"How would you prepare 150. mL of 0.1M sodium hydroxide given a stock solution of 3.0M NaOH?"

Question

I made calculations and figured the process would be to dilute 5 mL of the 3.0M NaOH to 150. mL. Is this correct?

For the ideal 150. mL of 0.1M NaOH, I calculated there to need .015 mol of NaOH in the solution, which is present in 5 mL of 3.0M NaOH.

Answer 1

You're right. Go to the head of the class.

Answer 2

3.0M in V ml
0.1M in 150 mL

dilution factor: 30x
Therefore, 150/30 = 5ml from the stock solution