At a certain temperature, the solubility of strontium arsenate, sr3(aso4)2, is 0.0590 g/l. what is the ksp of this salt at this temperature?

First, we have to change Sr3(ASO4)2 from 0.059 g/L to moles:

and when the molar mass of Sr3(ASO4)2 = 87.6*3 + 74.9*2 + 16*8
                                                                      = 540.6 g/mol

Molarity (X) = 0.059 g/L * 1 mol / 540.6 g/mol
                    = 1.1 x 10^-4 moles 

When the solubility equation is:

Sr3(ASO4)2 ↔ 3Sr++(aq)   +  2ASO-3(aq)

so, Ksp = [Sr++]^3[ASO-3]^2

from ICE table and the balanced equation:
 [Sr++] = 3 X
           = 3* 1.1 x 10^-4 = 3.3x10^-4 M
 [ASO-3] = 2X
               = 2* 1.1 x 10^-4 = 2.2 x 10^-4 M

So, by substitution:

∴Ksp = (3.3 x 10^-4)^3 * (2.2 x 10^-4) ^2
       = 1.74 x 10^-18
       

The density would be 1.622

Cis the answer ur welcome if i get it wrong u don't have to me