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Percentage by volume formula depicts the relationship between the solute and its solution in terms of volume of solute/volume of solution percentage. The percent by volume formula helps us to know the concentration of the solute in a solution. The mixture of a solute and a solvent results in a solution. The concentration of the solute in a solution plays a vital role in understanding chemical reactions. This formula deals with the concentration of liquid solutions and solutes.
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Key Takeaways: Percent by volume formula, mass by volume percentage, mass percentage, dilute and concentrated solution, quantitative expression of concentration, Raoult’s Law
Percentage by Volume Formula
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The percent of volume formula calculates the concentration of the solution by using the volume of the solute (the dissolved content in the solvent) and the volume of the solution (the total volume of the solute and the solvent).
Volume % = Volume of the soluteVolume of solution ×100
In the case of liquids, we can’t always assume that the solution is the mixture of solute and solvent. But while solving questions, we can assume the mixture of the solution. The volume percent is always denoted in percentage (%). It is important that the volume of solute and solution should have the same units (ml, cm3, L, etc).
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Determining Concentration of Solute
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Determining the concentration of a solute in a solution is an important step in studying chemical reactions. When a solute is added to the solvent, the molecules collide in the solution which indirectly determines the rate of a reaction and the conditions at equilibrium. Thus the volume percent formula is one among the many ways to determine the concentration of the solution.
Things to Remember
- By using the Volume percentage formula, it becomes easy to practically express the measure of volume of a solution rather than the mass of the solution.
- A solution can be termed as dilute or concentrated based on the measure of solute relative to the solvent.
- Semi-quantitatively a solution can also be called a saturated solution if no more addition of solute will dissolve in the solvent and unsaturated solution if the solvent can accommodate more dissolution of solute in it.
- The volume of a solution is temperature sensitive, that is, it varies depending on temperature.
- The volume of a solution is moderately affected by pressure. Change in pressure does not cause that much change in the volume of a solution as much as change in temperature can.
- The volume of solution is not equal to the sum of volume of solute and the volume of solvent. As the difference between the volume of solution and the sum of the volume of solvent and solute is pretty small, we generally assume that they both are the same.
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Sample Questions
Ques. How much ethanol is inside a 750 ml wine bottle that contains 12% v/v of alcohol? [2 marks]
Ans. If 12 % is the volume percentage of the 750 ml bottle contains ethanol,
We have volume % as 12 %, Volume of the solution as 750 ml, thus
Volume % = Volume of the soluteVolume of solution ×100
12 % = Volume of the solute750 ml ×100
Volume of ethanol = 12 ×750100 ml
Volume of ethanol = 90.0 ml
Ques. A solution of propanol is prepared by dissolving 67 ml in enough water to have a final volume of 250 ml. What is the volume percent of the ethanol? [2 marks]
Ans. Here the volume of the solvent (water) is not given separately instead the combined volume of the solution is given, thus
Volume of solute = 67 ml
Volume of solution = 250 ml
Volume % = Volume of the soluteVolume of solution ×100
Volume % = 67 ml250 ml ×100
Volume % = 26.8 %
Ques. How much volume of HNO3 concentrate do we need to prepare 250 mL of solution 4%? [2 marks]
Ans. Identifying the data, we need to find the volume of the solute HNO3 while the volume of solution is given as 250 ml and volume percentage is 4 % thus
Volume % = Volume of the soluteVolume of solution ×100
4 % = Volume of the solute250 ml ×100
Volume of HNO3 = 4 ×250100 ml
Volume of HNO3 = 10.0 ml
Thus we need 10 ml of HNO3 to prepare a solution with 4% concentration.
Ques. Rubbing Alcohol is commonly used as an antiseptic for small cuts. It is sold as a 70% volume percentage solution of isopropyl alcohol in water. What volume of isopropyl alcohol is used to make 500 ml of rubbing alcohol? [2 marks]
Ans. Here the solute is isopropyl alcohol, and the solvent is water. Rubbing Alcohol is the solution of isopropyl alcohol and water. The given volume of solution of rubbing alcohol is 500 ml. Thus the volume of isopropyl alcohol in the solution is –
Volume % = Volume of the soluteVolume of solution ×100
70 % = Volume of the solute500 ×100
Volume of isopropyl alcohol = 70 ×500100
Volume of isopropyl alcohol = 350 ml
Ques. What happens to the volume percentage when 25 ml of ethanol is combined with water to give a solution of 200 ml? [2 marks]
Ans. Here, the solute is ethanol with a volume of 25 ml
The solvent here is water. The total volume of the solution is given as 200 ml. Conversion is not required as both the terms have the same unit.
Volume % = Volume of the soluteVolume of solution ×100
Volume % = 25200 ×100
Volume % = 0.125 ×100
Volume % = 12.5 %
Ques. A solution of 3000 ml is made by adding hydrogen peroxide of 90 ml in it. The solution has a combination of water and hydrogen peroxide. What is the volume percentage of hydrogen peroxide added in the water? [2 marks]
Ans. Here, the solute is hydrogen peroxide with a volume of 90 ml
The solvent here is water. The total volume of the solution is given as 3000 ml. Conversion is not required as both the terms have the same unit.
Volume % = Volume of the soluteVolume of solution ×100
Volume % = 903000 ×100
Volume % = 3.0 %
Ques. What will be the mass percentage of urea if 6g of urea was dissolved in 500g of water? [2 marks]
Ans. Here the mass of the solute urea is 6g. The mass of the solvent water is 500g.
Mass percentage = Mass of the soluteMass of solution ×100
Mass percentage = Mass of the soluteMass of solute + Mass of solvent ×100
Mass percentage = 6 g500g +6g ×100
Mass percentage = 6 g506g ×100
Mass percentage = 1.186 %
Ques. What is the mass percentage of the glucose solution produced by the mixture of 34.2 g of glucose in 400 g of water? [2 marks]
Ans. Here the mass of the solute glucose is 34.2 g. The mass of the solvent water is 400g.
Mass percentage = Mass of the soluteMass of solution ×100
Mass percentage = Mass of the soluteMass of solute + Mass of solvent ×100
Mass percentage = 34.2 g400g +34.2 g ×100
Mass percentage = 34.2 g434.2 g ×100
Mass percentage = 7.87 %
Ques. 15 g of cane sugar is dissolved in 60 g of water to prepare a solution. calculate the mass percentage of each component [3 marks]
Ans. As per the mass percentage formula,
Mass of a solution = mass of solute + mass of solvent
Mass of a solution = 15 g + 60 g
Mass of a solution = 75 g
Mass percentage = Mass of the soluteMass of solution ×100
Mass percentage = Mass of the soluteMass of solute + Mass of solvent ×100
Thus mass percentage of solute (cane sugar) = 1575 ×100
Mass percentage of solute (cane sugar) = 20%
Mass percentage of solvent (water) = 100% - 20%
Mass percentage of solvent (water) = 80%
Ques. Identify which of the following are insoluble, partially soluble, and highly soluble in water? [3 marks]
Phenol, Toluene, Formic Acid, Ethylene Acid, Chloroform (trichloromethane), Pentanol, Vinegar, Oil
Ans.
| Solute | Solubility |
|---|---|
| Phenol | Soluble |
| Toluene | Insoluble |
| Formic Acid | Soluble |
| Ethylene Acid | Soluble |
| Chloroform (trichloromethane) | Partially soluble |
| Pentanol | Insoluble |
| Vinegar | Insoluble |
| Oil | Insoluble |
Ques. What is the volume percent of a solution made from 60.0 ml of water with 2.5 ml of pure acetic acid (CH3COOH) added? [5 marks]
Ans. Here, the solute is pure acetic acid with a volume of 2.5 ml
The solvent here is water with a volume of 60.0 ml. Conversion is not required as both the terms have the same unit.
The solution is a mixture of solute and solvent thus the volume of solution is –
= Volume of solvent + volume of solute
= Volume of water + volume of acetic acid
= (60.0 + 2.5) ml
= 62.5 ml
Volume % = Volume of the soluteVolume of solution ×100
Volume % = 2.560.0 ×100
Volume % = 0.04 ×100
Volume % = 4.0 %
Ques. Calculate the volume percentage of Benzene (C6H6) and Carbon Tetrachloride (CCl4) if 22 ml of benzene is dissolved in 122 ml of carbon tetrachloride. [5 marks]
Ans. Here, the solute is benzene with a volume of 22 ml
The solvent here is carbon tetrachloride with a volume of 122 ml. Conversion is not required as both the terms have the same unit.
The solution is a mixture of solute and solvent thus the volume of solution is –
= Volume of solvent + volume of solute
= Volume of water + volume of acetic acid
= (122+22) ml
= 144 ml
Volume % = Volume of the soluteVolume of solution ×100
Volume % = 22144 ×100
Volume % = 0.1528 ×100
Volume % = 15.28 %
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