The given molarity of sodium hydroxide solution = 2.0 M
The required concentration of sodium hydroxide is 65 mL of 0.6 M NaOH
Converting 65 mL to L:
[tex]65mL*\frac{1L}{1000mL} =0.065L[/tex]
Calculating the moles of NaOH in the final solution:
[tex]0.065L * \frac{0.6 mol}{L} =0.039mol[/tex]
Finding out the volume of 2.0 M solution taken to prepare the final solution:
[tex]0.039 mol * \frac{1L}{2.0mol}=0.0195L*\frac{1000mL}{1L} =19.5mL[/tex]
Therefore, 19.5 mL of 2.0 M NaOH solution and make it up to 65 mL to prepare 0.6 M NaOH solution.
