When 0.280 g of Na(s) reacts with excess F2(g) to form NaF(s), 6.90 kJ of heat is released to surroundings at standard-state conditions.?

Na(s) + (1/2)F₂ → NaF(s) …… ΔH

Molar mass of Na = 23.0 g/mol
No. of moles of Na reacted = (0.280 g) / (23.0 g/mol) = 0.01217 mol
No. of moles of NaF formed = (0.01217 mol) × (1/1) = 0.01217 mol

As heat is released to the surroundings, ΔHf[NaF(s)]
= (-6.9 kJ) / (0.01217 mol)
= -567 kJ/mol

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

here we have 0.280 g of Na or 0.280 / 23.0 mole of Na = 0.012174 mol of Na

heat per mole = 6.90 kJ * 1/ 0.012174 = 566 kJ / mol

since Na and F are in their standard states

the standard enthalpy of formation of NaF(s) = 566 kJ / mol ( 3 significant figures )