How many mL of a stock solution of 2M potassium nitrate would you need to prepare 100mL of .150 M potassium nitrate?

Use the dilution formula of c1V1 = c2v2.

c = concn
v = volume

To find out how many mL of a stock solution of 2M potassium nitrate are needed to prepare 100mL of a 0.150M potassium nitrate solution, we can use the equation:

C₁V₁ = C₂V₂

Where:
C₁ = concentration of the stock solution (2M)
V₁ = volume of the stock solution to be used (unknown)
C₂ = concentration of the final solution (0.150M)
V₂ = volume of the final solution (100 mL)

Rearranging the equation, we get:

V₁ = (C₂ * V₂) / C₁

Now, let's plug in the values:

V₁ = (0.150 M * 100 mL) / 2 M

V₁ = (0.150 * 100) / 2

V₁ = 7.5 mL

Therefore, you would need to measure out 7.5 mL of the stock solution of 2M potassium nitrate to prepare 100 mL of a 0.150M potassium nitrate solution.