Answer
2.97 x $10^{19}$ molecules $CCl_{4}$ are in 7.58 mg $CCl_{4}$.
Work Step by Step
Strategy: We have to convert mass $CCl_{4}$ to number of molecules. So the conversion factors needed are two:
conversion factor to mol ( 1 mol/Molar mass) and
conversion factor to number of molecules ( Avogadro's number/ 1 mol ).
First find Molar mass for carbon tetra-chloride $CCl_{4}$.
Molar mass $CCl_{4}$ = 1 x 12.01 + 4 x 35.45
Molar mass $CCl_{4}$ =153.81 g/mol
Find number of molecules that are in 7.58 mg $CCl_{4}$ ( or 7.58 x $10^{-3}$g $CCl_{4}$ :
7.58 x $10^{-3}$g $CCl_{4}$ x $\frac{1 mol (CCl_{4})}{ 153.81 g (CCl_{4})}$ x $\frac{6.02 \times 10^{23}(CCl_{4} molecules)}{1 mol(CCl_{4})}$ =
4.928 x $10^{-5}$mol $CCl_{4}$ x $\frac{6.02 \times 10^{23}(CCl_{4} molecules)}{1 mol(CCl_{4})}$ = 29.7 x $10^{18}$ molecules $CCl_{4}$.
Therefore 29.7 x $10^{18}$ molecules $CCl_{4}$ or 2.97 x $10^{19}$ molecules $CCl_{4}$ are in 7.58 mg $CCl_{4}$.
