To do this, use the concentration of stock solution A and the volume of sample added:

$$Molarity = \frac{Moles}{Litres}$$

$$Moles = Molarity \times Litres$$

$$n = \frac{(13.3g/132.14 g/mol)}{0.1L}$$

$$n = 0.1 \ mol/L$$

$$Moles \ of \ NH_4SO_4 \ in \ 12.00 \ mL = (0.1000 mol/L) \times (12.00 mL) = 0.001200 \ mol$$

Step 2: Calculate the total volume of solution after adding 12.00 mL of sample.

The total volume is 57.00 mL + 12.00 mL = 69.00 mL

Step 3: Calculate the new concentration of ammonium sulfate in the diluted solution.

The new concentration can be calculated using the formula:

$$M_1V_1 = M_2V_2$$

where M1 is the original concentration, V1 is the original volume, M2 is the new concentration, and V2 is the new volume.

$$(0.1000 mol/L) \times (100.0 mL) = M_2 \times (69.00 mL)$$

$$M_2 = \frac{(0.1000 mol/L) \times (100.0 mL)}{69.00 mL} = 0.0725 mol/L$$

Therefore, the new concentration of ammonium sulfate in the diluted solution is 0.0725 mol/L.