The number of moles of sodium hydroxide present in 2.5 L and 0.5 M aqueous solution will be:

1.25

The number of moles of sodium hydroxide present in a 2.5 L and 0.5 M aqueous solution will be:

\[ \text{Number of moles} = \text{Molarity (M)} \times \text{Volume (L)} \]

Given:
Volume of solution = 2.5 L
Molarity of solution = 0.5 M

Substituting the given values into the formula, we get:

\[ \text{Number of moles} = 0.5 \, \text{M} \times 2.5 \, \text{L} = 1.25 \, \text{moles} \]

Therefore, the number of moles of sodium hydroxide present in the 2.5 L and 0.5 M aqueous solution is 1.25 moles. Hence, the correct answer is (1) 1.25.