Answer
$2.77*10^{19}\text{ molecules Freon-12}$
$3.26\text{ mg Cl}$
Work Step by Step
$0.00556\text{ grams Freon-12}*\frac{1\text{ mol Freon-12}}{120.9135\text{ grams Freon-12}}*\frac{6.0221*10^{23}\text{ molecules Freon-12}}{1\text{ mol Freon-12}}=2.77*10^{19}\text{ molecules Freon-12}$
$\frac{\text{ Mass of Cl}}{\text{ Molar Mass}}=\frac{70.906}{120.9135}*0.00556=3.26\text{ mg Cl}$
