Chemistry solution needed

Question

83.4 g of potassium iodide is placed into enough water to make 425 mL of solution. Then, 25.0 mL of this solution is diluted to a final volume of 125.0 mL. What is the concentration of the final solution?

In the first blank: Determine the concentration of the original potassium iodide solution. (Before the dilution)

In the second blank: Determine the new concentration after the dilution has been performed.

Answer 1

Hello, Jack,

We need to determine the number of moles in 83.4g of KI. The molar mass of KI is 166 g/mole.

Moles KI = 83.4g KI/(166 g/mole) = 0.5024 mole KI

Concentration is Molar, and is defined as "Moles/Liter). The KI is added to 425ml of solution. We need liters, so divide by 1000ml/L, which is 0.425 L. Divide 0.5024 moles KI by 0.425L to get 1.18M KI. That is the "original concentration."

To find the concentration after dilution, use the relationship M1V1 = M2V2, where M is concentration and V is volume of the intital (1) and final (2) solutions.

M1V1 = M2V2

(1.18M KI)*(25.0ml) = M2*(125.0ml)

M2 = 0.236 M KI

Bob