$$C_1V_1 = C_2V_2$$
where $$C_1$$ is the initial concentration, $$V_1$$ is the initial volume, $$C_2$$ is the final concentration, and $$V_2$$ is the final volume.
We can rearrange this equation to solve for $$V_2$$:
$$V_2 = \frac{C_1V_1}{C_2}$$
In this case, $$C_1 = 0.911\ M$$, $$V_1 = 80.00\ mL$$, and $$C_2 = 0.160\ M$$. Substituting these values into the equation, we get:
$$V_2 = \frac{(0.911\ M)(80.00\ mL)}{0.160\ M} = 455.6\ mL$$
Therefore, the volume of solution that would result by diluting 80.00 mL of 0.911 M NaOH to a concentration of 0.160 M is 455.6 mL.
