$$C_1V_1 = C_2V_2$$
Where:
- \(C_1\) is the concentration of the stock solution (in moles per liter).
- \(V_1\) is the volume of the stock solution (in liters) that we want to find.
- \(C_2\) is the concentration of the diluted solution (in moles per liter).
- \(V_2\) is the volume of the diluted solution (in liters) that contains the same number of moles as the stock solution.
In this case, we know the concentration of the diluted solution (\(C_2 = 0.15 M\)) and the volume of the diluted solution (\(V_2 = 50 mL = 0.05 L\)). We also know the concentration of the stock solution (\(C_1 = 1.8 M\)).
Substituting these values into the formula, we get:
$$(1.8 M)V_1 = (0.15 M)(0.05 L)$$
Solving for \(V_1\), we get:
$$V_1 = \frac{(0.15 M)(0.05 L)}{1.8 M}$$
$$V_1 = 0.0042 L = 4.2 mL$$
Therefore, the volume of the stock solution (Part A) that contains the same number of moles present in the diluted solution (Part B) is 4.2 mL.
