Problems involving replacement of solution parts: Replacing solution parts can be a challenging but important skill in solving numerical problems. It involve...
Replacing solution parts can be a challenging but important skill in solving numerical problems. It involves adjusting the proportions of different components in a solution to achieve the desired concentration or ratio.
Key concepts:
Solution parts: The original amount of solute and solvent combined to form a solution.
Replacements: Adding or removing solution parts to achieve the target concentration or ratio.
Proportions: The relative amounts of solute and solvent used in a replacement.
Ratio: The mathematical relationship between the amounts of solute and solvent.
Examples:
Solution:
Start by subtracting the initial volume of water from the total volume to get the amount of water to add. 100 mL - 50 mL = 50 mL.
Add this amount of water to the solution to achieve the desired volume.
Solution:
Calculate the solute concentration by multiplying the concentration percentage by the total volume of solution. 10% x 500 mL = 50 mL.
Add 50 mL of the pure solute to the solution to reach the desired volume.
Solution:
Calculate the amount of water needed by subtracting the initial volume of water from the total volume. 100 mL - 20 mL = 80 mL.
Add this amount of water to the flask to reach the original volume