Final answer:
To prepare 1.0 L of 0.09000 N NaOH solution, 15 mL of a 6.0 M NaOH solution is required, using the concept of dimensional analysis.
Step-by-step explanation:
To calculate the volume (in mL) of 6.0 M NaOH solution required to prepare 1.0 L of 0.09000 N NaOH solution, we need to do a few conversions and apply the dimensional analysis technique. First, recall that molarity (M) is moles of solute per liter of solution, and normality (N) for NaOH is the same as molarity because NaOH provides one equivalent of OH- per molecule. So, 0.09000 N is also 0.09000 M.
To find out how much of the 6.0 M NaOH solution we need, we use the formula:
C1V1 = C2V2, where C1 and V1 are the concentration and volume of the concentrated solution, and C2 and V2 are the concentration and volume of the diluted solution. Plugging in our known values, we get:
6.0 M × V1 = 0.09000 M × 1.0 L
To find V1, divide both sides by the concentration of the stock solution (6.0 M):
V1 = 0.09000 M × 1.0 L / 6.0 M
V1 = 0.015 L
Now, convert the volume from liters to milliliters because the question asks for mL:
0.015 L × 1000 mL/L = 15 mL
Therefore, 15 mL of 6.0 M NaOH solution is required to prepare 1.0 L of 0.09000 N NaOH solution.