BMAT 2022 Q 26 solubility

hi
the answer is D)
this is my work: 4-3,63=0,37
n=0,37/74 = 5 * 10⁻³
Ca(OH)2 = Ca²⁺ + 2HO⁻
so produce 5 * 10⁻³ mols of Ca²⁺ and 10⁻² mols of HO⁻
Ksp = 5 * 10⁻³ * (10⁻²)² which is obviously incorrect

they asked for solubility what means we need to find n/L so i did 0.005n/0.2l
or you can just do 0.05/2 and you get 0.025 and then you get d so you get the n of moles that dissolved divided by the amount of water in the solution

Hi
mass of Ca(OH)2 dissolved = 0.37g which is in 0.2 dm3
= 0.074g in 1 dm3 of H2O

Molar mass of Ca(OH)2 = 74 gmol-1

mol of Ca(OH)2 dissolved = 1 * 10^-3 mol

Soulbility in (mol/dm3) = 1 *10^ -3mol/ 0.2 dm3
= 0.005 mol/dm3

hope this helps