We always discuss a solution being diluted or concentrated; this is a qualitative way of expressing the concentration of the solution. A dilute solution means the quantity of solute is relatively very small, and a concentrated solution implies that the solution has a large amount of solute. But these are relative terms and do not give us the quantitative concentration of the solution. So, to quantitatively describe the concentrations of various solutions around us, we commonly express levels in the following way: It is the amount of solute present in one litre of solution. It is denoted by C or S. \(\begin{array}{l}C or S = \frac{Weight\ of\ solute\ in\ grams}{Volume\ in\ liters}\end{array} \)
 1. Concentration in Parts Per Million (ppm)The parts of a component per million parts (106) of the solution. \(\begin{array}{l}ppm(A) = \frac{Mass\ of\ A}{Total\ mass\ of\ the\ solution}\times {10}^6\end{array} \) 2. Mass Percentage (w/w):When the concentration is expressed as the percent of one component in the solution by mass it is called mass percentage (w/w). Suppose we have a solution containing component A as the solute and B as the solvent, then its mass percentage is expressed as: Mass % of A = 3. Volume Percentage (V/V):Sometimes we express the concentration as a percent of one component in the solution by volume, it is then called as volume percentage and is given as: volume % of A = For example, if a solution of NaCl in water is said to be 10 % by volume that means a 100 ml solution will contain 10 ml NaCl. 4. Mass by Volume Percentage (w/V):This unit is majorly used in the pharmaceutical industry. It is defined as the mass of a solute dissolved per 100mL of the solution. % w/V = (Mass of component A in the solution/ Total Volume of the Solution)x 100 5. Molarity (M):One of the most commonly used methods for expressing the concentrations is molarity. It is the number of moles of solute dissolved in one litre of a solution. Suppose a solution of ethanol is marked 0.25 M, this means that in one litre of the given solution 0.25 moles of ethanol is dissolved. Molarity (M) = Moles of Solute/Volume of Solution in litres 6. Molality (m):Molality represents the concentration regarding moles of solute and the mass of solvent. It is given by moles of solute dissolved per kg of the solvent. The molality formula is as given- 7. NormalityIt is the number of gram equivalents of solute present in one litre of the solution and it is denoted by N. \(\begin{array}{l}N = \frac{Weight\ of\ solute\ in\ grams}{Equivalent\ mass \times Volume\ in\ liter}\end{array} \) The relation between normality and molarity.
8. FormalityIt is the number of gram formula units present in one litre of solution. It is denoted by F. \(\begin{array}{l}F = \frac{Weight\ of\ solute\ in\ gram}{Formula\ wt \times Volume\ in\ liter}\end{array} \) It is applicable in the case of ionic solids like NaCl. 9. Mole Fraction:If the solution has a solvent and the solute, a mole fraction gives a concentration as the ratio of moles of one component to the total moles present in the solution. It is denoted by x. Suppose we have a solution containing A as a solute and B as the solvent. Let nA and nB be the number of moles of A and B present in the solution respectively. So, mole fractions of A and B are given as: The above-mentioned methods are commonly used ways of expressing the concentration of solutions. All the methods describe the same thing that is, the concentration of a solution, each of them has its own advantages and disadvantages. Molarity depends on temperature while mole fraction and molality are independent of temperature. All these methods are used on the basis of the requirement of expressing the concentrations. Solutions of Solids in Liquids
Solubility of GasesThe solubility of gases is mostly expressed in terms of the absorption coefficient,k that is the volume of the gas dissolved by unit volume of solvent at 1 atm pressure and a specific temperature. The solubility of a gas in a liquid depends upon
Sometimes, by modifying the quantity of solvent, a worker would need to modify the concentration of a solution. Dilution is the addition of a solvent that reduces the solute concentration of the solution. Concentration is solvent elimination, which increases the solute concentration in the solution.
A concentration of persons means that in one place there are more of them. A high concentration of a material in a solution means there’s a lot of it compared to the volume: because of the high concentration of salt, the Great Salt Lake has very little fish.
In general, a mild condensed acid is more harmful than a solid diluted acid. Although concentrated acetic acid is much less reactive, you do not want to touch the skin or mucous membranes because it is corrosive.
A solution concentration is a measure of the quantity of solute that has been dissolved in a given quantity of solvent or solution. One that contains a relatively high volume of dissolved solute is a concentrated solution. That that contains a relatively minimal volume of dissolved solute is a dilute solution.
Solutions of known concentration can be prepared either by dissolving the known mass of the solvent solution and diluting it to the desired final volume or by diluting it to the desired final volume by diluting the acceptable volume of the more concentrated solution (the stock solution).
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out of 0 arewrong out of 0 are correct out of 0 are Unattempted View Quiz Answers and AnalysisIn 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 (see Figure 1). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified. 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. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions. 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. 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) 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: [latex]M = \frac{\text{mol solute}}{\text{L solution}}[/latex]
Calculating Molar Concentrations Solution [latex]M = \frac{\text{mol solute}}{\text{L solution}} = \frac{0.133 \;\text{mol}}{355 \;\text{mL} \times \frac{1 \;\text{L}}{1000 \;\text{mL}}} = 0.375 \; M[/latex] Check Your Learning
Deriving Moles and Volumes from Molar Concentrations Solution
[latex]M = \frac{\text{mol solute}}{\text{L solution}}[/latex] [latex]\text{mol solute} = 0.375 \;\frac{\text{mol sugar}}{\text{L}} \times (10 \;\text{mL} \times \frac{1 \text{L}}{1000 \;\text{mL}}) = 0.004 \;\text{mol sugar}[/latex] Check Your Learning
Calculating Molar Concentrations from the Mass of Solute Solution [latex]M = \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}} = 0.839 \;M[/latex] [latex]\begin{array}{r @{{}={}} l} M & \frac{\text{mol solute}}{\text{L solution}} = 0.839\;M \\[1em] M & \frac{0.839 \;\text{mol solute}}{1.00 \;\text{L solution}} \end{array}[/latex] Check Your Learning
Determining the Mass of Solute in a Given Volume of Solution Solution [latex]M = \;\frac{\text{mol solute}}{\text{L solution}}[/latex] [latex]\text{mol solute} = M \times \text{L solution}[/latex] [latex]\text{mol solute} = 5.30 \;\frac{\text{mol NaCl}}{\text{L}} \times 0.250 \;\text{L} = 1.325 \;\text{mol NaCl}[/latex] Finally, this molar amount is used to derive the mass of NaCl: [latex]1.325 \;\text{mol NaCl} \times \frac{58.44 \;\text{g NaCl}}{\text{mol NaCl}} = 77.4 \;\text{g NaCl}[/latex] Check Your Learning 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.
Determining the Volume of Solution Containing a Given Mass of Solute Solution [latex]\text{g solute} \times \frac{\text{mol solute}}{\text{g solute}} = \text{mol solute}[/latex] Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute: [latex]\text{mol solute} \times \frac{\text{L solution}}{\text{mol solute}} = \text{L solution}[/latex] Combining these two steps into one yields:
[latex]\text{g solute} \times \frac{\text{mol solute}}{\text{g solute}} \times \frac{\text{L solution}}{\text{mol solute}} = \text{L solution}[/latex][latex]75.6 \;\text{g CH}_3\text{CO}_2\text{H} (\frac{\text{mol CH}_3\text{CO}_2\text{H}}{60.05 \;\text{g}}) (\frac{\text{L solution}}{0.839 \;\text{mol CH}_3\text{CO}_2\text{H}}) = 1.50 \;\text{L solution}[/latex] Check Your Learning 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 increasingly 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). 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). 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 molar amount of solute in a solution is equal to the product of the solution’s molarity and its volume in liters: [latex]n = ML[/latex] Expressions like these may be written for a solution before and after it is diluted: [latex]n_1 = M_1L_1[/latex] [latex]n_2 = M_2L_2[/latex] 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,n1 = n2. Thus, these two equations may be set equal to one another: [latex]M_1L_1 = M_2L_2[/latex] 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 factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form: [latex]C_1V_1 = C_2V_2[/latex] where C and V are concentration and volume, respectively.
Use the simulation to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation.
Determining the Concentration of a Diluted Solution Solution [latex]C_1V_1 = C_2V_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: [latex]C_2 = \frac{0.850 \;\text{L} \times 5.00 \frac{\text{mol}}{\text{L}}}{1.80 \;\text{L}} = 2.36 \;M[/latex] This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M). Check Your Learning
Volume of a Diluted Solution Solution [latex]C_1V_1 = C_2V_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: [latex]V_2 = \frac{(0.45\;M)(0.011 \;\text{L})}{0.12 \; M}[/latex] 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
Volume of a Concentrated Solution Needed for Dilution Solution [latex]C_1V_1 = C_2V_2[/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: [latex]V_1 = \frac{(0.100\;M)(5.00 \;\text{L})}{1.59 \; M}[/latex] 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 |
When making up a solution of known concentration should you first dissolve the solute in a minimum volume of solvent?
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