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:
\(\frac{v}{v\%} = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\)
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.
\(70\% = \frac{\text{volume of solute}}{100\text{ mL}} \times 100\)
Volume of solute = 70 mL
Let’s consider a bottle of wine, listed as 12% ABV (alcohol by volume, which is another way of saying volume percent).
\(12\% = \frac{\text{volume of solute}}{750 \text{ mL}} \times 100\)
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).
\(\frac{v}{v\%} = \frac{40\text{ mL}}{120\text{ mL}} × 100=33.3\%\text{ ethanol }\frac{v}{v}\)
Now, you should be able to understand what you’re being asked and can easily make the calculations.
Thanks for watching, and happy studying!
Text Solution
0.09570.07590.05797.95%`
Answer : 4
Solution : `10%` by volume means in every `100 ml` of solution there is `10 ml` of ethanol. To get the mass percent we must divide the mass of ethanol by the mass of solution followed by multiplication with `100%`. Use densities to find the mass of ethanol and mass of solution. <br> According to volume percent `(10%)`, we have <br> Volume of ethanol `=10 mL` <br> Volume of solution `=100 mL` <br> Thus <br> `"Mass of ethanol" = "volume" xx "density"` <br> `=(10 mL)(0.785 g mL^(-1))` <br> `=7.85 g` <br> `"Mass of solution" = "volume" xx "density"` <br> `=(100 mL) (0.9866 g)` <br> `=98.66 g` <br> Thus, <br> Mass percent `=("Mass of component")/("Total mass of solution")xx100%` <br> `=(7.85g)/(98.66 g)xx100%` <br> `=7.95%`
Answer
Hint: Solution is represented in terms of relative percent concentration of solute in a solution. This calculation provides information about the concentration of the solution. The percentage expresses the number of parts of a particular substance per 100 total parts.
Complete Solution :
We know that the volume percent is used to express the concentration of a solution when the volume of a solute and the volume of a solution is given. And we write the equation as:$\text{Volume}\,\text{percentage=}\dfrac{\text{Volume}\,\text{of}\,\text{solute}}{\text{Volume}\,\text{of}\,\text{solution}}\text{ }\!\!\times\!\!\text{ 100 }\!\!%\!\!\text{ }$Since it is given that, 10% of the total solution has ethanol. Now take the total volume of solution as 100ml, then the volume of ethanol will be 10ml.Volume of ethanol = 10mlVolume of solution = 100ml$\text{Density =}\dfrac{\text{Mass}}{\text{Volume}}$By rearranging the above equation, the mass of ethanol and solution can be derived.And the equation becomes,Weight of ethanol = Volume$\times$density Weight of ethanol = $10\times 0.785=7.85g$Weight of solution = $100\times 0.9866=98.66g$$\text{Weight}\,\text{percent}\,\text{of}\,\text{ethanol=}\dfrac{\text{Weight}\,\text{of}\,\text{ethanol}}{\text{Weight}\,\text{of}\,\text{solution}}\text{ }\!\!\times\!\!\text{ 100 }\!\!%\!\!\text{ }$Weight percent of ethanol = $\dfrac{7.85}{98.66}\times 100%$Weight percent of ethanol = 7.95%So, the correct answer is “Option A”.
Note: We generally express concentration in terms of relative units. And the three different types of percentage concentration which is commonly used are as follows:
1. Mass percent: The mass percent is used to express the concentration of a solution when the mass of a solute and the mass of solution is given.2. Volume percent: The volume percent is commonly used to express the concentration of a solution when the volume of a solute and the volume of a solution is provided.3. Mass/Volume percent: This measures the mass or weight of solute in grams vs. the volume of solution.