As stated previously, chemists have defined several types of concentrations, which each use a different chemically-acceptable unit, or combination of units, to indicate the amount of solute that is dissolved in a given amount of solvent. The following paragraphs will present and apply the equation that is used to calculate a mass/volume percent, which is the final type of percent-based concentration that will be discussed in this chapter. The mass/volume percent of a solution is defined as the ratio of the mass of solute that is present in a solution, relative to the volume of the solution, as a whole. Because this type of concentration is expressed as a percentage, the indicated proportion must be multiplied by 100, as shown below. \(\text{Mass/Volume Percent}\) = \( \dfrac{ \rm{m_{solute} \; (\rm{g})}}{\rm{V_{solution} \; (\rm{mL})}} \) × \({100}\) As discussed in the previous two sections of this chapter, mass percents and volume percents can be calculated using an alternative equation, in which the masses or volumes, respectively, of the solute and the solvent that are contained in a solution are added to obtain the mass or volume, respectively, of that solution, as a whole. While mass percents are typically reported for solid- and liquid-phase solutions, and volume percents are usually determined for liquid- and gas-phase solutions, a mass/volume percent concentration is most often calculated for solutions that are specifically prepared by dissolving solid solutes in liquid solvents. In order to create this type of solution, the solid solute particles must overcome the attractive forces that exist between the liquid solvent molecules, in order to move throughout and occupy the "empty" spaces that are temporarily created during the solvation process. After the solute particles have dispersed throughout the solvent, the solvent molecules interact more strongly with the solvated solute particles than with other solvent molecules and, consequently, exist in closer physical proximity to those solute particles, relative to other solvent molecules. As a result of these solute-solvent interactions, the solvated solute particles occupy less space than they had prior to their solvation, which causes the volume of the solution, as a whole, to decrease, relative to the combined volumes of the individual solute and solvent. Because the magnitude of this volumetric contraction varies based on the solute and solvent that are utilized to prepare a solution, calculating the mass/volume percent of a solution by adding the volumes of its components is prohibitively challenging. Therefore, only the equation that is shown above can be applied to reliably determine the mass/volume percent of a solution.
In order to be incorporated into the equation that is shown above, the mass of the solute must be expressed in grams, the volume of the solution must be provided in milliliters, and the chemical formula of each component must be written as the secondary unit on its associated numerical quantity. Therefore, if either of these measurements is reported using an alternative unit, its value would need to be converted to the appropriate unit prior to being incorporated into the mass/volume percent equation. During the multiplication and division processes that are used to solve this equation, no unit cancelation occurs, because the units that are present in the numerator and denominator, "g" and "mL," respectively, do not match one another. Therefore, the unit that results from the division of the indicated quantities is "g/mL," which is a unit that is typically utilized to report the density of a substance. Because densities and mass/volume percent concentrations have unique definitions and are calculated using different equations, these measurements are distinctive quantities and, consequently, cannot be expressed using the same unit. Therefore, the mass and volume units are eliminated during the simplification of the mass/volume percent equation, even though "g" and "mL" do not cancel, mathematically, and the calculated concentration is expressed as a percentage. However, as stated previously, the quantity of solute that is present in a given solution can be expressed using three unique percent-based concentrations. In order to distinguish a mass/volume percent, which is calculated by simplifying a mass-to-volume ratio, from the other percent-based concentrations, the unit in which a mass/volume percent concentration is reported is "% m/v," and the chemical formula of the solute is written as the secondary unit on this calculated quantity. Finally, because mass/volume percents are not defined as exact quantities, their values should be reported using the correct number of significant figures. However, "100" is an exact number and, therefore, does not impact the significance of the final reported concentration. Exercise \(\PageIndex{1}\)Calculate the mass/volume percent of a 762.5 milliliter solution that is prepared by dissolving 289.15 grams of calcium azide, Ca(N3)2, in water. Answer In order to calculate the mass/volume percent of a solution, each substance that is referenced in the problem must first be classified as a solute or a solvent. Because the indicator word "in" is present in the given statement, the chemical that is mentioned after this word, water, H2O, is the solvent in this solution, and the remaining substance, calcium azide, Ca(N3)2, is the solute, "by default."Before this equation can be applied, the validity of the units that are associated with the given numerical values must be confirmed. As stated above, the mass of the solute must be expressed in grams, and the volume of the solution must be provided in milliliters. Therefore, the given quantities are both expressed in the appropriate unit and can be directly incorporated into the mass/volume percent equation, as shown below. The mass and volume units are eliminated during the simplification of this equation, even though "g" and "mL" do not cancel, mathematically, in order to avoid obtaining a density unit as a result of dividing the given quantities. In order to distinguish a mass/volume percent, which is calculated by simplifying a mass-to-volume ratio, from the other percent-based concentrations, the unit in which the resultant concentration is reported is "% m/v Ca(N3)2." The chemical formula of the solute is written as the secondary unit on the calculated quantity, and applying the correct number of significant figures to this value results in the final answer that is shown below. \(\text{Mass/Volume Percent}\) = \( \dfrac{289.15 \; \rm{g} \; \rm{Ca(N_3)_2}}{762.5 \; \rm{mL} \; \rm{solution}}\) × \({100}\) \(\text{Mass/Volume Percent}\) = \({37.92131... \%\ \rm{m/v} \; \rm{Ca(N_3)_2}} ≈ {37.92 \%\ \rm{m/v} \; \rm{Ca(N_3)_2}}\) Key Concepts⚛ Weight/Volume Percentage Concentration is a measurement of the concentration of a soluton. · Weight/Volume percentage concentration is also known as mass/volume percentage concentration. ⚛ weight/volume percentage concentration is usually abbreviated as w/v (%) or w/v% or (w/v)% or %(w/v) or %w/v · mass/volume percentage concentration is usually abbreviated as m/v (%) or m/v% or (m/v)% or %(m/v) or %m/v ⚛ w/v% (m/v%) is a useful concentration measure when dispensing reagents. ⚛ To calculate w/v% concentration (m/v% concentration):
⚛ Common units(1) for w/v% concentration are g/100 mL (grams of solute per 100 mL of solution) · example: 5%(w/v) = 5 g/100 mL · example: 12%(m/v) = 12 g/100 mL ⚛ Rearrange the equation for w/v% (m/v%) concentration to find: (i) mass of solute mass(solute) = [volume(solution) × (w/v)%]/100 (ii) volume of solution volume(solution) = [mass(solute)/(w/v)%] × 100 Please do not block ads on this website. Weight/Volume Percentage Concentration Calculations Weight/Volume percentage concentration (w/v%), or mass/volume percentage concentration (m/v%), is a measure of the concentration of a solution.
A percentage concentration tells us how many parts of solute are present per 100 parts of solution. In weight per volume terms (mass per volume terms), this means a percentage concentration tells us the parts of solute by mass per 100 parts by volume of solution. In SI units, w/v% (m/v%) concentration would be given in kg/100 L, but these units are far too large to be useful to Chemists in the lab, grams and milliltres are more convenient units for us. Recall that 1 kg = 1 000 g Recall that 1 L = 1 000 mL so kg/100 L = 1000 g/100 000 mL = g/100 mL so the units for w/v% concentration are most often given as g/100 mL Therefore the units for w/v% (m/v%) concentration are grams of solute per 100 mL of solution (g/100 mL). This means that the weight/volume percentage concentration (mass/volume percentage concentration) can be given in different, but equivalent, ways. Some examples are given in the table below:
To prepare a solution with a particular concentration, you will weigh out the solid and then dissolve it enough solvent to make a known volume of solution. Therefore you will know the value of two quantities:
which you can use to calculate the weight/volume percentage concentration (mass/volume percentage concentration). To calculate a weight/volume percentage concentration (mass/volume percentage concentration):
Worked Examples: w/v% (m/v%) CalculationsQuestion 1. What is the weight/volume percentage concentration of 250 mL of aqueous sodium chloride solution containing 5 g NaCl? Solution: Step 1: Write the equation: either w/v% = w/v × 100 or m/v% = m/v × 100 weight/volume (%) = (mass solute ÷ volume of solution) × 100 Step 2: Identify the solute and solvent (by name or chemical formula) solute = sodium chloride = NaCl solvent is water, H2O, because this is an aqueous solution. Step 3: Extract the data from the question (mass of solute, volume of solution) mass solute (NaCl) = 5 g volume of solution = 250 mL Step 4: Check the units for consistency and convert if necessary (mass in grams, volume in millilitres) mass solute (NaCl) = 5 g (no unit conversion needed) volume of solution = 250 mL (no unit conversion needed) Step 5: Substitute these values into the equation and solve. w/v (%) = (5 g ÷ 250 mL) × 100 = 2 g/100 mL (Note: only 1 significant figure is justified) Step 6: Write the answer w/v% = 2 g/100 mL = 2%(w/v) = 2%(m/v) Question 2. 10.00 g BaCl2 is dissolved in 90.00 g of water. The density of the solution is 1.090 g/mL (1.090 g mL-1). Calculate the mass/volume percentage concentration of the solution. Solution: Step 1: Write the equation: either m/v% = m/v × 100 or w/v% = w/v × 100 mass/volume (%) = (mass solute ÷ volume of solution) × 100 Step 2: Identify the solute and solvent (by name or chemical formula) solute = barium chloride = BaCl2 solvent = water = H2O(l) (This is an aqueous solution) Step 3: Extract the data from the question (mass of solute, volume of solution) mass of solute = mass(BaCl2) = 10.00 g mass of solvent = mass(H2O) = 90.00 g density of solution = 1.090 g/mL Step 4: Check the units for consistency and convert if necessary (mass in grams, volume in millilitres) mass of solute = mass(BaCl2) = 10.00 g (no unit conversion needed) volume of solution is unknown and needs to be calculated using the density of solution and the masses of solute and solvent: density = mass(solution) ÷ volume(solution) mass(solution) = mass(solute) + mass(solvent) mass(solution) = 10.00 g BaCl2 + 90.00 g water = 100.0 g volume(solution) = mass(solution) ÷ density(solution) volume(solution) = 100.0 g ÷ 1.090 g/mL = 91.74 mL Step 5: Substitute these values into the equation and solve. m/v (%) = (mass solute ÷ volume solution) × 100 m/v (%) = (10.00 g ÷ 91.74 mL) × 100 m/v (%) = 10.90 g/100 mL (Note: only 4 significant figures are justified) Step 6: Write the answer m/v% = 10.90 g/100 mL = 10.90 %(m/v) = 10.90 %(w/v) ↪ Back to top Conversion from Other Units to w/v % or m/v %The most common units for w/v% (m/v%) concentration are g/100 mL (grams of solute per 100 mL of solution). If the mass of the solute is not given in grams then you will need to convert the units to grams. If the volume of the solution is not given in millilitres then you will need to convert the units to millilitres. Worked Examples: w/v% (m/v%) Calculations Requiring Unit ConversionsQuestion 1. 2.00 L of an aqueous solution of potassium chloride contains 45.0 g of KCl. What is the weight/volume percentage concentration of this solution in g/100 mL? Solution: Step 1: Write the equation: either w/v% = w/v × 100 or m/v% = m/v × 100 Step 2: Identify the solute and solvent (by name or chemical formula) solute = potassium chloride = KCl solvent is water, H2O, because this is an aqueous solution. Step 3: Extract the data from the question (mass of solute, volume of solution) mass KCl = 45.0 g volume of solution = 2.00 L Step 4: Check the units for consistency and convert if necessary (mass in grams, volume in millilitres) mass KCl = 45.0 g (mass in g, no unit conversion needed) volume of solution = 2.00 L (need to convert to mL) Step 5: Substitute these values into the equation and solve. w/v (%) = [mass solute (g) ÷ volume solution (mL)] × 100 w/v (%) = [45.0 g ÷ 2000 mL] × 100 = 2.25 g/100 mL (Note: only 3 significant figures are justified) Step 6: Write the answer w/v% = 2.25 g/100 mL = 2.25 %(w/v) = 2.25 %(m/v) Question 2. 15 mL of an aqueous solution of sucrose contains 750 mg sucrose. What is the mass/volume percentage concentration of this solution in g/100 mL? Solution: Step 1: Write the equation: either m/v% = m/v × 100 or w/v% = w/v × 100 Step 2: Identify the solute and solvent (by name or chemical formula) solute = sucrose solvent is water, H2O, because this is an aqueous solution. Step 3: Extract the data from the question (mass of solute, volume of solution) mass solute (sucrose) = 750 mg volume solution = 15 mL Step 4: Check the units for consistency and convert if necessary (mass in grams, volume in millilitres) mass solute (sucrose) = 750 mg (need to convert to grams) volume solution = 15 mL (no conversion needed) Step 5: Substitute these values into the equation and solve. w/v (%) = [mass solute (g) ÷ volume solution (mL)] × 100 w/v (%) = (0.750 g ÷ 15 mL) × 100 = 5.0 g/100 mL (Note: only 2 significant figures are justified) Step 6: Write the answer m/v% = 5.0 g/100 mL = 5.0 %(m/v) = 5.0 %(w/v) Question 3. 186.4 L of aqueous sodium hydroxide solution contains 1.15 kg NaOH. What is the weight/volume percentage concentration of this solution in g/100 mL? Solution: Step 1: Write the equation: either w/v% = w/v × 100 or m/v% = m/v × 100 Step 2: Identify the solute and solvent (by name or chemical formula) solute = sodium hydroxide = NaOH solvent is water, H2O, because this is an aqueous solution. Step 3: Extract the data from the question (mass of solute, volume of solution) mass solute (NaOH) = 1.15 kg volume solution = 186.4 L Step 4: Check the units for consistency and convert if necessary (mass in grams, volume in millilitres)(2) mass solute (NaOH) = 1.15 kg (convert units to g) mass solute (NaOH) = 1.15 kg = 1.15 kg × 1000 g/kg = 1 150 g volume solution = 186.4 L (convert units to mL) volume solution = 186.4 L = 186.4 L × 1000 mL/L = 186 400 mL Step 5: Substitute these values into the equation and solve. w/v (%) = [mass solute (g) ÷ volume solution (mL)] × 100 w/v (%) = (1 150 g ÷ 186 400 mL) × 100 w/v (%) = 0.617 g/100 mL (Note: only 3 significant figures are justified) Step 6: Write the answer w/v% = 0.617 g/100 mL = 0.617 %(w/v) = 0.617 %(m/v) ↪ Back to top Reagent Volume and Mass CalculationsIn the sections above we calculated the w/v% concentration of solutions using the known mass of solute and volume of solution. When we come to use this solution in the lab, we are most likely to use a pipette or burette to deliver a volume of solution. If we know the volume of solution used, we can calculate the mass of solute present.
If we know the mass of solute we want to use, we can calculate the volume of solution we will need to use.
Question 1. A student must add 1.22 g of sodium chloride to a reaction vessel. The student is provided with an 11.78 g/100 mL aqueous sodium chloride solution (11.78 %m/v). What volume of this solution must be added to the reaction vessel? Solution: Step 1: Write the equation: either w/v% = w/v × 100 or m/v% = m/v × 100 Rearrange the equation to find volume of solution: w/v% = mass solute (g) ÷ volume solution (mL) × 100 Multiply both sides of the equation by volume w/v% × volume (mL) = (mass solute (g) ÷ volume solution (mL)) × volume (mL) × 100 w/v% × volume (mL) = mass solute (g) × 100 Divide both sides of the equation by w/v% [w/v% × volume (mL)]/w/v% = (mass solute (g) × 100)/w/v%
Step 2: Identify the solute and solvent (by name or chemical formula) solute = sodium chloride = NaCl solvent is water, H2O, because this is an aqueous solution. Step 3: Extract the data from the question (mass of solute, volume of solution) mass of solute required = mass(NaCl) = 1.22 g concentration of NaCl(aq) provided = w/v (%) = 11.78 g/100 mL volume of solution needed = ? mL Step 4: Check the units for consistency and convert if necessary (mass in grams, volume in millilitres) mass of solute required = mass(NaCl) = 1.22 g (no unit conversion needed) concentration of NaCl(aq) provided = w/v (%) = 11.78 g/100 mL (no unit conversion needed) volume of solution needed = ? mL Step 5: Substitute these values into the equation and solve.
(Note: only 3 significant figures are justified) Step 6: Write the answer volume of solution = 10.4 mL Question 2. What is the mass in grams of potassium iodide in 14.86 mL of a 32.44 g/100 mL aqueous potassium iodide solution (32.44 %w/v)? Solution: Step 1: Write the equation: either w/v% = w/v × 100 or m/v% = m/v × 100 Rearrange the equation to find mass of solute: w/v(%) = mass solute (g) ÷ volume solution (mL) × 100 Divide both sides of the equation by 100 w/v(%) ÷ 100 = mass solute (g) ÷ volume solution (mL) Multiple both sides of the equation by volume solution (mL) mass solute (g) = (w/v (%) ÷ 100) × volume solution (mL) Step 2: Identify the solute and solvent (by name or chemical formula) solute = potassium iodide = KCl solvent is water, H2O, because this is an aqueous solution. Step 3: Extract the data from the question (mass of solute, volume of solution) mass solute = mass(KI) = ? g volume(KI(aq)) = 14.86 mL concentration of solution = w/v%(KI(aq)) = 32.44 g/100 mL Step 4: Check the units for consistency and convert if necessary (mass in grams, volume in millilitres) volume(KI(aq)) = 14.86 mL (no unit conversion needed) w/v%(KI(aq)) = 32.44 g/100 mL (no unit conversion needed) Step 5: Substitute these values into the equation and solve. mass solute (g) = (w/v% ÷ 100) × volume solution (mL) mass(KI) = (32.44 ÷ 100) × 14.86 = 4.821 g (Note: 4 significant figures are justified) Step 6: Write the answer mass of solute (KI) = 4.821 g ↪ Back to top Sample Question: w/v% (m/v%) calculationsDetermine the mass in grams of potassium nitrate (KNO3) in 22.65 mL of a 2.15%(m/v) aqueous solution of potassium nitrate. ↪ Back to top
Footnotes: (1) Common units for w/v% (m/v%) are g/100 mL but other units are also possible, for example, kg/100 L, mg/100 μL We will restrict the following discussion to g/100 mL but you can apply the same logic and equations using other appropriate units. Note that solubilities are most often given as weight ratio percentage concentration. (2) You don't really need to convert these units because both the mass and volume are given in SI units, that is w/v% = (1.15 kg/186.4 L) × 100 = 0.617 kg/100 L = 617 g/100 000 mL = 0.617 g/100 mL but it's probably best to practice doing the unit conversions, at least until you have a full appreciation of what a w/v% concentration means. ↪ Back to top |
What is the concentration in mv of a solution prepared
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