A student prepares a solution by taking 5.0 cm³ of 0.1 M NaOH and diluting it with distilled water to a total volume of (a) 10.0 cm³, (b) 10.0 cm³, (c) 15.0 cm³, (d) 20.0 cm³, and (e) 25.0 cm³. Calcul

The initial volume of $0.1 \text{ M NaOH}$ solution is $5.0 \text{ cm}^3$. The volume of distilled water added is the total volume in the cylinder minus the initial volume of $\text{NaOH}$.

• For $V_1$: Total volume $= 10.0 \text{ cm}^3$. Volume of distilled water $= 10.0 \text{ cm}^3 - 5.0 \text{ cm}^3 = \boxed{5.0 \text{ cm}^3}$
• For $V_2$: Total volume $= 10.0 \text{ cm}^3$. Volume of distilled water $= 10.0 \text{ cm}^3 - 5.0 \text{ cm}^3 = \boxed{5.0 \text{ cm}^3}$
• For $V_3$: Total volume $= 15.0 \text{ cm}^3$. Volume of distilled water $= 15.0 \text{ cm}^3 - 5.0 \text{ cm}^3 = \boxed{10.0 \text{ cm}^3}$
• For $V_4$: Total volume $= 20.0 \text{ cm}^3$. Volume of distilled water $= 20.0 \text{ cm}^3 - 5.0 \text{ cm}^3 = \boxed{15.0 \text{ cm}^3}$
• For $V_5$: Total volume $= 25.0 \text{ cm}^3$. Volume of distilled water $= 25.0 \text{ cm}^3 - 5.0 \text{ cm}^3 = \boxed{20.0 \text{ cm}^3}$

We use the dilution formula $M_1V_1 = M_2V_2$, where $M_1 = 0.1 \text{ M}$ and $V_1 = 5.0 \text{ cm}^3$.