Since the solution is $2\%$, this means there are $2$ grams of silver nitrate in $100$ mL of solution.
Converting this to molarity, we have:
$2\% = \frac{2 \text{ g}}{100 \text{ mL}} = \frac{2 \text{ g}}{0.1 \text{ L}} = 0.02 \text{ mol/L}$
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