How many moles of calcium chloride, cacl2, can be added to 1.5 l of 0.020 m potassium sulfate, k2so4, before a precipitate is expected? assume that the volume of the solution is not changed significantly by the addition of calcium chloride?

When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]

when Ksp = 4.93 x 10^-5
and [SO4] = 0.02 M 
so by substitution we can get [Ca2+] 
4.93x10^-5 = [Ca2+] [0.02]
∴ [Ca2+] = 0.0025 mol/L

∴ the moles of calcium chloride = 0.0025 mol / L * 1.5 L
                                                      = 0.00167 mol

The answer options is d