At equilibrium:
Ag2CrO4 <----> 2 Ag+ + CrO42-
2s s
Ksp = [Ag+]^2[CrO42-]
8*10^-12=(2s)^2*(s)
8*10^-12= 4(s)^3
s = 1.26*10^-4 M
Molar mass of Ag2CrO4,
MM = 2*MM(Ag) + 1*MM(Cr) + 4*MM(O)
= 2*107.9 + 1*52.0 + 4*16.0
= 331.8 g/mol
Molar mass of Ag2CrO4= 331.8 g/mol
s = 1.26*10^-4 mol/L
To covert it to g/L, multiply it by molar mass
s = 1.26*10^-4 mol/L * 331.8 g/mol
s = 4.18*10^-2 g/L
Now use:
volume = mass / solubility
= 3.5 g / 4.18*10^-2 g/L
= 83.7 L
Answer: 83.7 L