First, we need to determine the moles of AgNO3 in the solution:
Moles of AgNO3 = Molarity x Volume
Moles of AgNO3 = 0.276 mol/L x 0.025 L
Moles of AgNO3 = 0.0069 moles
Since the reaction is 1:1 between Ag+ and Cl-, we need an equal number of moles of Cl- ions to completely precipitate the silver. Therefore, we need 0.0069 moles of NaCl.
Now, we need to calculate the mass of NaCl needed:
Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl)
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl = moles of NaCl x molar mass of NaCl
Mass of NaCl = 0.0069 mol x 58.44 g/mol
Mass of NaCl = 0.403 g
Therefore, 0.403 grams of solid NaCl must be added to 25.0 mL of 0.276 M AgNO3 solution to completely precipitate the silver.