I did try that but can't seem to find the right answer.

How much precipitate would be produced if the reation went to completion:

Moles of CH3COONa = Concentration x Volume = 0.200 mol/L x 0.050 L = 0.010 mol

Since the reaction is 1:1 between CH3COONa and AgCH3COO, the moles of AgCH3COO produced will also be 0.010 mol.

Calculate the mass of AgCH3COO produced:

Molar mass of AgCH3COO = 107.87 g/mol (Ag) + 12.01 g/mol (C) + 3 x 1.01 g/mol (H) + 2 x 16.00 g/mol (O) = 166.912 g/mol

Mass of AgCH3COO produced = Moles x Molar mass = 0.010 mol x 166.912 g/mol ≈ 1.669 g

Calculate the molar solubility (amount that stays dissolved) using the corrected Ksp:

Ksp for AgCH3COO is given as 1.94 x 10^-3 mol^2/L^2.

To find the molar solubility (S), we'll take the square root of Ksp.

S = √(Ksp) = √(1.94 x 10^-3 mol^2/L^2) ≈ 0.044 mol/L

Calculate the moles of AgCH3COO that will stay dissolved (molar solubility):

Moles of AgCH3COO not precipitated = Molar solubility x Volume = 0.044 mol/L x 0.250 L = 0.011 mol

Calculate the mass of AgCH3COO that will not precipitate:

Mass of AgCH3COO not precipitated = Moles x Molar mass = 0.011 mol x 166.912 g/mol ≈ 1.836 g

And then I subtracted the latter from the former as you said and 1.836 -1.669 g ≈ 0.167 g

But the correct answer is 0.484g