2}SO_{4}
Contributors
 Anonymous
Modified by Joshua Halpern, Scott Sinex and Scott Johnson
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