Volume percent or volume/volume percent (v/v%) is used when preparing solutions of liquids. It is very easy to prepare a chemical solution using volume percent, but if you misunderstand the definition of this unit of concentration, you'll experience problems. Volume percent is defined as:
Note that volume percent is relative to the volume of solution, not the volume of solvent. For example, wine is about 12% v/v ethanol. This means there is 12 ml ethanol for every 100 ml of wine. It is important to realize liquid and gas volumes are not necessarily additive. If you mix 12 ml of ethanol and 100 ml of wine, you will get less than 112 ml of solution. As another example, 70% v/v rubbing alcohol may be prepared by taking 700 ml of isopropyl alcohol and adding sufficient water to obtain 1000 ml of solution (which will not be 300 ml). Solutions made to a specific volume percent concentration typically are prepared using a volumetric flask. Volume percent (vol/vol% or v/v%) should be used whenever a solution is prepared by mixing pure liquid solutions. In particular, it's useful where miscibility comes into play, as with volume and alcohol. Acid and base aqueous reagents are usually described using weight percent (w/w%). An example is concentrated hydrochloric acid, which is 37% HCl w/w. Dilute solutions are often described using weight/volume % (w/v%). An example is 1% sodium dodecyl sulfate. Although it's a good idea to always cite the units used in percentages, it seems common for people to omit them for w/v%. Also, note "weight" is really mass. Hi, and welcome to this video on volume percent—the concentration of a solute in a solution calculated as a percentage of volume! Volume percent, represented as \(\frac{v}{v \text{%}}\), is commonly used to represent the concentration of binary mixtures of liquids, as liquids are often measured in volume. For example, when you buy isopropanol at the store (rubbing alcohol), you’ll notice that it says 70% (\(\frac{v}{v}\)), which means by volume. This tells you that the solution is 70% isopropanol. The equation for volume percent is pretty straightforward:
So, returning to our rubbing alcohol, 70% (\(\frac{v}{v}\)) means that for every 100 mL of solution, there are 70 mL of the solute, isopropanol.
Let’s consider a bottle of wine, listed as 12% ABV (alcohol by volume, which is another way of saying volume percent).
The bottle is 750 mL, which means the wine contains 90 mL of ethanol. We’ve demonstrated in these two examples that given the total volume of solution and the volume percent, it’s pretty simple to calculate the volume of solute. However, if you plan on making a solution with a certain volume percent, you should be aware of a slight caveat when you’re working in the laboratory. Let’s consider another example. Let’s say you want to make a 70% \(\frac{v}{v%}\) solution of ethanol in water. You prepare beakers of 70 mL ethanol and 30 mL water. By mixing them together, you believe this will give you the 70% \(\frac{v}{v%}\) mixture you desire. But when you pour the 70 mL of ethanol and 30 mL of water into your graduated cylinder, you actually find that you’ve come up short! Your graduated cylinder only has 96 mL of solution. How is that possible? As you mixed the two liquids, the ethanol and water molecules began to interact with each other, packing together in a different way than in either pure ethanol or pure water. This results in a new density of the solution and thus, a volume that does not equal the sum of the two original volumes. In other words, the volume of solution, the denominator of our volume percent equation, does not equal the volume of solute plus the volume of solvent. In the case of water and ethanol, their mixture has less volume than the sum of the pure liquids. So how do you actually prepare a 70% \(\frac{v}{v%}\) solution of ethanol in water? Instead of adding the premeasured liquids together, you would fill a graduated cylinder with 70 mL ethanol and then top up to 100 mL with water. This way, you’ve reached a total volume of 100 mL, 70 of which are ethanol, but you added a bit more water than you initially expected. The difference between the volume of solute plus the volume of solvent and the volume of the solution is typically pretty small, so you’ll often be instructed to assume that they are the same. For example, you might get a question like, what is the volume percent of ethanol in a mixture of 40 mL ethanol and 80 mL water (assume that the volume of solution is 120 mL).
Now, you should be able to understand what you’re being asked and can easily make the calculations. Thanks for watching, and happy studying!
The percentage concentration of any solution is most commonly expressed as mass percent: Mass % of any component of the solution = Other methods are: Volume % of a component = i.e. Mass by Volume percentage = Here's a point to be kept in mind : The concentration of a solution is most of the time expressed as the number of moles of solute present in 1 Liter of the solution (also called molarity ) (There are also other ways to express concentration. Please follow this link. ) EXAMPLE: (b) What is the molarity of a solution prepared by dissolving 15.0 g of sodium hydroxide in enough water to make a total of 225 mL of solution? Solution Moles of NaOH = 15.0 g NaOH × #(1"mol NaOH")/(40.00"g NaOH")# = 0.375 mol NaOH Volume = 225 mL × #(1"L")/(1000"mL")# = 0.225 L soln Molarity = #(0.375"mol")/(0.225"L")# = 1.67 mol/L
Let's address the question for both percent concentration by mass and for percent concentration by volume. Percent concentration by mass is defined as the mass of solute divided by the total mass of the solution and multiplied by 100%. So, #c% = m_(solute)/(m_(solution)) * 100%#, where #m_(solution) = m_(solvent) + m_(solute)# There are two ways to change a solution's concentration by mass
Let's take an example to better illustrate this concept. Say we dissolve 10.0g of a substance in 100.0g of water. Our concentration by mass will be #c% = (10.0g)/(10.0g + 100.0g) * 100% = 9.09%# Now let's try doubling the mass of the solute; the new concentration will be #c% = (2 * 10.0g)/(2*10.0g + 100.0g) * 100% = 16.7%# However, if we keep the mass of the solute at 10.0g and doubled the mass of the solvent (in this case, water), the concentration will be #c% = (10.0g)/(10.0g + 2*100.0g) * 100% = 4.76%# The same is true for percent concentration by volume, which is defined as the volume of the solute divided by the total volume of the solution and multiplied by 100%. #c_(volume)% = V_(solute)/(V_(solute) + V_(solvent)) * 100%# It's easy to see that manipulating either the volume of the solute or the volume of the solvent (or both) would change the solution's percent concentration by volume.
There are two types of percent concentration: percent by mass and percent by volume. PERCENT BY MASS Percent by mass (m/m) is the mass of solute divided by the total mass of the solution, multiplied by 100 %. Percent by mass = #"mass of solute"/"total mass of solution"# × 100 % Example What is the percent by mass of a solution that contains 26.5 g of glucose in 500 g of solution? Solution Percent by mass = #"mass of glucose"/"total mass of solution" × 100 % = (26.5"g")/(500"g")# × 100 % = 5.30 % PERCENT BY VOLUME Percent by volume (v/v) is the volume of solute divided by the total volume of the solution, multiplied by 100 %. Percent by volume = #"volume of solute"/"total volume of solution"# × 100 % Example How would you prepare 250 mL of 70 % (v/v) of rubbing alcohol Solution 70 % = #"volume of rubbing alcohol"/"total volume of solution" × 100 %# × 100 % So Volume of rubbing alcohol = volume of solution × #"70 %"/"100 %"# = 250 mL × #70/100# = 175 mL You would add enough water to 175 mL of rubbing alcohol to make a total of 250 mL of solution.
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