how to calculate dilution factor from given concentration

How to Calculate Dilution Factor from Given Concentration

Chemicals in concentrated forms are seldom useful. Hence, they are diluted using solvents in order to reduce their 'ph' value. This Buzzle article shows how to calculate the dilution factor of solutions, given their initial and final concentrations.

Did You Know? The forces responsible for the binding of a solute and solvent to form a solution are 'van der waals force' or 'hydrogen bonds'.
According to chemistry principles, a solute and solvent combine to form a solution. An example can be salt dissolved in water. In order to understand how to calculate the dilution factor from a given concentration value, we need to first understand a few terms. Dilution Reducing the concentration of any chemical (solution, gas, vapor) is called dilution. The concentration level of many solutions is much more than the desired level. Therefore, a solvent is added to such solutions to make them diluted. This process is carried out in day-to-day life. Consider concentrated orange squash. It is diluted with water before consumption. Dilution reduces the 'ph' level of the chemical. Concentration The amount of solute dissolved in any solvent is termed as the concentration. Concentration is the inverse of dilution. Its units are M (molar), mg/L, and ppm. Molarity of a solution is higher when it is concentrated. The number of moles of solute remain the same before and after dilution. In order to dilute a given solution, solvent is added to the original concentrated solution. On addition of the solvent, the quantity of the solution increases. Consider the initial volume and concentration of the solution as Vinitial and Cinitial, respectively. On dilution, the final volume and concentration of the solution changes to Vfinal and Cfinal, respectively. Dilution Factor The ratio of volume of the initial solution (concentrated) to the volume of the final solution (diluted) is called dilution factor. Consider the formula:
Cinitial × Vinitial = Cfinal × Vfinal
If you have three out of four values in the above equation, you can go about calculating the fourth value. Serial Dilution The process of bringing about step-by-step dilution of a solution is called serial dilution. In this process, the dilution factor remains constant throughout. Serial dilution finds its applications in physics, pharmacology, biochemistry, etc.
Sample Problems
Problem I
PPM (Parts Per Million) PPM = (mass of solute) (gm) ÷ (mass of solute and solvent) (gm) ) ÷ 1,000,000 ppm Consider you have 2L of an aqueous solution, with molarity 0.50M. You want to dilute this solution to molarity 0.20M. What needs to be done? Data given: Cinitial = 0.50M, Cfinal = 0.20M, Vinitial = 2L, Vfinal = ? Cinitial × Vinitial = Cfinal × Vfinal 0.50 × 2 = 0.20 × Vfinal Vfinal = (0.50 X 2) ÷ 0.20 Vfinal = 5L The result implies that the original solution needs to be diluted more by 3L, such that the final volume of the solution becomes 5L. That is, as the concentration of the solution decreases, the volume increases (inverse proportion).
Problem II
Pulverized iron is added to a vessel filled with water. Its molarity is 5ppm. You take 8mL of water from this vessel and add 2mL of distilled water to it. What is the molarity of the resulting solution? Data Given: Cinitial = 5ppm, Vinitial = 8mL, Vfinal = 10mL, Cfinal = ? Cinitial × Vinitial = Cfinal × Vfinal 5 × 8 = Cfinal × 10 Cfinal = 40 ÷ 10 Cfinal = 4ppm
Applications of Dilutions
Diluted hydrogen peroxide (H2O2) is stored in concentrated form in laboratories, because its diluted form is voluminous. It is sold at pharmacies to be used as a disinfectant or a bleaching agent. However, it is used in its concentrated form in rocket propellants. Dishwashing liquids need dilution as they are too concentrated. Diluting them doesn't affect their effectiveness in cleaning dishes or cookware.
Dilution is a commonly followed practice for diluting any solution, gas, or vapor. For human beings, it is difficult to breathe pure oxygen. However, it gets mixed with other gases in the atmosphere, which makes it suitable for the breathing process. Therefore, the significance of this process is crystal clear.

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