$$2.330 \text{ g KCl} \times \frac{1 \text{ mol KCl}}{74.55 \text{ g KCl}} = 0.03130 \text{ mol KCl}$$

To calculate the volume of 0.1250 M KCl solution required, we can use the formula:

$$V = \frac{n}{C}$$

where:

\(V\) is the volume in liters

\(n\) is the number of moles of solute

\(C\) is the concentration of the solution in M

So, substituting the values we have:

$$V = \frac{0.03130 \text{ mol KCl}}{0.1250 \text{ M KCl}} = 0.2504 \text{ L}$$

Converting to milliliters, we get:

$$V = 0.2504 \text{ L} \times \frac{1000 \text{ mL}}{1 \text{ L}} = 250.4 \text{ mL}$$

Therefore, 250.4 mL of 0.1250 M KCl solution contain 2.330 g KCl.