20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is _______ $$\times 10^{-2}M$$.(Nearest integer)

We need to find the final concentration when 20 mL of 2 M NaOH is added to 400 mL of 0.5 M NaOH.

When two solutions of the same solute are mixed, the total moles of solute are conserved:

$$C_{final} = \frac{n_1 + n_2}{V_{total}}$$

where $$n = C \times V$$ (moles = concentration times volume).

Solution 1: $$n_1 = C_1 \times V_1 = 2 \times 20 = 40$$ mmol (using mL for volume)

Solution 2: $$n_2 = C_2 \times V_2 = 0.5 \times 400 = 200$$ mmol

$$V_{total} = 20 + 400 = 420 \text{ mL}$$

$$C_{final} = \frac{n_1 + n_2}{V_{total}} = \frac{40 + 200}{420} = \frac{240}{420} = \frac{4}{7} \text{ M}$$

$$C_{final} = \frac{4}{7} \approx 0.5714 \text{ M} = 57.14 \times 10^{-2} \text{ M}$$

Rounding to the nearest integer: $$57 \times 10^{-2}$$ M.

The answer is 57.