Question

What volume in liters will be required to dissolve 60.9 grams of Mg(OH)2? Ksp = 1.8×10-11

After doing the math, i got 1.58e-4, but its not right, so i have no idea what to do now.

Answer 1

1.58e-4 what? mL? L? Does it sound reasonable that you can dissolve almost 61 g Mg(OH)2 in less than 1 mL (even less than 1 L)?
If you had shown your work I could tell instantly what you did wrong. As it is I sit here and type all this out and let you look and see if you can find the error. Wasted time on my part. I could have used that time to help others while YOU did the typing.
...........Mg(OH)2 ==> Mg^2+ + 2OH^-
I..........solid.......0.........0
C..........solid.......x.........2x
E..........solid.......x.........2x

Ksp = 1.8E-11 = (x)(2x)^3 = 4x^3
x = ? = M = mols/L
Then grams = mols x molar mass = grams/L which is approx 0.01 g but you need to do that more accurately.
That's about 0.01g/L so how many L will it take to dissolve 60.9g? That's 1 L x (60.9/0.01) = ? L. Remember this is approximate.