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.