To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms with meanings that depend on various factors. Concentration is the measure of how much of a given substance is mixed with another substance. Solutions are said to be either dilute or concentrated. When we say that vinegar is \(5\%\) acetic acid in water, we are giving the concentration. If we said the mixture was \(10\%\) acetic acid, this would be more concentrated than the vinegar solution. A concentrated solution is one in which there is a large amount of solute in a given amount of solvent. A dilute solution is one in which there is a small amount of solute in a given amount of solvent. A dilute solution is a concentrated solution that has been, in essence, watered down. Think of the frozen juice containers you buy in the grocery store. To make juice, you have to mix the frozen juice concentrate from inside these containers with three or four times the container size full of water. Therefore, you are diluting the concentrated juice. In terms of solute and solvent, the concentrated solution has a lot of solute versus the dilute solution that would have a smaller amount of solute. The terms "concentrated" and "dilute" provide qualitative methods of describing concentration. Although qualitative observations are necessary and have their place in every part of science, including chemistry, we have seen throughout our study of science that there is a definite need for quantitative measurements in science. This is particularly true in solution chemistry. In this section, we will explore some quantitative methods of expressing solution concentration.
There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100: \[\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%} \nonumber \] mass of solution = mass of solute + mass solvent If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration. Suppose that a solution was prepared by dissolving \(25.0 \: \text{g}\) of sugar into \(100.0 \: \text{g}\) of water.
mass of solution = 25.0g sugar + 100.0g water = 125.0 g
\[\text{Percent by mass} = \dfrac{25.0 \: \text{g sugar}}{125.0 \: \text{g solution}} \times 100\% = 20.0\% \: \text{sugar} \nonumber \]
A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution?
We can substitute the quantities given in the equation for mass/mass percent: \(\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}\)
A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution? Answer7.90 %
Sometimes you may want to make up a particular mass of solution of a given percent by mass and need to calculate what mass of the solute to use. Using mass percent as a conversion can be useful in this type of problem. The mass percent can be expressed as a conversion factor in the form \(\dfrac{g \; \rm{solute}}{100 \; \rm{g solution}}\) or \(\dfrac{100 \; \rm g solution}{g\; \rm{solute}}\) For example, if you need to make \(3000.0 \: \text{g}\) of a \(5.00\%\) solution of sodium chloride, the mass of solute needs to be determined.
Given: 3000.0 g NaCl solution 5.00% NaCl solution Find: mass of solute = ? g NaCl Other known quantities: 5.00 g NaCl is to 100 g solution
To solve for the mass of NaCl, the given mass of solution is multiplied by the conversion factor. \[g NaCl = 3,000.0 \cancel{g \: NaCl \:solution} \times \dfrac{5.00 \:g \: NaCl}{100\cancel{g \: NaCl \: solution}} = 150.0g \: NaCl \nonumber \] You would need to weigh out \(150 \: \text{g}\) of \(\ce{NaCl}\) and add it to \(2850 \: \text{g}\) of water. Notice that it was necessary to subtract the mass of the \(\ce{NaCl}\) \(\left( 150 \: \text{g} \right)\) from the mass of solution \(\left( 3000 \: \text{g} \right)\) to calculate the mass of the water that would need to be added.
What is the amount (in g) of hydrogen peroxide (H2O2) needed to make a 6.00 kg, 3.00 % (by mass) H2O2 solution? Contributors and Attributions LICENSED UNDER 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 mass of water must be added to 10g of sugar to make 5 percent mass by mass percentage of the solution?
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