Answer
Please see the work below.
Work Step by Step
We know that $1.0gKCl\times \frac{1mol\space KCl}{74.55g}\times \frac{1mol\space Cl^-}{1mol\space KCl}=0.01341mol\space Cl^-$
$1.0g\space CaCl_2\times \frac{1mol\space CaCl_2}{110.98g}\times \frac{2molCl^-}{1mol\space CaCl_2}=0.01802mol\space Cl^-$
