Answer is: mass of xenon tetrafluoride is 47.98 grams.
1) m(OF₂) = 25.0 g; mass of oxygen difluoride.
n(OF₂) = m(OF₂) ÷ M(OF₂).
n(OF₂) = 25 g ÷ 54 g/mol.
n(OF₂) = 0.462 mol; amount of substance.
In one oxygen difluoride (OF₂) there are two fluorine atoms: n(F) : n(OF₂) = 2 : 1.
n(F) = 0.462 mol · 2.
n(F) = 0.926 mol.
N(F) = n(F) · Na(Avogadro constant).
N(F) = 0.926 mol · 6.022·10²³ 1/mol.
N(F) = 5.576·10²³; number of fluorine atoms in oxygen difluoride.
2) In one molecule of xenon tetrafluoride there are four fluorine atoms:
n(XeF₄) : n(F) = 1 :4.
n(XeF₄) = 0.926 mol ÷ 4.
n(XeF₄) = 0.2315 mol; amount of xenon tetrafluoride.
m(XeF₄) = 0.2315 mol · 207.28 g/mol.
m(XeF₄) = 47.98 g.