5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol$$^{-1}$$ respectively

We need to find the volume of stock NaOH solution required to prepare 500 mL of 0.1 M NaOH solution.

The molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles of NaOH = $$\frac{5}{40} = 0.125$$ mol, and the volume of the stock solution is 450 mL = 0.450 L, giving a molarity of $$\frac{0.125}{0.450} = \frac{25}{90} = \frac{5}{18}$$ M.

Using the dilution formula $$M_1V_1 = M_2V_2$$, we have $$\frac{5}{18} \times V_1 = 0.1 \times 500$$, which yields $$V_1 = \frac{50 \times 18}{5} = 180$$ mL.

Thus, 180 mL of the stock solution is required.