= 5.79 × 10^19 molecules
The molar mass of the compound is 312 g/mol
Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)
We are required to calculate the number of molecules present
We will use the following steps;
[tex]Moles=\frac{mass}{molar mass}[/tex]
Therefore;
Moles of the compound will be;
[tex]=\frac{0.030}{312g/mol}[/tex]
= 9.615 × 10⁻5 mole
Using the Avogadro's constant, 6.022 × 10^23
1 mole of a compound contains 6.022 × 10^23 molecules
Therefore;
9.615 × 10⁻5 moles of the compound will have ;
= 9.615 × 10⁻5 moles × 6.022 × 10^23 molecules
= 5.79 × 10^19 molecules
Therefore the compound contains 5.79 × 10^19 molecules