To what volume should you dilute 50.0 mL of a 5.00 M KI solution so that 25.0 mL of the diluted solution contains 3.05 g of KI?

Question

Accepted Answer

Hello everyone today. With the following problem, determine the volume. We should dilute 78 mL of a 3.50 molar potassium chloride solution. So that 50 mL of the diluted solution contains 5.09 g of potassium chloride. So recall that dilution can be calculated using the following equation that the initial molarity multiplied by the initial volume is equal to the final molarity multiplied by the final volume. So first, we need to determine that malaria the final solution. So we need our moles of potassium chloride to do this. So our moles of potassium chlorine can be found using the mass of our potassium chloride in grams divided by the molar mass of our potassium chloride. So we have our moles of potassium chloride is equal to, we have our five 0.09 g of our potassium chloride. And we divide that by the molar mass by taking the individual masses. So we have the mass of potassium and the mass of chlorine. Adding those up gives us a mass of 74.55 g per mole. Our uno grams will cancel out and we will be left with 0.0683 moles of potassium chloride. No. In doing this, we can take the moles for potassium cord that we could solve for. And we can have this will be our second step. We will have our molarity equaling the moles of ours potassium chloride divided by solu the liters of the solution. Our multi calculated was 0.0683. And then the volume that we had was 50 mL, but we have to convert it into units of years. So use a conversion factor that 1 mL is equal to 10 to the negative third liters where our units cancel out. We are left with a molarity of 1.37. And we can then apply that to our equation. So we have the initial molarity or M one. We can say that M one is our 3.5 molar. Our initial volume is a 78 middle leaders legal that we don't. We have our second molarity, our final molarity which is 1.37 mole that we solved from the previous step. And then we multiply by our final volume which we don't have dividing all size by 1.37 molar. We arrive at a final volume of 199.27 mL or answer twice. C overall. I hope this helped. And until next time.

To what volume should you dilute 50.0 mL of a 5.00 M KI solution so that 25.0 mL of the diluted solution contains 3.05 g of KI?

Verified Solution

Key Concepts

Molarity (M)

Dilution Equation

Mass to Moles Conversion