How many atoms of Fluorine are in 595.85g of SeO2F2?

A: 6.02 x 10^23

B: 1.204 x 10^24

C: 2.408 x 10^24

D: 4.816 x 10^24

Molar mass of SeO₂F₂=78.96 + (2*15.99) + (2*18.99) = 148.936 g/mol

No. of molecules of SeO₂F₂ = (given mass/molar mass) * avogadro's no.

No. of molecules of SeO₂F₂ = (595.85/148.936) * (6.02*10²³)

No. of molecules of SeO₂F₂=2.408*10²⁴

As,

one molecule contains 2 fluorine atoms.

So, total no. of fluorine atoms are 2 * (2.408*10²⁴) = 4.8168*10²⁴ atoms

Answer D is correct.