Find moles of NaOH in each solution

Use: $\text{moles} = M \times V$ where volume must be in litres.

• For the first solution: $V_1 = 20\text{ mL} = 0.020\text{ L}$ $M_1 = 2\text{ M}$ $n_1 = 2 \times 0.020 = 0.040\text{ mol}$

• For the second solution: $V_2 = 400\text{ mL} = 0.400\text{ L}$ $M_2 = 0.5\text{ M}$ $n_2 = 0.5 \times 0.400 = 0.200\text{ mol}$

Total moles after mixing

$n_{\text{total}} = 0.040 + 0.200 = 0.240\text{ mol}$

Total volume after mixing

$V_{\text{total}} = 20 + 400 = 420\text{ mL} = 0.420\text{ L}$

Final concentration

$M_{\text{final}} = \frac{n_{\text{total}}}{V_{\text{total}}} = \frac{0.240}{0.420} = 0.5714\text{ M}$

Express in the required form

We need: $\underline{\hspace{1cm}} \times 10^{-2}\text{ M}$

Since $0.5714\text{ M} = 57.14 \times 10^{-2}\text{ M}$

Nearest integer: $57$

Comparison with stored answer

Derived answer = $57$

Stored correct answer = $57$

Hence, they agree.