A solution is labeled as ~8M NaOH. How many mL of this solution would be needed to prepare 750mL of a ~0.2M NaOH solution?
mL x M = mL x M
To determine the volume of the ~8M NaOH solution needed to prepare a ~0.2M NaOH solution, we need to use the concept of molarity, which relates the amount of solute (in moles) to the volume of solution (in liters). The equation for molarity is:
Molarity (M) = moles of solute / volume of solution (in liters)
First, let's calculate the moles of NaOH needed to prepare the ~0.2M solution. We can use the formula:
moles = Molarity × volume (in liters)
moles = 0.2M × 0.75L = 0.15 moles
Now, we'll use the molarity of the ~8M NaOH solution to calculate the volume of this solution required. Rearranging the equation for molarity:
Volume of solution (in liters) = moles of solute / Molarity
Volume = 0.15 moles / 8M
Volume = 0.01875 liters
To convert this volume to milliliters, we multiply by 1000:
Volume = 0.01875 liters × 1000 mL/liter
Volume ≈ 18.75 mL
Therefore, approximately 18.75 mL of the ~8M NaOH solution would be needed to prepare 750 mL of a ~0.2M NaOH solution.