Answer
$1.00 \space L$ of that saline solution contains 0.154 mole of $Na^+$ and 0.154 mole of $Cl^-$.
Work Step by Step
1. Calculate the amount of equivalents.
$$1.00 \space L \times \frac{154 \space mEq} L \times \frac{1 \space Eq}{1000 \space mEq} = 0.154 \space Eq$$
2. Since both $Na^+$ and $Cl^-$ contain 1 charge per ion, $1 \space Eq = 1 \space mole$
$$0.154 \space Eq \times \frac{1 \space mole}{1 \space Eq} = 0.154 \space mole \space of \space each$$
