- Given: Volume \( V = 200 \, \text{mL} = 0.2 \, \text{L} \), Molarity \( M = 0.15 \, \text{mol/L} \)
- Molar mass of sodium hydroxide (NaOH) \( = 23 + 16 + 1 = 40 \, \text{g/mol} \)
- Number of moles required:
\[
n = M \times V = 0.15 \times 0.2 = 0.03 \, \text{mol}
\]
- Mass of NaOH needed:
\[
m = n \times \text{Molar mass} = 0.03 \times 40 = 1.2 \, \text{g}
\]
- Therefore, to prepare 200 mL of 0.15 M NaOH solution, dissolve 1.2 g of sodium hydroxide in water and make up the volume to 200 mL.