Answer
$1.50 \space L$ of that solution contains 0.060 mole of $K^+$ and 0.060 mole of $Cl^-$.
Work Step by Step
1. Calculate the amount of equivalents.
$$1.5 \space L \times \frac{40. \space mEq} L \times \frac{1 \space Eq}{1000 \space mEq} = 0.060 \space Eq$$
2. Since both $K^+$ and $Cl^-$ contain 1 charge per ion, $1 \space Eq = 1 \space mole$
$$0.060 \space Eq \times \frac{1 \space mole}{1 \space Eq} = 0.060 \space mole \space of \space each$$
