Answer
There are 5.569 x $10^{19}$ molecules $ClF_{3}$ in 7.58 mg $ClF_{3}$ .
Work Step by Step
Strategy: We have to convert mass $ClF_{3}$ 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 chlorine trifluoride $ClF_{3}$.
Molar mass $ClF_{3}$ = 1 x 35.45 + 3 x 18.99
Molar mass $ClF_{3}$ = 92.42 g/mol
Find number of molecules that are in8.55 mg $ClF_{3}$ ( or 8.55 x $10^{-3}$g $ClF_{3}$ :
8.55 x $10^{-3}$g $ClF_{3}$ x $\frac{1 mol (ClF_{3})}{ 92.42 g (ClF_{3})}$ x $\frac{6.02 \times 10^{23}(ClF_{3} molecules)}{1 mol(ClF_{3})}$ =
9.251 x $10^{-5}$mol $ClF_{3}$ x $\frac{6.02 \times 10^{23}(ClF_{3} molecules)}{1 mol(ClF_{3})}$ = 55.69 x $10^{18}$ molecules $ClF_{3}$.
Therefore 55.69 x $10^{18}$ molecules $ClF_{3}$ or 5.569 x $10^{19}$ molecules $ClF_{3}$ are in 7.58 mg $ClF_{3}$.
