Calculate the solubility in grams per liter?

Question

(note that Ksp is in terms of moles and liters) of Ag2CO3
Ksp = 8.10 x 10−12 (Neglect the basic properties of CO3−2)

Answer 1

Ag2CO3 --> 2Ag+ + CO3 2-

Ksp= 8.10e-12= [Ag+]^2[CO3 2-]
since there are 2 Ag+ ions for every CO3 2- ion, the concentrations in the Ksp equation are:
[Ag+]= [2x] which then gets raised to the second power
[CO3 2-]= [x]

therefore:

8.10e-12= [2x]^2[x]
8.10e-12= 4x^3
x^3= 2.025e -12
x=1.265e-4

1.265e-4 mol/L x 275.75 g Ag2CO3/mol = .034885 g/L

Answer 2

to go to mols to grams you need to multiply by the molar mass of the compound... (as a method i find it usefull - when in doubt of wheater u need to divide or multiply- to try eliminate the units you don't need).
solubility = mol / liter = mol / liter * (g / mol) = g / liter
Mm = 2*108 +12 + (16*3) = 276 g/mol
therefore solubility = 8.10*10^-12 * 276 = 2*10^-9 g/liter

hope it helps=)