Final answer:
The new concentration of the KCl solution after dilution, when 10 mL of a 0.1 M solution is added to 990 mL of water, is 0.001 M.
Step-by-step explanation:
The question at hand involves the dilution of a chemical solution, which is a fundamental concept in chemistry, particularly in the subject of solution concentration. The concentration of a solution is determined by the amount of solute dissolved in a given volume of solvent.
In this case, we are looking at a dilution problem where a known volume and concentration of KCl solution is being added to a larger volume of water, thus changing the concentration.
To calculate the new concentration after dilution, we can use the formula M₁V₁ = M₂V₂, where M₁ and V₁ are the molarity and volume of the initial solution, and M₂ and V₂ are the molarity and volume of the final solution, respectively. Knowing that 10 mL of a 0.1 M KCl solution is added to 990 mL of water:
- The initial amount of KCl (in moles) is given by M₁V₁, which is 0.1 M × 0.01 L = 0.001 moles.
- The final volume V₂ of the diluted solution is the sum of the volumes of the initial KCl solution and the water, which is 0.01 L + 0.99 L = 1.00 L.
- The final concentration M₂ can then be found using M₂ = (M₁V₁) / V₂, which yields M₂ = 0.001 moles / 1.00 L = 0.001 M.
Therefore, the new concentration of the KCl solution after dilution is 0.001 M.